XDOJ
1000.a+b。
#include<bits/stdc++.h>
using namespace std; int a,b; int main()
{
ios::sync_with_stdio(false);
while(~scanf("%d%d",&a,&b)) printf("%d\n",a+b);
return ;
}
1001.不知道n和m大小，可以用一维数组处理位置，或者直接使用vector。
#include<bits/stdc++.h>
using namespace std; int n,m,q,a[],r[],c[];
string s; int main()
{
ios::sync_with_stdio(false);
int z = ;
while(cin >> n >> m)
{
cout << "Case #" << ++z << ":" << endl;
for(int i = ;i <= n;i++) r[i] = i;
for(int i = ;i <= m;i++) c[i] = i;
int cnt = ;
for(int i = ;i <= n;i++)
{
cin >> s;
for(int j = ;j < m;j++)
{
if(s[j] == 'T') a[++cnt] = ;
else if(s[j] == 'i') a[++cnt] = ;
else a[++cnt] = ;
}
}
cin >> q;
while(q--)
{
int x,y,z;
cin >> x >> y >>z;
if(x == )
{
int t = (r[y]-)*m+c[z];
switch(a[t])
{
case : cout << "Empty" << endl;break;
case : cout << "Tree" << endl;break;
case : cout << "Phone" << endl;break;
}
}
else if(x == ) swap(r[y],r[z]);
else swap(c[y],c[z]);
}
} return ;
}
#include<bits/stdc++.h>
using namespace std; int n,m,q,r[],c[];
vector<int> v[];
string s; int main()
{
ios::sync_with_stdio(false);
int z = ;
while(cin >> n >> m)
{
cout << "Case #" << ++z << ":" << endl;
for(int i = ;i <= n;i++) r[i] = i;
for(int i = ;i <= m;i++) c[i] = i;
for(int i = ;i <= n;i++)
{
v[i].clear();
v[i].push_back();
}
for(int i = ;i <= n;i++)
{
cin >> s;
for(int j = ;j < m;j++)
{
if(s[j] == 'T') v[i].push_back();
else if(s[j] == 'i') v[i].push_back();
else v[i].push_back();
}
}
cin >> q;
while(q--)
{
int x,y,z;
cin >> x >> y >>z;
if(x == )
{
switch(v[r[y]][c[z]])
{
case : cout << "Empty" << endl;break;
case : cout << "Tree" << endl;break;
case : cout << "Phone" << endl;break;
}
}
else if(x == ) swap(r[y],r[z]);
else swap(c[y],c[z]);
}
} return ;
}
1002.dp[x][y][z]代表前x段，y个绿，z个蓝造成的最大伤害，复杂度O(n^3)。
可以优化空间，优化到二维状态。
继续优化时间，可以贪心发现红放最末为最优，于是一开始全放红，dp[x][y]表示x个绿，y个蓝的最大伤害，复杂度O(n^2)。
#include<bits/stdc++.h>
#define LL long long
using namespace std; LL n,x,y,z,t;
LL dp[][][]; int main()
{
while(cin >> n)
{
cin >> x >> y >> z >> t;
memset(dp,,sizeof(dp));
dp[][][] = x*t;
for(int i = ;i <= n;i++)
{
for(int g = ;g <= i;g++)
{
int b = ;
for(b = ;b+g < i;b++) dp[i][g][b] = dp[i-][g][b]+(g*y+x)*(b*z+t);
if(g > ) dp[i][g][b] = max(dp[i][g][b],dp[i-][g-][b]+y*(g-)*(b*z+t));
if(b > ) dp[i][g][b] = max(dp[i][g][b],dp[i-][g][b-]+g*y*((b-)*z+t));
}
}
LL ans = ;
for(int i = ;i <= n;i++)
{
for(int j = ;j+i <= n;j++) ans = max(ans,dp[n][i][j]);
}
cout << ans << endl;
}
return ;
}
#include<bits/stdc++.h>
using namespace std; long long n,x,y,z,t,dp[][]; int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
cin >> x >> y >> z >> t;
dp[][] = x*t;
dp[][] = ;
dp[][] = ;
for(int i = ;i <= n;i++)
{
for(int g = i;g >= ;g--)
{
for(int b = i-g;b >= ;b--)
{
if(b+g != i) dp[g][b] = dp[g][b]+(g*y+x)*(b*z+t);
else dp[g][b] = ;
if(g > ) dp[g][b] = max(dp[g][b],dp[g-][b]+y*(g-)*(b*z+t));
if(b > ) dp[g][b] = max(dp[g][b],dp[g][b-]+g*y*((b-)*z+t));
}
}
}
long long ans = ;
for(int i = ;i <= n;i++)
{
for(int j = ;j+i <= n;j++) ans = max(ans,dp[i][j]);
}
cout << ans << endl;
}
return ;
}
#include<bits/stdc++.h>
using namespace std; long long n,x,y,z,t,dp[][]; int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
cin >> x >> y >> z >> t;
memset(dp,,sizeof(dp));
dp[][] = x*t*n;
for(int g = ;g <= n;g++)
{
for(int b = ;b+g <= n;b++)
{
if(g == && b == ) continue;
if(g > ) dp[g][b] = max(dp[g][b],dp[g-][b]-x*(b*z+t)+y*(b*z+t)*(n-g-b));
if(b > ) dp[g][b] = max(dp[g][b],dp[g][b-]-x*((b-)*z+t)+(g*y+x)*z*(n-g-b));
}
}
long long ans = ;
for(int i = ;i <= n;i++)
{
for(int j = ;j+i <= n;j++) ans = max(ans,dp[i][j]);
}
cout << ans << endl;
}
return ;
}
1003.高精度模拟。
#include<bits/stdc++.h>
using namespace std; int change1(char a)
{
if(a <= '') return a-'';
return a-'A'+;
} char change2(int a)
{
if(a <= ) return a+'';
else return a-+'A';
} string s1,s2;
int n,ans[]; int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
cin >> s1 >> s2;
int p1 = s1.length()-,p2 = s2.length()-,cnt = ,temp = ;
while(p1 >= && p2 >= )
{
int now = change1(s1[p1--])+change1(s2[p2--])+temp;
ans[++cnt] = now%n;
temp = now/n;
}
while(p1 >= )
{
int now = change1(s1[p1--])+temp;
ans[++cnt] = now%n;
temp = now/n;
}
while(p2 >= )
{
int now = change1(s2[p2--])+temp;
ans[++cnt] = now%n;
temp = now/n;
}
if(temp) ans[++cnt] = temp;
for(int i = cnt;i > ;i--) cout << change2(ans[i]);
cout << endl;
}
return ;
}
1004.每次固定一个对应点，旋转5次，暴力比较。
#include<bits/stdc++.h>
using namespace std; int a[],b[],c[],d[];
int point[][] = {{,},{,},{,},{,},{,},{,},{,},{,},{,},{,},{,},{,}};
int x[][] = {{,,,,,,,,,},{,,,,,,,,,},{,,,,,,,,,},{,,,,,,,,,},
{,,,,,,,,,},{,,,,,,,,,},{,,,,,,,,,},{,,,,,,,,,},
{,,,,,,,,,},{,,,,,,,,,},{,,,,,,,,,},{,,,,,,,,,}}; int ok()
{
for(int i = ;i < ;i++)
{
if(c[i] != d[i]) return ;
}
return ;
} int main()
{
ios::sync_with_stdio();
int T;
cin >>T;
while(T--)
{
for(int i = ;i <= ;i++) cin >> a[i];
for(int i = ;i <= ;i++) cin >> b[i];
for(int i = ;i < ;i++) c[i] = b[x[][i]];
int flag = ;
for(int i = ;i < ;i++)
{
if(a[point[i][]] == b[] && a[point[i][]] == b[])
{
for(int j = ;j < ;j++) d[j] = a[x[i][j]];
for(int j = ;j < ;j++)
{
if(ok()) flag = ;
int t = d[];
for(int k = ;k > ;k--) d[k] = d[k-];
d[] = t;
t = d[];
for(int k = ;k > ;k--) d[k] = d[k-];
d[] = t;
}
}
}
if(flag) cout << "Identical" << endl;
else cout << "Different" << endl;
}
return ;
}
1005.每次除2向上取整，直到最后一根。
#include<bits/stdc++.h>
using namespace std; long long n; int main()
{
while(cin >> n)
{
int cnt = ;
while(n > )
{
n = n/+n%;
cnt++;
}
cout << cnt << endl;
}
return ;
}
1006.按位运算的dp，a[i]，b[i]，c[i]分别对应^，|，&，表示从头到i当前位的累和。
#include<bits/stdc++.h>
using namespace std; int bin[][],two[],n;
double a[],b[],c[]; int main()
{
two[] = ;
for(int i = ;i <= ;i++) two[i] = two[i-]*;
while(cin >> n)
{
memset(bin,,sizeof(bin));
int maxcnt = ;
double ans1 = ,ans2 = ,ans3 = ;
for(int i = ,x;i <= n;i++)
{
cin >> x;
ans1 -= x;
ans2 -= x;
ans3 -= x;
int cnt = ;
while(x)
{
bin[i][++cnt] = x%;
x /= ;
}
maxcnt = max(cnt,maxcnt);
}
for(int j = ;j <= maxcnt;j++)
{
int last0 = ,last1 = ;
for(int i = ;i <= n;i++)
{
if(bin[i][j] == )
{
if(last1 > ) a[i] = a[last1-]+i-last1;
else a[i] = i;
b[i] = i-last0;
c[i] = i;
last1 = i;
}
else
{
if(i > ) a[i] = a[i-];
else a[i] = ;
b[i] = ;
c[i] = last1;
last0 = i;
}
ans1 += a[i]*two[j];
ans2 += b[i]*two[j];
ans3 += c[i]*two[j];
}
}
cout << fixed << setprecision() << ans1/n/n << " " << fixed << setprecision() << ans2/n/n << " " << fixed << setprecision() << ans3/n/n << endl;
}
return ;
}
1007.dp[i][j]表示i个蛋，j层需要的最多次数，另外可以发现k层最多需要logn+1个蛋，复杂度O(n^2*min(logn,k))。
发现对于每一个dp[i][j]的状态，并不需要枚举每一层，因为这是一个先递减，再递增的函数，二分找最低点即可，复杂度O(nlogn*min(logn,k))。
还可以继续优化很多，具体请参考这篇论文https://pan.baidu.com/s/1htXipWsA9mYKaSzhSNYlqw
#include<bits/stdc++.h>
using namespace std; int dp[][]; int main()
{
int n,k;
while(cin >> n >> k)
{
memset(dp,0x3f,sizeof(dp));
for(int i=;i<=n;i++)
{
dp[i][]=;
dp[i][]=;
}
for(int i=;i<=k;i++) dp[][i]=i;
for(int j=;j<=k;j++)
{
for(int i=;i<=n;i++)
{
for(int t=;t<=j;t++) dp[i][j]=min(dp[i][j],max(dp[i-][t-],dp[i][j-t])+);
}
}
cout << dp[n][k] << endl;
}
return ;
}
#include<bits/stdc++.h>
using namespace std; int n,k,dp[][]; int main()
{
ios::sync_with_stdio();
while(cin >> n >> k)
{
for(int i = ;i <= n;i++)
{
dp[i][] = ;
dp[i][] = ;
}
n = min(n,int(log(k)/log()+));
for(int i = ;i <= k;i++) dp[][i] = i;
for(int j = ;j <= k;j++)
{
for(int i = ;i <= n;i++)
{
int l = ,r = j;
while(l < r)
{
int mid = (l+r+)/;
if(dp[i-][mid-] <= dp[i][j-mid]) l = mid;
else r = mid-;
}
dp[i][j] = min(dp[i][j-l],dp[i-][l])+;
}
}
cout << dp[n][k] << endl;
}
return ;
}
1008.约瑟夫环。
#include<bits/stdc++.h>
using namespace std; int n,k;
int nextt[],lastt[]; int main()
{
while(cin >> n >> k)
{
for(int i = ;i < n;i++) nextt[i] = i+;
for(int i = ;i <= n;i++) lastt[i] = i-;
nextt[n] = ;
lastt[] = n;
cout << k;;
nextt[lastt[k]] = nextt[k];
lastt[nextt[k]] = lastt[k];
int cnt = n-,now = nextt[k];
while(cnt)
{
int left = (k-)%cnt;
while(left--) now = nextt[now];
cout << " " << now;
nextt[lastt[now]] = nextt[now];
lastt[nextt[now]] = lastt[now];
now = nextt[now];
cnt--;
}
cout << endl;
}
return ;
}
1009.约瑟夫环线段树实现，查找和更新一起做了。
#include<bits/stdc++.h>
using namespace std; struct segtree
{
int left,right,sum;
}tree[]; void build(int pos,int l,int r)
{
tree[pos].left = l;
tree[pos].right = r;
if(l == r)
{
tree[pos].sum = ;
return;
}
int mid = (l+r)/;
build(pos*,l,mid);
build(pos*+,mid+,r);
tree[pos].sum = tree[pos*].sum+tree[pos*+].sum;
} int update(int pos,int num)
{
tree[pos].sum--;
if(tree[pos].left == tree[pos].right) return tree[pos].left;
if(num <= tree[pos*].sum) return update(pos*,num);
else return update(pos*+,num-tree[pos*].sum);
} int main()
{ int n,m;
while(cin >> n >> m)
{
build(,,n);
cout << m;
int now = m;
update(,now);
for(int i = ;i <= n;i++)
{
now = (now+m-)%tree[].sum;
if(now == ) now = tree[].sum;
cout << " " << update(,now);
}
cout << endl;
}
return ;
}
1010.分为两段ab和cd，预处理，枚举中点，求两段的和的最大值。
#include<bits/stdc++.h>
using namespace std; int lmax[],rmax[],a[],n; int main()
{
ios::sync_with_stdio(false);
int T;
cin >> T;
while(T--)
{
cin >> n;
for(int i = ;i <= n;i++) cin >> a[i];
int minn = a[];
lmax[] = -1e9;
for(int i = ;i <= n;i++)
{
lmax[i] = max(lmax[i-],a[i]-minn);
minn = min(minn,a[i]);
}
int maxx = a[n];
rmax[n] = -1e9;
for(int i = n-;i >= ;i--)
{
rmax[i] = max(rmax[i+],maxx-a[i]);
maxx = max(maxx,a[i]);
}
int ans = -1e9;
for(int i = ;i < n-;i++) ans = max(lmax[i]+rmax[i+],ans);
cout << ans << endl;
}
return ;
}
1011.规律。
#include<bits/stdc++.h>
using namespace std; long long n; int main()
{
ios::sync_with_stdio(false);
while(cin >> n)
{
long long y = sqrt(n);
if(y*y <= n) y++;
if(y%) cout << "even" << endl;
else cout << "odd" << endl;
}
return ;
}
1012.KMP的运用，先将串反转。
#include<bits/stdc++.h>
using namespace std; int nextt[];
string s; void get_next(int len)
{
int i = ,j = -;
nextt[] = -;
while(i < len)
{
if(j == - || s[len-i-] == s[len-j-]) nextt[++i] = ++j;
else j = nextt[j];
}
} int main ()
{
while(cin >> s)
{
int endd = s.length();
get_next(endd);
int maxlen = ,maxcnt = ;
for(int i = ;i <= endd;i++)
{
int len = i-nextt[i];
if(len && i%len == )
{
int cnt = i/len;
if(cnt >= maxcnt)
{
maxcnt = cnt;
maxlen = len;
}
}
}
if(maxcnt > ) cout << s.substr(endd-maxlen) << " " << maxcnt << endl;
else cout << - << endl;
}
return ;
}
1013.求一个累和。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std; int n,m,a[],b[]; int main()
{
while(cin >> n)
{
long long sum = ;
memset(b,,sizeof(b));
for(int i = ;i <= n;i++) cin >> a[i];
cin >> m;
for(int i = ,l,r,k;i <= m;i++)
{
cin >> l >> r >> k;
b[l+] -= k;
b[r+] += k;
}
for(int i = ;i <= n;i++)
{
b[i] += b[i-];
a[i] += b[i];
sum += a[i];
}
int ave = sum/n;
for(int i = ;i <= n;i++) a[i] -= ave;
sum = ;
for(int i = ;i <= n;i++) sum += (long long)a[i]*a[i];
cout << sum << endl;
}
return ;
}
1014.求最大值期望。
先将l，r和210转化为1-n的数，问题就成了求k次取1-n中的数，求最大值的数学期望。
参考：http://www.zhihu.com/question/30463955?sort=created
#include<bits/stdc++.h>
using namespace std; int l,r;
double x[]; int main()
{
while(cin >> l >> r)
{
if(l < && r <= )
{
cout << "stupid" << endl;
continue;
}
if(l >= )
{
cout << << endl;
continue;
}
int m = r-l+,ok = -l+;
for(int i = ;i <= m;i++) x[i] = ;
double now = ;
int ans = ;
while(m - now <= ok-1e-)
{
now = ;
for(int i = ;i <= m-;i++)
{
x[i] *= (double)i/m;
now += x[i];
}
ans++;
}
cout << ans << endl;
}
return ;
}
1015.先确定最多的位数，然后从高位开始贪心。
#include<bits/stdc++.h>
using namespace std; int n,w[]; int main()
{
ios::sync_with_stdio();
while(~scanf("%d",&n))
{
int minn = 1e9;
for(int i = ;i <= ;i++)
{
scanf("%d",&w[i]);
minn = min(minn,w[i]);
}
if(n < minn)
{
printf("-1\n");
continue;
}
int cnt = n/minn,t = n%minn,now = ;
while(n >= minn)
{
while(t+minn < w[now]) now--;
printf("%d",now);
n -= w[now];
t += minn-w[now];
}
printf("\n");
}
return ;
}
1016.a+b。
#include<bits/stdc++.h>
using namespace std; int main()
{
ios::sync_with_stdio(false);
long long a,b;
while(cin >> a >> b) cout << a+b << endl;
return ;
}
1017.排序。
#include<bits/stdc++.h>
using namespace std; int n;
struct aa
{
string name;
int rating;
friend bool operator <(aa x,aa y)
{
if(x.rating != y.rating) return x.rating > y.rating;
return x.name < y.name;
}
}a[]; int main()
{
ios::sync_with_stdio(false);
while(cin >> n && n)
{
for(int i = ;i < n;i++) cin >> a[i].name >> a[i].rating;
sort(a,a+n);
for(int i = ;i < n;i++) cout << a[i].name << " " << a[i].rating << endl;
}
return ;
}
1018.约瑟夫环数学结论。
#include<bits/stdc++.h>
using namespace std; int n,k,t; int fun(int x)
{
int a = (n-x+k)%(n-x+);
for(int i = ;i <= x;)
{
int temp = min(x+-i,(n-x+i-a-)/k+);
a = (a+k*temp)%(n-x+i+temp-);
i += temp;
}
return a;
}
int main()
{
while(cin >> n >> k >> t)
{
int x;
cin >> x;
cout << fun(x)+;
while(--t)
{
cin >> x;
cout << " " << fun(x)+;
}
cout << endl;
}
return ;
}
1019.末尾0由2*5构成，2足够多，看5的数量就可以了。注意25是2个5，125是3个5...先用等比公式推出趋于极限时，n = m*4，但实际可能比这个数大一点，再向上找。
或者直接二分找5的个数大于等于n的最小值，再判断。
#include<bits/stdc++.h>
using namespace std; int n; int f(int x)
{
int now = ,ans = ;
while(now <= x)
{
ans += x/now;
now *= ;
}
return ans;
} int main()
{
ios::sync_with_stdio(false);
int T;
cin >> T;
while(T--)
{
cin >> n;
int ans = n*;
while(f(ans) < n) ans++;
if(f(ans) == n) cout << ans << endl;
else cout << "No solution" << endl;
}
return ;
}
1020.概率dp，dp[i][j]表示前i题对j题的概率。
#include<bits/stdc++.h>
using namespace std; int n,k;
double a[],dp[][]; int main()
{
ios::sync_with_stdio(false);
int T;
cin >> T;
while(T--)
{
cin >> n >> k;
for(int i = ;i <= n;i++) cin >> a[i];
dp[][] = -a[];
dp[][] = a[];
for(int i = ;i <= n;i++)
{
dp[i][] = dp[i-][]*(-a[i]);
for(int k = ;k <= n;k++) dp[i][k] = dp[i-][k]*(-a[i])+dp[i-][k-]*a[i];
}
double ans = ;
for(int i = k;i <= n;i++) ans += dp[n][i];
cout << fixed << setprecision() << ans << endl;
}
return ;
}
1021.分割成三角形，求面积和。
#include<bits/stdc++.h>
#define PI atan(1)*4
using namespace std; int main()
{
int T;
cin >> T;
while(T--)
{
int n;
double x;
cin >> n >> x;
cout << fixed << setprecision() << 0.25*n*x*x/tan(PI/n) <<endl;
}
return ;
}
1022.统计每个数被加了多少次，两次前缀和。
#include<bits/stdc++.h>
using namespace std; int sum[];
int main()
{
ios::sync_with_stdio(false);
for(int i = ;i <= ;i++)
{
for(int j = i;j <= ;j += i) sum[j]++;
}
for(int i = ;i <= ;i++) sum[i] = (sum[i]+sum[i-])%;
for(int i = ;i <= ;i++) sum[i] = (sum[i]+sum[i-])%;
int a,b;
while(cin >> a >> b) cout << (sum[b]-sum[a-]+)% << endl;
return ;
}
1023.区间不交叉，直接排序，二分按r中找r满足最左段，判断是否在该段中。
#include<bits/stdc++.h>
using namespace std; int n,m;
struct city
{
int l,r,id;
city(){};
city(int ll,int rr,int idd):l(ll),r(rr),id(idd){};
friend bool operator <(city x,city y)
{
return x.r < y.r;
}
}a[]; int main()
{
ios::sync_with_stdio(false);
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i = ;i <= n;i++) scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].id);
sort(a+,a+n+);
scanf("%d",&m);
while(m--)
{
int ip;
scanf("%d",&ip);
int ans = lower_bound(a+,a++n,city(,ip,))-a;
if(ans == n+ || ip < a[ans].l) printf("-1\n");
else printf("%d\n",a[ans].id);
}
}
return ;
}
1024.树状数组维护一下前缀和。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; int n,tree[]; inline int lowbit(int x)
{
return x&-x;
} void add(int pos,int x)
{
while(pos <= )
{
tree[pos] += x;
pos += lowbit(pos);
}
} int getsum(int pos)
{
int sum = ;
while(pos > )
{
sum += tree[pos];
pos -= lowbit(pos);
}
return sum;
} int main()
{
ios::sync_with_stdio(false);
int T;
scanf("%d",&T);
while(T--)
{
memset(tree,,sizeof(tree));
scanf("%d",&n);
long long ans = ;
for(int i = ;i <= n;i++)
{
int x;
scanf("%d",&x);
add(x+,);
ans = (ans+i-getsum(x+))%MOD;
}
printf("%lld\n",ans);
}
return ;
}
1025.记忆化dp。
#include<bits/stdc++.h>
using namespace std; vector<int> make[];
int v[],dp[],n,m,w; int dfs(int x)
{
if(dp[x] != -) return dp[x];
dp[x] = v[x];
if(!make[x].empty())
{
int sum = w;
for(int i = ;i < make[x].size();i++) sum += dfs(make[x][i]);
dp[x] = min(dp[x],sum);
}
return dp[x];
} int main()
{
int T;
scanf("%d",&T);
while(T--)
{
memset(dp,-,sizeof(dp));
scanf("%d%d%d",&n,&m,&w);
for(int i = ;i < n;i++) make[i].clear();
for(int i = ,cnt;i < n;i++)
{
scanf("%d%d",&v[i],&cnt);
for(int j = ;j <= cnt;j++)
{
int x;
scanf("%d",&x);
make[i].push_back(x);
}
}
int ans = ;
for(int i = ;i <= m;i++)
{
int x;
scanf("%d",&x);
ans += dfs(x);
}
printf("%d\n",ans);
}
return ;
}
1026.快速幂模版。
#include<bits/stdc++.h>
using namespace std; long long a,b,c; long long qpower(long long a, long long b, long long c)
{
long long ans = ;
a = a%c;
while(b)
{
if(b%) ans = ans*a%c;
a = a*a%c;
b /= ;
}
return ans;
} int main()
{
ios::sync_with_stdio();
while(cin >> a >> b >> c) cout << qpower(a,b,c) << endl;
return ;
}
1027.矩阵快速幂模版。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; int n;
struct matrix
{
long long m[][];
}; matrix one = { ,,
, };
matrix base = { ,,
, }; matrix mul(matrix a, matrix b)
{
matrix tmp;
for(int i = ; i < ;i++)
{
for(int j = ; j < ;j++)
{
tmp.m[i][j] = ;
for(int k = ; k < ;k++) tmp.m[i][j] = (tmp.m[i][j]+a.m[i][k]*b.m[k][j])%MOD;
}
}
return tmp;
} long long fast_mod(int n)
{
matrix ans = one,y = base;
while(n)
{
if(n&) ans = mul(ans,y);
y = mul(y,y);
n /= ;
}
return ans.m[][];
} int main()
{
while(cin >> n) cout << fast_mod(n-) << endl;
return ;
}
1028.素数筛，dp求最小花费。
#include<bits/stdc++.h>
using namespace std; int n,cnt = ,vis[] = {},prime[],dp[]; void getprime(int MAXN)
{
for(int i = ;i <= MAXN;i++)
{
if(!vis[i]) prime[++cnt] = i;
for(int j = ;j <= cnt && (long long)i*prime[j] <= MAXN;j++)
{
vis[prime[j]*i] = ;
if(!(i%prime[j])) break;
}
}
} int main()
{
getprime();
memset(dp,0x3f,sizeof(dp));
dp[] = ;
for(int i = ;i <= ;i++)
{
dp[i+] = min(dp[i]+,dp[i+]);
for(int j = ;j <= cnt && i*prime[j] <= ;j++)
{
int t = i*prime[j];
dp[t] = min(dp[i]+,dp[t]);
}
}
while(~scanf("%d",&n)) printf("%d\n",dp[n]);
return ;
}
1029.首位化log再化回来，末尾快速幂模10。
#include<bits/stdc++.h>
using namespace std; int n; long long PowerMod(int a, int b, int c)
{
long long ans = ;
a = a%c;
while(b > )
{
if(b % == ) ans = (ans*a)%c;
b = b/;
a = (a*a)%c;
}
return ans;
} int main()
{
int T;
cin >> T;
while(T--)
{
cin >> n;
double x = log10()*n;
int ans = pow(,x-(int)x)+1e-;
cout << ans << " " << PowerMod(,n,) << endl;
}
return ;
}
1030.排序后枚举前两个，二分第三个。
#include<bits/stdc++.h>
using namespace std; int a[],n; int main()
{
ios::sync_with_stdio(false);
while(cin >> n)
{
int flag = ;
for(int i = ;i <= n;i++) cin >> a[i];
sort(a+,a++n);
for(int i = ;i <= n-;i++)
{
for(int j = i+;j <= n-;j++)
{
int sum = -a[i]-a[j];
int pos = lower_bound(a+j+,a+n+,sum)-a;
if(pos != n+ && a[pos] == sum)
{
flag = ;
cout << a[i] << " " << a[j] << " " << a[pos] << endl;
}
}
}
if(flag) cout << "Ren Chou Jiu Gai Duo Du Shu!" << endl;
}
return ;
}
1031.规律，a[i][j] = (i+j)/gcd(i,j)。
#include<bits/stdc++.h>
using namespace std; int n,m; int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> m)
{
int ans = ;
for(int i = ;i <= n;i++)
{
for(int j = ;j <= m;j++)
{
ans += (i+j)/__gcd(i,j);
}
}
cout << ans << endl;
}
return ;
}
1032.规律，组合数，lacus模版。
#include<bits/stdc++.h>
#define LL long long
using namespace std; LL n,m; LL PowerMod(LL a, LL b, LL c)
{
LL ans = ;
a = a % c;
while(b)
{
if(b&) ans = (ans*a)%c;
b = b>>;
a = (a*a)%c;
}
return ans;
} LL c(LL m,LL n)
{
if(m < n) return ;
if(m == n) return ;
if(n > m-n) n = m-n;
LL mm = ,nn = ;
for(LL i = ;i < n;i++)
{
mm = mm*(m-i)%;
nn = nn*(n-i)%;
}
return mm*PowerMod(nn,,)%;
} LL lucas(LL m,LL n)
{
LL ans = ;
while(m && n && ans)
{
ans = ans%*c(m%,n%)%;
n /= ;
m /= ;
}
return ans;
} int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> m) cout << lucas(m,n) << endl;
return ;
}
1033.直接比较。
#include<bits/stdc++.h>
using namespace std; int main()
{
ios::sync_with_stdio(false);
int T;
cin >> T;
while(T--)
{
int a;
cin >> a;
if(a <= ) cout << "YES" << endl;
else cout << "NO" << endl;
}
return ;
}
1034.直接模拟。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
int n,T;
cin >> T;
while(T--)
{
cin >> n;
int sub = ,ans = ;
while(n > sub)
{
n -= sub;
ans++;
sub *= ;
}
if(n <= sub/) cout << ans- << " " << sub/ << endl;
else cout << ans << " " << n << endl;
}
return ;
}
1035.set去重判断。
#include<bits/stdc++.h>
using namespace std; int a[][];
int x[] = {,,,,,,,,},y[] = {,,,,,,,,}; set<int> s;
int main()
{
ios::sync_with_stdio(false);
int T;
cin >> T;
while(T--)
{
int flag = ;
for(int i = ;i <= ;i++)
{
s.clear();
for(int j = ;j <= ;j++)
{
cin >> a[i][j];
if(a[i][j] == || a[i][j] > ) flag = ;
s.insert(a[i][j]);
}
if(s.size() != ) flag = ;
}
for(int j = ;j <= ;j++)
{
s.clear();
for(int i = ;i <= ;i++) s.insert(a[i][j]);
if(s.size() != ) flag = ;
}
for(int k = ;k < ;k++)
{
s.clear();
for(int i = x[k],cnt1 = ;cnt1 < ;i++,cnt1++)
{
for(int j = y[k],cnt2 = ;cnt2 < ;j++,cnt2++) s.insert(a[i][j]);
}
if(s.size() != ) flag = ;
}
there:
if(flag) cout << "yes" << endl;
else cout << "no" << endl;
}
return ;
}
1036.容量为sum/2的01背包。
#include<bits/stdc++.h>
using namespace std; int n,w[],dp[]; int main()
{
int T;
cin >> T;
while(T--)
{
memset(dp,,sizeof(dp));
int sum = ;
cin >> n;
for(int i = ;i <= n;i++)
{
cin >> w[i];
sum += w[i];
}
int endd = sum/;
for(int i = ;i <= n;i++)
{
for(int j = endd;j >= w[i];j--)
{
dp[j] = max(dp[j],dp[j-w[i]]+w[i]);
}
}
cout << sum-dp[endd]* << endl;
}
return ;
}
1037.构造逆向逆数b，然后找a的后缀和b的前缀相同的部分。
#include<bits/stdc++.h>
using namespace std; int a[],b[];
string s; int main()
{
int T;
cin >> T;
while(T--)
{
cin >> s;
int n = s.length();
for(int i = ;i <= n;i++)
{
a[i] = s[i-]-'';
b[n-i+] = a[i]^;
}
int pos;
for(pos = ;pos <= n;pos++)
{
int flag = ;
for(int i = pos,j = ;i <= n;i++,j++)
{
if(a[i] != b[j])
{
flag = ;
break;
}
}
if(flag) break;
}
for(int i = ;i < pos;i++) cout << a[i];
for(int i = ;i <= n;i++) cout << b[i];
cout << endl;
}
return ;
}
1038.状压dp。
#include<bits/stdc++.h>
using namespace std; int a[][],sta[],dp[][],sum[][],n,m;
int main()
{
ios::sync_with_stdio(false);
int T;
cin >> T;
while(T--)
{
memset(dp,,sizeof(dp));
memset(sum,,sizeof(sum));
cin >> n >> m;
for(int i = ;i <= n;i++)
{
for(int j = ;j <= m;j++) cin >> a[i][j];
}
int cnt = ,endd = <<(m-);
for(int i = ;i < endd;i++)
{
if(i & (i<<)) continue;
sta[++cnt] = i;
}
for(int i = ;i < n;i++)
{
for(int j = ;j <= cnt;j++)
{
for(int k = ;k <= m-;k++)
{
if(<<(k-) & sta[j])
{
if(a[i][k] && a[i][k+] && a[i+][k] && a[i+][k+]) sum[i][j]++;
}
}
}
}
for(int i = ;i <= cnt;i++) dp[][i] = sum[][i];
for(int i = ;i < n;i++)
{
for(int j = ;j <= cnt;j++)
{
for(int k = ;k <= cnt;k++)
{
dp[i+][k] = max(dp[i+][k],dp[i][j]+sum[i+][k]);
if(sta[j] & sta[k]) continue;
if(sta[j] & (sta[k]<<)) continue;
if(sta[j] & (sta[k]>>)) continue;
dp[i+][k] = max(dp[i+][k],dp[i][j]+sum[i+][k]);
}
}
}
int ans = ;
for(int i = ;i <= cnt;i++) ans = max(ans,dp[n-][i]);
cout << ans << endl;
}
return ;
}
1039.大模拟。
#include<bits/stdc++.h>
using namespace std; struct yy
{
int num,y;
friend bool operator<(yy A,yy B)
{
return A.y > B.y;
}
}y;
struct zz
{
int num,z;
friend bool operator<(zz A,zz B)
{
return A.z > B.z;
}
}z; struct xx
{
int x,last,next;
}a[];
int eat[],die[],n,m,live,turn;
priority_queue<yy> qy;
priority_queue<zz> qz; void fun(int i)
{
a[i].x--;
if(a[i].x == )
{
die[i] = ;
eat[i] = ;
live--;
a[a[i].last].next = a[i].next;
a[a[i].next].last = a[i].last;
}
} void eatt(int num)
{
if(!eat[num]) fun(num);
for(int i = ;i <= n;i++)
{
if(eat[i]) fun(i);
}
} int main()
{
int T;
cin >> T;
while(T--)
{
while(!qy.empty()) qy.pop();
while(!qz.empty()) qz.pop();
memset(die,,sizeof(die));
memset(eat,,sizeof(eat));
cin >> n >> m;
for(int i = ;i <= n;i++)
{
a[i].last = i-;
a[i].next = i+;
cin >> a[i].x;
}
a[].last = n;
a[n].next = ;
for(int i = ;i <= n;i++)
{
y.num = i;
cin >> y.y;
qy.push(y);
}
for(int i = ;i <= n;i++)
{
z.num = i;
cin >> z.z;
qz.push(z);
}
turn = ,live = n;
while(m--)
{
int card,now;
cin >> card;
if(n == ) break;
switch(card)
{
case :
now = qy.top().num;
while(die[now])
{
qy.pop();
now = qy.top().num;
}
eatt(now);
break;
case :
now = qz.top().num;
while(die[now])
{
qz.pop();
now = qz.top().num;
}
eatt(now);
break;
case :
eat[turn] = ;
break;
case :
eatt(a[turn].last);
break;
case :
eatt(a[turn].next);
break;
case :
eatt(turn);
break;
}
if(live == ) break;
turn = a[turn].next;
while(die[turn]) turn = a[turn].next;
if(live == ) break;
}
for(int i = ;i <= m;i++)
{
int card;
cin >> card;
}
switch(live)
{
case :
cout << << endl;
break;
case :
cout << turn << endl;
break;
default:
cout << - << endl;
break;
}
}
return ;
}
1040.解方程组判断。
#include<bits/stdc++.h>
using namespace std; int a,b,c; int main()
{
int T;
cin >> T;
while(T--)
{
cin >> a >> b >> c;
int x = (a+b-c)/,y = (a+c-b)/,z = (b+c-a)/;
if((x+y+z)*==(a+b+c)&&<=x&&x<=&&<=y&&y<=&&<=z&&z<=) cout << x << " " << y << " " << z << endl;
else cout << "Impossible" << endl;
}
return ;
}
1041.奇偶性规律。
#include<bits/stdc++.h>
using namespace std; int n,m; int main()
{
int T;
cin >> T;
while(T--)
{
cin >> n >> m;
if(n% || m%) cout << << endl;
else cout << << endl;
}
return ;
}
1042.每次操作最小的两个数。
#include<bits/stdc++.h>
using namespace std; priority_queue<double,vector<double>,greater<double> > q;
int n; int main()
{
int T;
cin >> T;
while(T--)
{
cin >> n;
for(int i = ;i <= n;i++)
{
double temp;
cin >> temp;
q.push(temp);
}
for(int i = ;i < n;i++)
{
double a = q.top();
q.pop();
double b = q.top();
q.pop();
q.push((a+b)/);
}
cout << fixed << setprecision() << q.top()+0.00001 << endl;
q.pop();
}
return ;
}
1043.概率dp，dp[i][j]表示前i天上j节课的概率，等比求和求期望。
#include<bits/stdc++.h>
using namespace std; double p[],dp[][];
int n,k; int main()
{
int T;
cin >> T;
while(T--)
{
cin >> n >> k;
for(int i = ;i <= n;i++) cin >> p[i];
dp[][] = ;
for(int i = ;i <= n;i++)
{
dp[i][] = dp[i-][]*(-p[i]);
for(int j = ;j <= n;j++) dp[i][j] = dp[i-][j-]*p[i]+dp[i-][j]*(-p[i]);
}
double x = ;
for(int i = k;i <= n;i++) x += dp[n][i];
cout << x/(-x) << endl;
}
return ;
}
1044.枚举所有情况就可以了，注意每种牌最多只有4张。
#include<bits/stdc++.h>
using namespace std; int dp[][][] = {},a[]; void init()
{
for(int i = ;i >= ;i--) dp[i][i][i] = dp[i+][i+][i+]+;
dp[][][] = ;
for(int i = ;i >= ;i--) dp[i][i+][i+] = dp[i+][i+][i+]+;
int x = ,y = ,z = ;
dp[][][] -= ;
for(int i = ;i >= ;i--)
{
int j;
for(j = ;j > i;j--)
{
dp[i][i][j] = dp[x][y][z]+;
x = i;
y = i;
z = j;
}
for(j--;j >= ;j--)
{
dp[j][i][i] = dp[x][y][z]+;
x = j;
y = i;
z = i;
}
}
dp[][][] += ;
dp[][][] += ;
for(int k = ;k >= ;k--)
{
for(int j = k-;j >= ;j--)
{
for(int i = j-;i >= ;i--)
{
if(dp[i][j][k]) continue;
dp[i][j][k] = dp[x][y][z]+;
x = i;
y = j;
z = k;
}
}
}
dp[][][] -= ;
dp[][][] += ;
} int main()
{
ios::sync_with_stdio(false);
init();
int T;
cin >> T;
while(T--)
{
cin >> a[] >> a[] >> a[];
sort(a,a+);
cout << dp[a[]][a[]][a[]] << endl;
}
return ;
}
1045.博弈dfs，注意dfs过程中还原原图。
#include<bits/stdc++.h>
using namespace std; string a[];
int dir[][] = {-,,,-,,,,}; bool dfs(int x,int y)
{
a[x][y] = '';
for(int i = ;i < ;i++)
{
int xx = x+dir[i][],yy = y+dir[i][];
if(xx > || xx < || yy > || yy < || a[xx][yy] == '') continue;
if(dfs(xx,yy))
{
a[x][y] = '';
return ;
}
}
a[x][y] = '';
return ;
} int main()
{
int T;
cin >> T;
while(T--)
{
for(int i = ;i <= ;i++)
{
cin >> a[i];
a[i] = " "+a[i];
}
int flag = ;
for(int i = ;i <= ;i++)
{
for(int j = ;j <= ;j++)
{
if(a[i][j] == '' && dfs(i,j))
{
flag = ;
break;
}
}
if(flag) break;
}
if(flag) cout << "win" << endl;
else cout << "lose" << endl;
}
return ;
}
1046.string类型的高精度。
#include<bits/stdc++.h>
using namespace std; int compare(string str1,string str2)
{
if(str1.length() > str2.length()) return ;
else if(str1.length() < str2.length()) return -;
else return str1.compare(str2);
} string add(string str1,string str2)
{
string str;
int len1 = str1.length(),len2 = str2.length();
if(len1 < len2)
{
for(int i = ;i <= len2-len1;i++) str1 = ""+str1;
}
else
{
for(int i = ;i <= len1-len2;i++) str2 = ""+str2;
}
int cf = ,temp;
for(int i = str1.length()-;i >= ;i--)
{
temp = str1[i]-''+str2[i]-''+cf;
cf = temp/;
temp %= ;
str = char(temp+'')+str;
}
if(cf != ) str = char(cf+'')+str;
return str;
} string sub(string str1,string str2)
{
string str;
int flag = ;
if(compare(str1,str2) < )
{
flag = ;
swap(str1,str2);
}
int tmp = str1.length()-str2.length(),cf = ;
for(int i = str2.length()-;i >= ;i--)
{
if(str1[tmp+i] < str2[i]+cf)
{
str = char(str1[tmp+i]-str2[i]-cf+''+)+str;
cf = ;
}
else
{
str = char(str1[tmp+i]-str2[i]-cf+'')+str;
cf = ;
}
}
for(int i = tmp-;i >= ;i--)
{
if(str1[i]-cf >= '')
{
str = char(str1[i]-cf)+str;
cf = ;
}
else
{
str = char(str1[i]-cf+)+str;
cf = ;
}
}
str.erase(,str.find_first_not_of(''));
if(str.empty()) str = "";
if(flag) str = "-"+str;
return str;
} string mul(string str1,string str2)
{
string str;
int len1 = str1.length();
int len2 = str2.length();
string tempstr;
for(int i = len2-;i >= ;i--)
{
tempstr = "";
int temp = str2[i]-'',t = ,cf = ;
if(temp != )
{
for(int j = ;j <= len2--i;j++) tempstr += "";
for(int j = len1-;j >= ;j--)
{
cf = (temp*(str1[j]-'')+cf);
t = cf%;
cf /= ;
tempstr = char(t+'')+tempstr;
}
if(cf != ) tempstr = char(cf+'')+tempstr;
str=add(str,tempstr);
}
}
str.erase(,str.find_first_not_of(''));
if(str.empty()) str = "";
return str;
} void div(string str1,string str2,string "ient,string &residue)
{
quotient = "";
residue = "";
if(str2 == "")
{
quotient = "ERROR";
residue = "ERROR";
return;
}
if(str1 == "")
{
quotient = "";
residue = "";
return;
}
int res = compare(str1,str2);
if(res < )
{
quotient = "";
residue = str1;
return;
}
else
{
int len1 = str1.length();
int len2 = str2.length();
string tempstr;
tempstr.append(str1,,len2-);
for(int i = len2-;i < len1;i++)
{
tempstr = tempstr+str1[i];
tempstr.erase(,tempstr.find_first_not_of(''));
if(tempstr.empty()) tempstr = "";
for(char ch = '';ch >= '';ch--)
{
string str,tmp;
str = str+ch;
tmp = mul(str2,str);
if(compare(tmp,tempstr) <= )
{
quotient = quotient+ch;
tempstr = sub(tempstr,tmp);
break;
}
}
}
residue = tempstr;
}
quotient.erase(,quotient.find_first_not_of(''));
if(quotient.empty()) quotient = "";
} int main()
{
string str1,str2;
string str3,str4;
while(cin >> str1 >> str2)
{
div(str1,str2,str3,str4);
cout << add(str1,str2) << ' ' << sub(str1,str2) << ' ' << mul(str1,str2) << ' ' << str3 << ' ' << str4 << endl;
}
return ;
}
1048.二分匹配模版。
#include<bits/stdc++.h>
using namespace std; int n,m,l;
int linker[];
bool used[];
vector<int> v[]; bool dfs(int u)
{
for(int i = ;i < v[u].size();i++)
{
int t = v[u][i];
if(used[t]) continue;
used[t] = ;
if(linker[t] == - || dfs(linker[t]))
{
linker[t] = u;
return ;
}
}
return ;
} int MaxMatch()
{
int ans = ;
memset(linker,-,sizeof(linker));
for(int i = ;i < n;i++)
{
memset(used,,sizeof(used));
if(dfs(i)) ans++;
}
return ans;
} int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> m)
{
for(int i = ;i < n;i++) v[i].clear();
cin >> l;
for(int i = ,a,b;i < l;i++)
{
cin >> a >> b;
v[a].push_back(b);
}
cout << MaxMatch() << endl;
}
return ;
}
1053.直接判断正负。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
while(cin >> n)
{
if(n > ) cout << "yes" << endl;
else if(n < ) cout << "no" << endl;
else cout << "light" << endl;
}
return ;
}
1054.权值累加。
#include<bits/stdc++.h>
using namespace std; int n,a[] = {,,,,,,,,,}; int main()
{
ios::sync_with_stdio(false);
while(cin >> n)
{
int ans = ;
while(n)
{
ans += a[n%];
n /= ;
}
cout << ans << endl;
}
return ;
}
1055.模拟遍历。
include<bits/stdc++.h>
using namespace std; int a[][],ans[],n,m; int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> m)
{
int total = n*m;
for(int i = ;i <= n;i++)
{
for(int j = ;j <= m;j++) cin >> a[i][j];
}
int now = ,x = ,y = ;
ans[] = a[][];
while(now < total)
{
if(x < n) ans[++now] = a[++x][y];
else ans[++now] = a[x][++y];
while(x > && y < m) ans[++now] = a[--x][++y];
if(y < m) ans[++now] = a[x][++y];
else ans[++now] = a[++x][y];
while(x < n && y > ) ans[++now] = a[++x][--y];
}
cout << ans[];
for(int i = ;i <= total;i++) cout << " " << ans[i];
cout << endl;
}
return ;
}
1056.dp。
#include<bits/stdc++.h>
using namespace std; int dp[][] = {},n; int main()
{
ios::sync_with_stdio(false);
dp[][] = ;
for(int i = ;i <= ;i++)
{
dp[i][] = ;
for(int j = ;j < i;j++) dp[i][j] = (dp[i-][j]+dp[i][j-])%;
dp[i][i] = dp[i][i-];
}
while(cin >> n) cout << dp[n][n] << endl;
return ;
}
1057.可重复组合，lucas模版。
#include<bits/stdc++.h>
#define LL long long
using namespace std; LL n,m; LL PowerMod(LL a, LL b, LL c)
{
LL ans = ;
a = a % c;
while(b)
{
if(b&) ans = (ans*a)%c;
b = b>>;
a = (a*a)%c;
}
return ans;
} LL c(LL m,LL n)
{
if(m < n) return ;
if(m == n) return ;
if(n > m-n) n = m-n;
LL mm = ,nn = ;
for(LL i = ;i < n;i++)
{
mm = mm*(m-i)%;
nn = nn*(n-i)%;
}
return mm*PowerMod(nn,,)%;
} LL lucas(LL m,LL n)
{
LL ans = ;
while(m && n && ans)
{
ans = ans%*c(m%,n%)%;
n /= ;
m /= ;
}
return ans;
} int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> m) cout << lucas(m+n-,m) << endl;
return ;
}
1058.把边的贡献算到点上，贡献大于0的点在子图内，小于0的点在子图外。
#include<bits/stdc++.h>
#define LL long long
using namespace std; LL a[];
int n,m,u,v,w; int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> m)
{
memset(a,,sizeof(a));
for(int i = ;i <= m;i++)
{
cin >> u >> v >> w;
a[u] += w;
a[v] += w;
}
LL ans = ;
for(int i = ;i <= n;i++) ans += a[i] > ?a[i]:-a[i];
cout << ans/ << endl;
}
return ;
}
1059.按时间以ms为单位模拟，可以用优先队列优化，好像还可以用数学的方法快速算。
#include<bits/stdc++.h>
using namespace std; double ha,hb,maxha,maxhb,aa,ab,adda,addb,sub2a,sub2b,sub1a,sub1b,ara,arb;
int ta,tb,cool1a,cool1b,cool2a,cool2b; bool cmp(double x,double y)
{
return x-y > 1e-;
} int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
cin >> ha >> maxha >> aa >> adda >> sub2a >> sub1a >> ara >> ta >> cool1a >> cool2a;
cin >> hb >> maxhb >> ab >> addb >> sub2b >> sub1b >> arb >> tb >> cool1b >> cool2b;
cool2a *= ;
cool2b *= ;
int stopa = /*(-ta),stopb = /*(-tb);
double hita = aa*/(+arb*(-sub1a)/-sub2a),hitb = ab*/(+ara*(-sub1b)/-sub2b);
int ta_att = ,tb_att = ,ta_mag = ,tb_mag = ,buffa = ,buffb = ,ta_buff = ,tb_buff = ,ta_stop = -,tb_stop = -;
int t = ;
while()
{
if(ta_stop == t)
{
ta_att = max(ta_att,t+stopa);
ta_mag = max(ta_mag,t+stopa);
}
if(tb_stop == t)
{
tb_att = max(tb_att,t+stopb);
tb_mag = max(tb_mag,t+stopb);
}
if(--ta_buff <= ) buffa = ;
if(--tb_buff <= ) buffb = ;
if(ta_att == t && tb_att == t)
{
double hitatob = min(hita*(+0.125*buffa),hb);
double hitbtoa = min(hitb*(+0.125*buffb),ha);
hb -= hitatob;
ha -= hitbtoa;
if(!cmp(ha,) && !cmp(hb,)) break;
if(!cmp(ha,))
{
hb = min(maxhb,hb+hitbtoa*addb/);
break;
}
if(!cmp(hb,))
{
ha = min(maxha,ha+hitatob*adda/);
break;
}
hb = min(maxhb,hb+hitbtoa*addb/);
ha = min(maxha,ha+hitatob*adda/);
buffa = min(buffa+,);
buffb = min(buffb+,);
ta_buff = ;
tb_buff = ;
ta_att += cool1a;
tb_att += cool1b;
}
else if(ta_att == t)
{
double hit = min(hita*(+0.125*buffa),hb);
ha = min(maxha,ha+hit*adda/);
hb -= hit;
if(!cmp(hb,)) break;
buffa = min(buffa+,);
ta_buff = ;
ta_att += cool1a;
}
else if(tb_att == t)
{
double hit = min(hitb*(+0.125*buffb),ha);
hb = min(maxhb,hb+hit*addb/);
ha -= hit;
if(!cmp(ha,)) break;
buffb = min(buffb+,);
tb_buff = ;
tb_att += cool1b;
}
if(ta_mag == t)
{
ta_mag = t+cool2a;
tb_stop = t+;
}
if(tb_mag == t)
{
tb_mag = t+cool2b;
ta_stop = t+;
}
t++;
}
if(t% >= ) t = t/+;
else t = t/;
int t1 = t/,t2 = t%;
cout << t1 << "." << setw() << setfill('') << t2 << " ";
cout << fixed << setprecision() << ha+1e- << " " << hb+1e- << endl;
}
return ;
}
1060.不知道如何直接输出，手动处理整数。
#include<bits/stdc++.h>
using namespace std; int a[];
long long x; int main()
{
while(cin >> x)
{
int cnt = ;
if(x% >= ) x = x/+;
else x = x/;
if(x == )
{
cout << "0.00" << endl;
continue;
}
while(x)
{
a[++cnt] = x%;
x /= ;
}
if(cnt > )
{
for(;cnt > ;cnt--) cout << a[cnt];
}
else
{
cout << "";
}
cout << ".";
if(cnt == ) cout << "";
for(;cnt > ;cnt--) cout << a[cnt];
cout << endl;
}
return ;
}
1061.通分后计算，再约分。
#include<bits/stdc++.h>
using namespace std; int a1,a2,b1,b2;
char c; int main()
{ while(cin >> a1 >> c >> a2 >> b1 >> c >> b2)
{
int x = a1*b2+a2*b1,y = a2*b2,g = __gcd(x,y);;
cout << x/g << "/" << y/g << endl;
}
return ;
}
1062.最大连续k段和，注意k小于n的情况。
#include<bits/stdc++.h>
using namespace std; int n,t,a[]; int main()
{
while(cin >> n >> t)
{
for(int i = ;i <= n;i++) cin >> a[i];
int sum = ;
if(n <= t)
{
for(int i = ;i <= n;i++) sum += a[i];
cout << sum << endl;
continue;
}
for(int i = ;i <= t;i++) sum += a[i];
int ans = sum;
for(int l = ,r = t+;r <= n;l++,r++)
{
sum += a[r];
sum -= a[l];
ans = max(ans,sum);
}
cout << ans << endl;
}
return ;
}
1063.数学推导，参考https://pan.baidu.com/s/1i3jljoBZWL74nf6_3aVz-A
#include<bits/stdc++.h>
#define LL long long
#define MOD 1000000007
using namespace std; LL PowerMod(LL a, LL b, LL c)
{
LL ans = ;
a = a % c;
while(b)
{
if(b&) ans = (ans*a)%c;
b = b>>;
a = (a*a)%c;
}
return ans;
} int main()
{
int n;
while(cin >> n)
{
if(n == ) cout << << endl;
else if(n%) cout << PowerMod(,n-,MOD) << endl;
else cout << (PowerMod(,n-,MOD)+PowerMod(,n/-,MOD))%MOD << endl;
}
return ;
}
1064.将坐标旋转45度，注意长度的变化，然后处理一下重叠面积和方格数量的转化。
#include<bits/stdc++.h>
#define LL long long
using namespace std; int main()
{
LL x1,y1,x2,y2,r,X1,Y1,X2,Y2;
while(cin >> x1 >> y1 >> x2 >> y2 >> r)
{
X1 = y1+x1;
Y1 = y1-x1;
X2 = y2+x2;
Y2 = y2-x2;
LL a = max(0LL,*r+-abs(X1-X2)),b = max(0LL,*r+-abs(Y1-Y2));
LL sub = a*b?(a*b-a-b+)/:;
cout << *(*r*r+*r+)-sub << endl;
}
return ;
}
1065.动态开点线段树，维护连续最长段。
#include<bits/stdc++.h>
using namespace std; int n,q,cnt = ,a[];
struct segtree
{
int l,r,lazy,len,x,lx,rx;
segtree *lson,*rson;
}*root,tree[]; segtree *newnode(int l,int r)
{
segtree *t = &tree[++cnt];
t->l = l;
t->r = r;
t->lson = NULL;
t->rson = NULL;
t->lazy = -;
t->len = r-l+;
t->x = t->len;
t->lx = t->len;
t->rx = t->len;
return t;
} segtree *newlson(segtree *pos)
{
int mid = (pos->l+pos->r)/;
return newnode(pos->l,mid);
} segtree *newrson(segtree *pos)
{
int mid = (pos->l+pos->r)/;
return newnode(mid+,pos->r);
} void pushup(segtree *pos)
{
pos->lx = pos->lson->lx;
pos->rx = pos->rson->rx;
pos->x = max(pos->lson->x,pos->rson->x);
if(pos->lx == pos->lson->len) pos->lx += pos->rson->lx;
if(pos->rx == pos->rson->len) pos->rx += pos->lson->rx;
pos->x = max(pos->x,pos->lson->rx+pos->rson->lx);
} void pushdown(segtree *pos)
{
if(!pos->lson) pos->lson = newlson(pos);
if(!pos->rson) pos->rson = newrson(pos);
if(pos->lazy != -)
{
pos->lson->lazy = pos->lazy;
pos->rson->lazy = pos->lazy;
pos->lson->x = pos->lson->lx = pos->lson->rx = pos->lazy*pos->lson->len;
pos->rson->x = pos->rson->lx = pos->rson->rx = pos->lazy*pos->rson->len;
pos->lazy = -;
}
} void update(segtree *pos,int l,int r,int x)
{
if(r < pos->l || pos->r < l) return;
if(l <= pos->l && pos->r <= r)
{
pos->x = pos->lx = pos->rx = pos->len*x;
pos->lazy = x;
return;
}
pushdown(pos);
update(pos->lson,l,r,x);
update(pos->rson,l,r,x);
pushup(pos);
} int getl(segtree *pos,int x)
{
if(pos->x < x) return 2e9;;
if(pos->lx >= x) return pos->l;
pushdown(pos);
int minn = 2e9;
if(pos->lson->rx+pos->rson->lx >= x) minn = pos->lson->r-pos->lson->rx+;
minn = min(minn,getl(pos->lson,x));
minn = min(minn,getl(pos->rson,x));
return minn;
} int main()
{
while(~scanf("%d%d",&n,&q))
{
set<int> s;
map<int,int> mp;
cnt = ;
root = newnode(,n);
while(q--)
{
char ss[];
int x;
scanf("%s",ss);
if(ss[] == 'm')
{
sscanf(ss,"malloc(%d)",&x);
int t = getl(root,x);
if(t == 2e9) printf("0\n");
else
{
printf("%d\n",t);
update(root,t,t+x-,);
s.insert(t);
mp[t] = t+x-;
}
}
else
{
sscanf(ss,"free(%d)",&x);
auto it = s.lower_bound(x);
if(it == s.end() || *it != x) continue;
update(root,*it,mp[*it],);
s.erase(it);
}
}
}
return ;
}
1066.a^b%c = a^(b%phi[c]+phi[c])%c。
#include<bits/stdc++.h>
#define LL long long
#define MOD 1000000007
using namespace std; LL a,b,n;
LL PowerMod(LL a, LL b, LL c)
{
LL ans = ;
a = a % c;
while(b)
{
if(b&) ans = (ans*a)%c;
b = b>>;
a = (a*a)%c;
}
return ans;
} int main()
{
int T;
cin >> T;
while(T--)
{
cin >> a >> n >> b;
int ans = b;
for(int i = ;i <= n;i++)
{
b = (b*b+MOD-)%(MOD-);
ans = ans*b%(MOD-);
}
cout << PowerMod(a,ans,MOD) << endl;
}
return ;
}
1067.dp[i]表示i个人的方案数，更新每个dp[i]时，枚举i的所有约数x，对于每一个x，可以有A(n-1,x-1)的情况的环，再根据乘法原理，乘上dp[i-x]。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; int n,k,cnt;
long long dp[],mul[],inv[],fac[]; long long qpower(long long a, long long b, long long c)
{
long long ans = ;
a = a%c;
while(b)
{
if(b%) ans = ans*a%c;
a = a*a%c;
b /= ;
}
return ans;
} int main()
{
ios::sync_with_stdio();
mul[] = ;
mul[] = ;
inv[] = ;
inv[] = ;
for(int i = ;i <= ;i++) mul[i] = mul[i-]*i%MOD;
for(int i = ;i <= ;i++) inv[i] = inv[i-]*qpower(i,MOD-,MOD)%MOD;
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
cnt = ;
int endd = sqrt(k+0.1);
for(int i = ;i <= endd;i++)
{
if(k%i == )
{
fac[++cnt] = i;
if(k/i != i) fac[++cnt] = k/i;
}
}
sort(fac+,fac+cnt+);
memset(dp,,sizeof(dp));
dp[] = ;
for(int i = ;i <= n;i++)
{
for(int j = ;j <= cnt && fac[j] <= i;j++) dp[i] = (dp[i]+dp[i-fac[j]]*mul[i-]%MOD*inv[i-fac[j]])%MOD;
}
printf("%lld\n",dp[n]);
}
return ;
}
1068.二分答案，根据单调性O(n)判断。
#include<bits/stdc++.h>
#define LL long long
using namespace std; LL a[],b[];
int n,m,k; int f(LL x)
{
int ans = ,now = m;
for(int i = ;i <= n;i++)
{
while(now)
{
if(a[i]*b[now] > x) now--;
else break;
}
ans += now;
}
return n*m-ans;
}
int main()
{
int T;
cin >> T;
while(T--)
{
cin >> n >> m >> k;
k--;
for(int i = ;i <= n;i++) cin >> a[i];
for(int i = ;i <= m;i++) cin >> b[i];
sort(a+,a++n);
sort(b+,b++m);
LL l = a[]*b[],r = a[n]*b[m];
while(l < r)
{
LL mid = (l+r)/;
int t = f(mid);
if(t <= k) r = mid;
else l = mid+;
}
cout << l << endl;
}
return ;
}
1069.建一个虚点连接所有人，再连接人和人之间的关系，求最小生成树。
#include<bits/stdc++.h>
using namespace std; struct line
{
int to,w;
line(int a,int b):to(a),w(b){}
friend bool operator<(line X,line Y)
{
return X.w > Y.w;
}
}; int n,m,k,x,dis[],vis[];
vector<line> v[]; int prim()
{
memset(vis,,sizeof(vis));
memset(dis,0x3f,sizeof(dis));
int ans = ,cnt = ;
priority_queue<line> q;
q.push(line(,));
int endd = n+m;
while(cnt <= endd && !q.empty())
{
while(!q.empty() && vis[q.top().to]) q.pop();
ans += q.top().w;
int now = q.top().to;
vis[now] = ;
cnt++;
for(int i = ;i < v[now].size();i++)
{
if(vis[v[now][i].to]) continue;
if(dis[v[now][i].to] <= v[now][i].w) continue;
dis[v[now][i].to] = v[now][i].w;
q.push(line(v[now][i].to,v[now][i].w));
}
}
return ans;
} int main()
{
int T;
cin >> T;
while(T--)
{
cin >> m >> n >> k >> x;
int endd = n+m;
for(int i = ;i <= endd;i++) v[i].clear();
for(int i = ;i <= endd;i++) v[].push_back(line(i,k));
while(x--)
{
int a,b,c;
cin >> a >> b >> c;
a += ;
b += m+;
c = k-c;
v[a].push_back(line(b,c));
v[b].push_back(line(a,c));
}
cout << prim() << endl;
}
return ;
}
1070.树形dp。
#include<bits/stdc++.h>
using namespace std; int dp[][],n,m;
vector<int> v[]; void dfs(int pos,int pre)
{
for(int i = ;i < v[pos].size();i++)
{
if(v[pos][i] == pre) continue;
int x = v[pos][i];
dfs(x,pos);
for(int j = m;j > ;j--)
{
for(int k = ;k <= j;k++)
{
dp[pos][j] = max(dp[pos][j],dp[x][j-k]+dp[pos][k]);
}
}
}
} int main()
{
int T;
cin >> T;
while(T--)
{
cin >> n >> m;
memset(dp,,sizeof(dp));
for(int i = ;i < n;i++) v[i].clear();
for(int i = ;i < n;i++) cin >> dp[i][];
for(int i = ;i < n;i++)
{
int a,b;
cin >> a >> b;
v[a].push_back(b);
v[b].push_back(a);
}
dfs(,-);
int ans = ;
for(int i = ;i < n;i++) ans = max(ans,dp[i][m]);
cout << ans << endl;
}
return ;
}
1071.枚举长宽。
#include<bits/stdc++.h>
using namespace std; int n,m; int main()
{
int T;
cin >> T;
while(T--)
{
cin >> n >> m;
int ans =;
for(int i = ;i <= n;i++)
{
for(int j = ;j <= m;j++) ans += i*j;
}
cout << ans << endl;
}
return ;
}
1072.求关键边的数量，即强连通分量-1，tarjan模版。
#include<bits/stdc++.h>
using namespace std; int n,m,dfn[],low[],num,cnt;
vector<int> v[]; void tarjan(int pos,int pre)
{
dfn[pos] = low[pos] = ++num;
for(int i = ;i < v[pos].size();i++)
{
int x = v[pos][i];
if(x == pre) continue;
if(dfn[x] == )
{
tarjan(x,pos);
low[pos] = min(low[pos],low[x]);
if(dfn[pos] < low[x]) cnt++;
}
else low[pos] = min(low[pos],dfn[x]);
}
} int main()
{
int T;
cin >> T;
while(T--)
{
num = ;
cnt = ;
memset(dfn,,sizeof(dfn));
memset(low,,sizeof(low));
cin >> n >> m;
for(int i = ;i < n;i++) v[i].clear();
while(m--)
{
int x,y;
cin >> x >> y;
v[x].push_back(y);
v[y].push_back(x);
}
tarjan(,-);
cout << cnt << endl;
}
return ;
}
1073.容量为sum/2的01背包，先判断奇偶。
#include<bits/stdc++.h>
using namespace std; int n,w[],dp[]; int main()
{
int T;
cin >> T;
while(T--)
{
memset(dp,,sizeof(dp));
cin >> n;
int sum = ;
for(int i = ;i <= n;i++)
{
cin >> w[i];
sum += w[i];
}
if(sum&)
{
cout << "No" << endl;
continue;
}
int endd = sum/;
for(int i = ;i <= n;i++)
{
for(int j = endd;j >= w[i];j--) dp[j] = max(dp[j],dp[j-w[i]]+w[i]);
}
if(dp[endd] == endd) cout << "Yes" << endl;
else cout << "No" << endl;
}
return ;
}
*1074.待补。
*1075.待补。
1076.类似于三进制，输入数据中有空行，用gets会出错。
#include<bits/stdc++.h>
using namespace std; char s[]; int f(char x)
{
if(x == '') return ;
if(x == '') return ;
if(x == '') return ;
} int main()
{
while(~scanf("%s",s))
{
int len = strlen(s);
long long ans = ;
for(int i = ;i < len;i++) ans = ans*+f(s[i]);
printf("%lld\n",ans);
}
return ;
}
1077.循环节仅看分母，先把分母2，5除完后为m，求欧拉数，循环节长度为欧拉数约数中最小数x的使10^x%m == 1的值。
#include<bits/stdc++.h>
#define LL long long
using namespace std; LL PowerMod(LL a, LL b, LL c)
{
LL ans = ;
a = a % c;
while(b)
{
if(b&) ans = (ans*a)%c;
b = b>>;
a = (a*a)%c;
}
return ans;
} int euler(int n)
{
int res = n,a = n;
for(int i = ;i*i <= a;i++)
{
if(a%i == )
{
res = res/i*(i-);
while(a%i == ) a /= i;
}
}
if(a > ) res = res/a*(a-);
return res;
} int gcd(int a,int b)
{
return b?gcd(b,a%b):a;
} int main()
{
int p,q;
while(cin >> p >> q)
{
q /= gcd(p,q);
while(q% == ) q /= ;
while(q% == ) q /= ;
if(q == )
{
cout << << endl;
continue;
}
int x = euler(q),now = ;
int xx = sqrt(x),flag = ;
for(int i = ;i <= xx;i++)
{
if(x%i) continue;
if(PowerMod(,i,q) == )
{
cout << i << endl;
flag = ;
break;
}
}
if(flag) continue;
for(int i = xx;i >= ;i--)
{
if(x%i) continue;
if(PowerMod(,x/i,q) == )
{
cout << x/i << endl;
break;
}
}
}
return ;
}
1078.分层图最短路模版，建k+1层图。
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std; struct edge
{
int to,cost;
friend bool operator>(edge x,edge y)
{
return x.cost > y.cost;
}
}e;
vector<edge> v[];
int n,m,k,a[][],dis[],vis[];
priority_queue< edge,vector<edge>,greater<edge> > q; int dij()
{
while(!q.empty()) q.pop();
memset(dis,0x3f,sizeof(dis));
memset(vis,,sizeof(vis));
dis[] = ;
e.to = ;
e.cost = ;
q.push(e);
while(!q.empty())
{
int now = q.top().to,cost = q.top().cost;
if(now == (k+)*n) return dis[(k+)*n];
q.pop();
if(vis[now]) continue;
vis[now] = ;
for(int i = ;i < v[now].size();i++)
{
e = v[now][i];
if(e.cost+dis[now] < dis[e.to])
{
dis[e.to] = e.cost+dis[now];
e.cost = dis[e.to];
q.push(e);
}
} }
} int main()
{
while(cin >> n >> m >> k)
{
for(int i = ;i <= ;i++) v[i].clear();
memset(a,0x3f,sizeof(a));
for(int i = ;i <= m;i++)
{
int u,v,w;
cin >> u >> v >> w;
a[u][v] = min(a[u][v],w);
a[v][u] = min(a[v][u],w);
}
for(int i = ;i <= n;i++)
{
for(int j = ;j <= n;j++)
{
if(i == j || a[i][j] == INF) continue;
for(int l = ;l <= k;l++)
{
e.to = l*n+j;
e.cost = a[i][j];
v[l*n+i].push_back(e);
if(l != k)
{
e.to = (l+)*n+j;
e.cost = ;
v[l*n+i].push_back(e);
}
}
}
}
cout << dij() << endl;
}
return ;
}
1079.枚举两个数，之后的处理类似于最大连续子序列和。
#include<bits/stdc++.h>
using namespace std; int a[],n,m,has[]; int main()
{
int T;
cin >> T;
while(T--)
{
memset(has,,sizeof(has));
cin >> n >> m;
int cnt = ;
for(int i = ;i <= n;i++)
{
cin >> a[i];
if(!has[a[i]]) has[a[i]] = ;
}
int ans = ;
for(int i = ;i <= m;i++)
{
for(int j = ;j <= m;j++)
{
if(i == j) continue;
if(!has[i] || !has[j]) continue;
int sum = ,flag = ;
for(int k = ;k <= n;k++)
{
if(a[k] == i) sum++;
else if(a[k] == j)
{
sum--;
flag = ;
}
if(sum < )
{
sum = ;
flag = ;
}
if(flag) ans = max(ans,sum);
else ans = max(ans,sum-);
}
}
}
cout << ans << endl;
}
return ;
}
1080.规律。
#include<bits/stdc++.h>
using namespace std; int main()
{
int n;
while(cin >> n) cout << n*n* << endl;
return ;
}
1081.判断两个范围是否有交集。
#include<bits/stdc++.h>
using namespace std; long long a1,a2,b1,b2,c1,c2,d1,d2; int main()
{
while(cin >> a1 >> a2 >> b1 >> b2 >> c1 >> c2 >> d1 >> d2)
{
long long x1 = a1*d1,x2 = a2*d2,y1 = b1*c1,y2 = b2*c2;
if(x1 > y2 || x2 < y1) cout << "NO" << endl;
else cout << "YES" << endl;
}
return ;
}
1082.若场数最多的人比其余的和要大，则ans为其余的和，否则ans为sum/2。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
int T;
cin >> T;
while(T--)
{
long long maxx = ,sum = ,temp;
cin >> n;
while(n--)
{
cin >> temp;
maxx = max(maxx,temp);
sum += temp;
}
cout << min(sum/,sum-maxx) << endl;
}
return ;
}
1083.从左到右扫一遍，记录左括号个数，有右括号时更新答案就可以了。
#include<bits/stdc++.h>
using namespace std; int main()
{
int T;
string s;
cin >> T;
while(T--)
{
cin >> s;
int len = s.length(),cnt = ,ans = ;
for(int i = ;i < len;i++)
{
if(s[i] == '(') cnt++;
else if(cnt)
{
cnt--;
ans += ;
}
}
cout << ans << endl;
}
return ;
}
1084.暴力搜索，不会超时。
#include<bits/stdc++.h>
using namespace std; int n,u,v,vis[],sco[],ans,k;
vector<int> pre[]; void dfs(int x,int sum)
{
if(x == )
{
ans = max(ans,sum);
return;
}
for(int i = ;i <= ;i++)
{
if(vis[i]) continue;
int flag = ;
for(int j = ;j < pre[i].size();j++)
{
if(!vis[pre[i][j]])
{
flag = ;
break;
}
}
if(!flag) continue;
vis[i] = ;
dfs(x-,sum+sco[i]);
vis[i] = ;
}
}
int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
memset(vis,,sizeof(vis));
for(int i = ;i <= ;i++) pre[i].clear();
while(n--)
{
cin >> u >> v;
pre[v].push_back(u);
}
for(int i = ;i <= ;i++) cin >> sco[i];
cin >> k;
ans = ;
dfs(k/,);
if(ans >= ) cout << ans << endl;
else cout << "I chose to die" << endl;
}
return ;
}
*1085.待补。
1086.简单模拟。
#include<bits/stdc++.h>
using namespace std; int n,m,h,op[],a[],hh[]; int main()
{
int T;
cin >> T;
while(T--)
{
memset(a,-,sizeof(a));
cin >> n;
for(int i = ;i < n;i++) cin >> op[i];
cin >> m;
for(int i = ;i <= m;i++) cin >> hh[i];
int pos;
for(int i = ;i <= m;i++)
{
cin >> pos;
if(pos <= n) a[pos] = hh[i];
}
cin >> h;
int flag = ,now = ;
while(now < n)
{
if(op[now] == ) h++;
else if(op[now] == ) h--;
if(h == || a[now] != - && h >= a[now])
{
flag = ;
break;
}
now++;
}
if(flag) cout << "V8Orz" << endl;
else cout << now << endl;
}
return ;
}
1087.贪心单位体积的价值。
#include<bits/stdc++.h>
using namespace std; int n,w,V;
struct aa
{
int v;
double a;
}a[]; bool cmp(aa x,aa y)
{
return x.a < y.a;
}
int main()
{
int T;
cin >> T;
while(T--)
{
cin >> n;
for(int i = ;i <= n;i++)
{
cin >> a[i].v >> w;
a[i].a = (double)w/a[i].v;
}
cin >> V;
sort(a+,a++n,cmp);
int now = n;
double ans = ;
while(V && now >= )
{
if(V > a[now].v)
{
ans += a[now].a*a[now].v;
V -= a[now].v;
now--;
}
else
{
ans += V*a[now].a;
break;
}
}
cout << fixed << setprecision() << ans << endl;
}
return ;
}
1088.按字典序从大到小排序。
#include<bits/stdc++.h>
using namespace std; string s[];
int n; bool cmp(string x,string y)
{
string X = x+y,Y = y+x;
return X > Y;
} int main()
{
while(cin >> n)
{
for(int i = ;i <= n;i++) cin >> s[i];
sort(s+,s++n,cmp);
for(int i = ;i <= n;i++) cout << s[i];
cout << endl;
}
return ;
}
1089.时间顺序的bfs，注意多个水珠同时到一点的情况。
#include<bits/stdc++.h>
using namespace std; int n,m,nn,tt,a[][],sta[][],t[][],dir[][] = {-,,,-,,,,};
struct water
{
int x,y,d,t;
water(int _x,int _y,int _d,int _t):x(_x),y(_y),d(_d),t(_t){};
}; void bfs(int x,int y)
{
queue<water> q;
q.push(water(x,y,,));
q.push(water(x,y,,));
q.push(water(x,y,,));
q.push(water(x,y,,));
while(!q.empty())
{
water temp = q.front();
q.pop();
if(temp.t > tt) return;
int xx = temp.x+dir[temp.d][],yy = temp.y+dir[temp.d][];
if(xx < || xx > n || yy < || yy > m) continue;
if(t[xx][yy] == temp.t) continue;
if(sta[xx][yy] == ) q.push(water(xx,yy,temp.d,temp.t+));
else if(sta[xx][yy] == )
{
sta[xx][yy] = ;
t[xx][yy] = temp.t;
q.push(water(xx,yy,,temp.t+));
q.push(water(xx,yy,,temp.t+));
q.push(water(xx,yy,,temp.t+));
q.push(water(xx,yy,,temp.t+));
}
else sta[xx][yy]++;
}
} int main()
{
while(cin >> n >> m >> nn >> tt)
{
memset(sta,,sizeof(sta));
memset(t,,sizeof(t));
for(int i = ;i <= nn;i++)
{
int x;
cin >> a[i][] >> a[i][] >> x;
sta[a[i][]][a[i][]] = x;
}
int x,y;
cin >> x >> y;
bfs(x,y);
for(int i = ;i <= nn;i++)
{
if(t[a[i][]][a[i][]] == ) cout << << " " << sta[a[i][]][a[i][]] << endl;
else cout << << " " << t[a[i][]][a[i][]] << endl;
}
}
return ;
}
1090.树形dp。
#include<bits/stdc++.h>
using namespace std; struct line
{
int next,w;
}l;
vector<line> v[];
int n; int dfs(int x)
{
int sum = ;
for(int i = ;i < v[x].size();i++) sum = max(sum,dfs(v[x][i].next)+v[x][i].w);
return sum;
} int main()
{
int T;
cin >> T;
while(T--)
{
cin >> n;
for(int i = ;i <= ;i++) v[i].clear();
for(int i = ;i <= n;i++)
{
int u;
cin >> u >> l.next >> l.w;
v[u].push_back(l);
}
cout << dfs() << endl;
}
return ;
}
1091.按结束时间排序，贪心。
#include<bits/stdc++.h>
using namespace std; struct vedio
{
int l,r;
}a[];
int n; bool cmp(vedio x,vedio y)
{
return x.r < y.r;
} int main()
{
while(cin >> n && n)
{
for(int i = ;i <= n;i++) cin >> a[i].l >> a[i].r;
sort(a+,a++n,cmp);
int ans = ,time = a[].r,now = ;
for(;now <= n;now++)
{
if(time > a[now].l) continue;
time = a[now].r;
ans++;
}
cout << ans << endl;
}
return ;
}
1092.模拟，每次移动后，不断下落，消去，直到不变化为止。
#include<bits/stdc++.h>
using namespace std; int n,a[][],ans[][]; int fmove()
{
int flag = ;
for(int i = ;i < ;i++)
{
for(int j = ;j < ;j++)
{
if(!a[i][j]) continue;
int t = j;
while(t > && a[i][t-] == )
{
flag = ;
swap(a[i][t],a[i][t-]);
t--;
}
}
}
return flag;
} void fdel()
{
memset(ans,,sizeof(ans));
for(int i = ;i < ;i++)
{
int l = ,cnt = ;
for(int j = ;j <= ;j++)
{
if(a[i][j] == a[i][j-]) cnt++;
else
{
if(cnt >= )
{
for(int k = l;k < j;k++) ans[i][k] = ;
}
l = j;
cnt = ;
}
}
}
for(int j = ;j < ;j++)
{
int l = ,cnt = ;
for(int i = ;i <= ;i++)
{
if(a[i][j] == a[i-][j]) cnt++;
else
{
if(cnt >= )
{
for(int k = l;k < i;k++) ans[k][j] = ;
}
l = i;
cnt = ;
}
}
}
for(int i = ;i < ;i++)
{
for(int j = ;j < ;j++)
{
if(ans[i][j]) a[i][j] = ;
}
}
}
int main()
{
while(~scanf("%d",&n))
{
int flag = ;
memset(a,,sizeof(a));
for(int i = ;i < ;i++)
{
for(int j = ;j < ;j++)
{
scanf("%d",&a[i][j]);
if(!a[i][j]) break;
}
}
while(n--)
{
int x,y,d;
scanf("%d%d%d",&x,&y,&d);
if(!a[x][y] || d+x > || d+x < )
{
printf("Runtime Error\n");
flag = ;
}
else
{
swap(a[x][y],a[x+d][y]);
fmove();
fdel();
while(fmove()) fdel();
}
}
if(flag) continue;
for(int i = ;i < ;i++)
{
for(int j = ;j < ;j++)
{
if(a[i][j]) flag = ;
}
}
if(!flag) printf("Accepted\n");
else printf("Wrong Answer\n");
}
return ;
}
1093.确定两个区间，二分答案。
#include<bits/stdc++.h>
using namespace std;
const double eps=0.0001;
double a,b,c,d; double ans[];
int sum; inline double calc(double x)
{
return a*x*x*x+b*x*x+c*x+d;
} void find(double l,double r)
{
while()
{
double mid = (l+r)/;
if(abs(calc(mid)) < eps)
{
ans[++sum] = mid;
return;
}
if(calc(mid)*calc(l) < ) r = mid;
else l = mid;
}
} int main()
{
while(cin >> a >> b >> c >> d)
{
sum = ;
for(int i=-;i<=;i++)
{
double tmp1 = calc(i),tmp2 = calc(i+);
if(abs(tmp1) < eps) ans[++sum] = i;
else if(abs(tmp2) < eps)
{
ans[++sum] = i+;
i++;
}
else if(tmp1*tmp2 < ) find(i,i+);
if(sum == ) break;
}
sort(ans+,ans+);
cout << fixed << setprecision() << ans[] << " " << ans[] << " " << ans[] << endl;
}
return ;
}
1094.递归输出。
#include<bits/stdc++.h>
using namespace std; int n; void printff(int x)
{
if(x == )
{
cout <<"2(0)";
return;
}
if(x == )
{
cout << "";
return;
}
int sum = ,b = -;
while(sum <= x)
{
sum *= ;
b++;
}
if(sum/ < x)
{
if(sum/ == )
{
cout << "2+";
printff(x-sum/);
}
else
{
cout << "2(";
printff(b);
cout << ")+";
printff(x-sum/);
}
}
else
{
cout << "2(";
printff(b);
cout << ")";
}
} int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
printff(n);
cout << endl;
}
return ;
}
1095.暴力搜索，不会超时。
#include<bits/stdc++.h>
using namespace std; int vis[],a[],n,k,ans; void dfs(int num,int v)
{
if(num == n)
{
if(abs(v-a[]) <= k) ans++;
return;
}
for(int i = ;i <= n;i++)
{
if(vis[i]) continue;
if(abs(v-a[i]) > k) continue;
vis[i] = ;
dfs(num+,a[i]);
vis[i] = ;
}
} int main()
{
ios::sync_with_stdio();
while(cin >> n >> k)
{
memset(vis,,sizeof(vis));
ans = ;
for(int i = ;i <= n;i++) cin >> a[i];
dfs(,a[]);
cout << ans << endl;
}
return ;
}
1096.dp[i][j]表示i拆分成最大为j的数。
#include<bits/stdc++.h>
using namespace std; int dp[][],n; int main()
{
ios::sync_with_stdio();
for(int i = ;i <= ;i++)
{
dp[i][] = ;
dp[][i] = ;
}
for(int i = ;i <= ;i++)
{
for(int j = ;j <= ;j++)
{
if(i == j) dp[i][j] = dp[i][j-]+;
else if(i < j) dp[i][j] = dp[i][j-];
else dp[i][j] = dp[i-j][j]+dp[i][j-];
}
}
while(cin >> n) cout << dp[n][n]- << endl;
return ;
}
1097.dp[i] = dp[i-1]+dp[i-2]+dp[i-3]。
#include<bits/stdc++.h>
using namespace std; int n;
long long dp[]; int main()
{
dp[] = ;
dp[] = ;
dp[] = ;
for(int i = ;i <= ;i++) dp[i] = dp[i-]+dp[i-]+dp[i-];
while(cin >> n) cout << dp[n] << endl;
return ;
}
1098.求欧拉函数模版。
#include<bits/stdc++.h>
using namespace std; int n; int euler(int n)
{
int res = n,a = n;
for(int i = ;i*i <= a;i++)
{
if(a%i == )
{
res = res/i*(i-);
while(a%i == ) a /= i;
}
}
if(a > ) res = res/a*(a-);
return res;
} int main()
{
ios::sync_with_stdio();
while(cin >> n) cout << euler(n) << endl;
return ;
}
1099.双指针，线性扫一遍。
#include<bits/stdc++.h>
using namespace std; int a[],n,k; int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
cin >> n >> k;
for(int i = ;i <= n;i++) cin >> a[i];
int l = ,now = ;
long long ans = ;
for(int i = ;i <= n;i++)
{
now += a[i];
while(now > k) now -= a[++l];
ans += i-l;
}
cout << ans << endl;
}
return ;
}
1100.即求最大矩阵面积，单调栈模版。
#include<bits/stdc++.h>
using namespace std; int n;
stack<int> s;
long long a[],b[]; int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
while(!s.empty()) s.pop();
long long ans = ;
b[] = ;
for(int i = ;i <= n;i++) cin >> a[i];
for(int i = ;i <= n;i++)
{
cin >> b[i];
b[i] += b[i-];
}
a[n+] = ;
for(int i = ;i <= n+;i++)
{
while(!s.empty() && a[s.top()] > a[i])
{
int temp = s.top();
s.pop();
long long len = s.empty()?b[i-]:b[i-]-b[s.top()+];
ans = max(ans,len*a[temp]);
}
s.push(i);
}
cout << ans << endl;
}
return ;
}
1101.模拟进制加法。
#include<bits/stdc++.h>
using namespace std; int k,n; int main()
{
ios::sync_with_stdio();
while(cin >> k >> n)
{
long long ans = ,temp = ;
while(n)
{
ans += temp*(n%);
n /= ;
temp *= k;
}
cout << ans << endl;
}
return ;
}
1102.dp[i][j]表示i个钩子，j个灯笼的方案数，先求一个钩子的情况，再O(n^3)求多个钩子的情况。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; int n,m;
long long dp[][]; int main()
{
ios::sync_with_stdio();
for(int i = ;i <= ;i++) dp[i][] = ;
for(int i = ;i <= ;i++)
{
for(int j = ;j < i;j++) dp[][i] = (dp[][i]+dp[][j]*dp[][i-j-])%MOD;
}
for(int i = ;i <= ;i++)
{
for(int j = ;j <= ;j++)
{
for(int k = ;k <= j;k++) dp[i][j] = (dp[i][j]+dp[i-][k]*dp[][j-k])%MOD;
}
}
while(cin >> n >> m) cout << dp[n][m] << endl;
return ;
}
*1103.待补。
1104.dfs两侧点的个数，相乘。
#include<bits/stdc++.h>
using namespace std; int n,m,from[],to[];
vector<int> v[]; int dfs(int s,int t)
{
int ans = ;
for(int i = ;i < v[s].size();i++)
{
if(v[s][i] == t) continue;
ans += dfs(v[s][i],s);
}
return ans;
} int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
cin >> n >> m;
for(int i = ;i <= n;i++) v[i].clear();
for(int i = ;i < n;i++)
{
int a,b;
cin >> a >> b;
from[i] = a;
to[i] = b;
v[a].push_back(b);
v[b].push_back(a);
}
for(int i = ;i <= m;i++)
{
int x;
cin >> x;
int t = dfs(to[x],from[x]);
cout << t*(n-t) << endl;
}
}
return ;
}
1105.约数个数。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
int T;
cin >> T;
while(T--)
{
cin >> n;
int ans = ;
for(int i = ;i <= sqrt(n+0.1);i++)
{
if(n%i == )
{
if(i == n/i) ans++;
else ans += ;
}
}
cout << ans << endl;
}
return ;
}
1106.求n!/(每个数字个数)!，先统计个数，然后对于每种个数分解质因数，得到最后剩余的所有质因数和其个数，再计算答案。
#include<bits/stdc++.h>
#define MAXN 1000000
using namespace std; int n,a[];
int cnt,prime[MAXN+] = {},mi[MAXN+],vis[MAXN+] = {};
long long b[]; void getprime()
{
cnt = ;
for(int i = ;i <= MAXN;i++)
{
if(!vis[i])
{
prime[++cnt] = i;
mi[i] = i;
}
for(int j = ;j <= cnt && (long long)i*prime[j] <= MAXN;j++)
{
vis[prime[j]*i] = ;
mi[prime[j]*i] = prime[j];
if(!(i%prime[j])) break;
}
}
}
int main()
{
ios::sync_with_stdio();
getprime();
while(~scanf("%d",&n))
{
memset(a,,sizeof(a));
memset(b,,sizeof(b));
for(int i = ;i <= n;i++)
{
int x;
scanf("%d",&x);
a[x]++;
}
for(int i = ;i <= n;i++) b[i]++;
for(int i = ;i <= ;i++)
{
for(int j = ;j <= a[i];j++) b[j]--;
}
for(int i = ;i >= ;i--)
{
if(b[i] == ) continue;
int cnt = b[i],t = i;
b[i] = ;
while(t != )
{
b[mi[t]] += cnt;
t /= mi[t];
}
}
long long now = ;
int ok = ;
for(int i = ;i <= ;i++)
{
while(b[i]--)
{
if(5e18/i > now && now*i <= 1e18) now *= i;
else ok = ;
}
}
if(ok) cout << now << endl;
else cout << "Look, shability!" << endl;
}
return ;
}
1107.按x排序好后，按y求最长上升子序列。
#include<bits/stdc++.h>
using namespace std; struct point
{
int x,y;
}a[];
int b[],n; bool cmp(point X,point Y)
{
if(X.x == Y.x) return X.y > Y.y;
return X.x < Y.x;
} int LIS()
{
int len = ,j;
b[] = a[].y;
for(int i = ;i <= n;i++)
{
if(a[i].y > b[len]) j = ++len;
else j = lower_bound(b+,b++len,a[i].y)-b;
b[j] = a[i].y;
}
return len;
}
int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
for(int i = ;i <= n;i++) cin >> a[i].x;
for(int i = ;i <= n;i++) cin >> a[i].y;
sort(a+,a++n,cmp);
cout << LIS() << endl;
}
return ;
}
1108.打表，前缀和。
#include<bits/stdc++.h>
using namespace std; int x,y,a[] = {}; int main()
{
a[] = a[] = ;
for(int i = ;i <= ;i++)
{
if(i%) a[i] = a[i-]+;
else a[i] = a[i/]+;
}
for(int i = ;i <= ;i++) a[i] += a[i-];
while(cin >> x >> y) cout << a[y]-a[x-] << endl;
return ;
}
1109.dp[i] = dp[i-1]+dp[i-2]+dp[i-3]。
#include<bits/stdc++.h>
using namespace std; int n,a[]; int main()
{
ios::sync_with_stdio();
a[] = ;
a[] = ;
a[] = ;
for(int i = ;i <= ;i++) a[i] = (a[i-]+a[i-]+a[i-])%;
while(cin >> n) cout << a[n] << endl;
return ;
}
1111.f(n)=f(n-1)+f(n-2)+n-2推出f(n)=3*f(n-1)-2*f(n-2)-f(n-3)+f(n-4)，矩阵快速幂。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; int n,anss[] = {,,,,,};
struct matrix
{
long long m[][];
}; matrix ans = { ,,,,
,,,,
,,,,
,,,,};
matrix base = { ,,,,
MOD-,,,,
MOD-,,,,
,,,,}; matrix mul(matrix a, matrix b)
{
matrix tmp;
for(int i = ; i < ;i++)
{
for(int j = ; j < ;j++)
{
tmp.m[i][j] = ;
for(int k = ; k < ;k++) tmp.m[i][j] = (tmp.m[i][j]+a.m[i][k]*b.m[k][j])%MOD;
}
}
return tmp;
} long long fast_mod(int n)
{
matrix x = ans,y = base;
while(n)
{
if(n & )
{
x = mul(x,y);
}
y = mul(y,y);
n >>= ;
}
return (anss[]*x.m[][]+anss[]*x.m[][]+anss[]*x.m[][]+anss[]*x.m[][])%MOD;
} int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
if(n <= ) cout << anss[n] << endl;
else cout << fast_mod(n-) << endl;
}
}
1112.dp[i][j]表示i个人j派的方案数。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; long long dp[][] = {},ans[] = {};
int n; int main()
{
ios::sync_with_stdio();
dp[][] = ;
for(int i = ;i <= ;i++)
{
for(int j = ;j <= i;j++) dp[i][j] = (dp[i-][j-]+dp[i-][j]*j)%MOD;
}
for(int i = ;i <= ;i++)
{
for(int j = ;j <= i;j++) ans[i] = (ans[i]+dp[i][j])%MOD;
}
while(cin >> n) cout << ans[n] << endl;
return ;
}
1113.dp[i][j]表示前i位起余数为j的个数。
#include<bits/stdc++.h>
using namespace std; string s;
int k;
long long dp[][]; int main()
{
ios::sync_with_stdio();
while(cin >> s >> k)
{
memset(dp,,sizeof(dp));
dp[][(s[]-'')%k]++;
for(int i = ;i < s.length();i++)
{
for(int j = ;j < k;j++) dp[i][(j*+s[i]-'')%k] += dp[i-][j];
dp[i][(s[i]-'')%k]++;
}
long long ans = ;
for(int i = ;i < s.length();i++) ans += dp[i][];
cout << ans << endl;
}
return ;
}
1114.先离线，按所有点的和k1,k2的大小排序，再树状数组维护一下。
#include<bits/stdc++.h>
using namespace std; struct arr
{
int o,x,pos,l,r;
}a[];
int n,m,tree[] = {},ans1[],ans2[]; bool cmp(arr X,arr Y)
{
if(X.x == Y.x) return X.o < Y.o;
return X.x < Y.x;
} inline int lowbit(int x)
{
return x & (-x);
} void update(int pos,int x)
{
while(pos <= n)
{
tree[pos] += x;
pos += lowbit(pos);
}
} int getsum(int pos)
{
int sum = ;
while(pos > )
{
sum += tree[pos];
pos -= lowbit(pos);
}
return sum;
} int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(tree,,sizeof(tree));
for(int i = ;i <= n;i++)
{
scanf("%d",&a[i].x);
a[i].pos = i;
a[i].o = ;
}
int cnt = n;
for(int i = ;i <= m;i++)
{
scanf("%d%d%d%d",&a[cnt+].l,&a[cnt+].r,&a[cnt+].x,&a[cnt+].x);
a[cnt+].o = ;
a[cnt+].o = ;
a[cnt+].pos = a[cnt+].pos = i;
a[cnt+].l = a[cnt+].l;
a[cnt+].r = a[cnt+].r;
cnt += ;
}
sort(a+,a++cnt,cmp);
for(int i = ;i <= cnt;i++)
{
if(a[i].o == ) update(a[i].pos,);
else if(a[i].o == ) ans1[a[i].pos] = getsum(a[i].r)-getsum(a[i].l-);
else ans2[a[i].pos] = getsum(a[i].r)-getsum(a[i].l-);
}
for(int i = ;i <= m;i++) printf("%d\n",ans2[i]-ans1[i]);
}
return ;
}
1115.x^n+y^n = (x^(n-1)+y^(n-1))*(x+y)-(x^(n-2)+y^(n-2))*x*y。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; int a,b,n;
long long ans[]; int main()
{
ios::sync_with_stdio();
ans[] = ;
while(cin >> a >> b >> n)
{
ans[] = a;
for(int i = ;i <= n;i++) ans[i] = ((ans[i-]*a-ans[i-]*b%MOD)+MOD)%MOD;
cout << ans[n] << endl;
}
return ;
}
1116.求n!/(每一个重要城市子树种的节点数)。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; int n,m,ok[],a[];
vector<int> v[]; long long qpower(long long a,long long b,long long c)
{
long long ans = ;
a %= c;
while(b)
{
if(b%) ans = ans*a%c;
a = a*a%c;
b /= ;
}
return ans;
} void dfs(int now,int pre)
{
a[now] = ;
for(int i = ;i < v[now].size();i++)
{
int t = v[now][i];
if(t == pre) continue;
dfs(t,now);
a[now] += a[t];
}
} int main()
{
ios::sync_with_stdio();
while(cin >> n >> m)
{
memset(ok,,sizeof(ok));
for(int i = ;i <= n;i++) v[i].clear();
for(int i = ;i < n;i++)
{
int x,y;
cin >> x >> y;
v[x].push_back(y);
v[y].push_back(x);
}
dfs(,-);
long long ans = ;
for(int i = ;i <= n;i++) ans = (ans*i)%MOD;
for(int i = ;i <= m;i++)
{
int x;
cin >> x;
if(ok[x]) continue;
ans = (ans*qpower(a[x],MOD-,MOD))%MOD;
ok[x] = ;
}
cout << ans << endl;
}
return ;
}
1117.状压dp。
#include<bits/stdc++.h>
#define LL long long
#define MOD 1000000007
using namespace std; int n,m;
LL dp[][<<];
vector<int> v[]; LL dfs(int left,int sta)
{
if(dp[left][sta] != -) return dp[left][sta];
if(left == ) return !sta;
dp[left][sta] = ;
for(int i = ;i < v[left].size();i++)
{
dp[left][sta] = (dp[left][sta]+dfs(left-,sta^v[left][i]))%MOD;
}
return dp[left][sta];
} int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(dp,-,sizeof(dp));
for(int i = ;i <= m;i++) v[i].clear();
for(int i = ;i <= m;i++)
{
int t,endd = <<n;
scanf("%d",&t);
for(int j = ;j < endd;j++)
{
int sum = ;
for(int k = ;k < n;k++)
{
if(j&(<<k)) sum++;
}
if(sum == t) v[i].push_back(j);
}
}
int a = ;
for(int i = ;i < n;i++)
{
int t;
scanf("%d",&t);
a |= (<<i)*t;
}
printf("%lld\n",dfs(m,a));
}
return ;
}
1118.暴力搜索边的组成顺序，每次判断只要延长边，判断是否正三角形。
#include<bits/stdc++.h>
#define PI (atan(1)*4)
using namespace std; int a[],b[],vis[] = {};
double ans; void dfs(int cnt)
{
if(cnt == )
{
if(b[]+b[]+b[] == b[]+b[]+b[] && b[]+b[]+b[] == b[]+b[]+b[])
{
ans = sqrt()/*((b[]+b[]+b[])*(b[]+b[]+b[])-b[]*b[]-b[]*b[]-b[]*b[]);
}
}
for(int i = ;i < ;i++)
{
if(vis[i]) continue;
b[cnt] = a[i];
vis[i] = ;
dfs(cnt+);
vis[i] = ;
}
}
int main()
{
ios::sync_with_stdio(false);
while(cin >> a[] >> a[] >> a[] >> a[] >> a[] >> a[])
{
ans = ;
dfs();
if(ans > 1e-) cout << fixed << setprecision() << ans << endl;
else cout << << endl;
}
return ;
}
1119.递增枚举最小公倍数。
#include<bits/stdc++.h>
using namespace std; int a[],n,k; bool ok(int x)
{
int cnt = ;
for(int i = ;i <= n;i++)
{
if(x%a[i] == ) cnt++;
}
if(cnt >= k) return ;
return ;
} int main()
{
ios::sync_with_stdio();
while(cin >> n >> k)
{
for(int i = ;i <= n;i++) cin >> a[i];
int ans = ;
for(;ans <=;ans++)
{
if(ok(ans)) break;
}
if(ans <= ) cout << ans << endl;
else cout << "Orz" << endl;
}
return ;
}
1120.贪心，时间后往前，大的先考虑，可以用优先队列优化一下。
#include<bits/stdc++.h>
using namespace std; struct task
{
int t,v;
}a[];
int n,vis[] = {}; bool cmp(task x,task y)
{
return x.v > y.v;
}
int main()
{
ios::sync_with_stdio();
cin >> n;
int t = ,sum = ,ans = ;
for(int i = ;i <= n;i++)
{
cin >> a[i].t;
t = max(a[i].t,t);
}
for(int i = ;i <= n;i++)
{
cin >> a[i].v;
sum += a[i].v;
}
sort(a+,a++n,cmp);
for(int i = t;i >= ;i--)
{
for(int j = ;j <= n;j++)
{
if(!vis[j] && i <= a[j].t)
{
ans += a[j].v;
vis[j] = ;
break;
}
}
}
cout << sum-ans << endl;
return ;
}
1121.60个刚好超1e18，60之前直接递归算，60之后的用60的来算。
#include<bits/stdc++.h>
using namespace std; long long a[],n,k; int f(long long x,long long y)
{
if(x == ) return ;
if(y <= a[x-]) return f(x-,y)^;
if(y == a[x-]+) return ;
return f(x-,a[x]-y+);
} int main()
{
ios::sync_with_stdio();
a[] = ;
for(int i = ;i <= ;i++) a[i] = a[i-]*+;
while(cin >> n >> k)
{
int ans;
if(n <= ) ans = f(n,k);
else
{
ans = f(,k);
if(n&) ans ^= ;
}
if(ans) cout << "srO" << endl;
else cout << "Orz\n" << endl;
}
return ;
}
1123.二分查找。
#include<bits/stdc++.h>
using namespace std; struct orz
{
int id,x;
friend bool operator<(struct orz X,struct orz Y)
{
if(X.x == Y.x) return X.id < Y.id;
return X.x < Y.x;
}
}a[],t;
int n,m; int main()
{
ios::sync_with_stdio();
t.id = ;
while(cin >> n >> m)
{
for(int i = ;i <= n;i++)
{
cin >> a[i].x;
a[i].id = i;
}
sort(a+,a++n);
int flag = ;
while(m--)
{
cin >> t.x;
int ans = lower_bound(a+,a++n,t)-a;
if(flag)
{
if(ans > n) cout << -;
else cout << a[ans].id;
flag = ;
}
else
{
if(ans > n) cout << " -1";
else cout << " " << a[ans].id;
}
}
cout << endl;
}
return ;
}
1124.找第一个可以替代的数，用后面最大且最后的数替代。
#include<bits/stdc++.h>
using namespace std; char a[]; int main()
{
while(cin >> a)
{
int len = strlen(a),first = ,last = ,now = ;
while(now < len)
{
first = last = now;
char maxx = a[now];
for(int i = len-;i > now;i--)
{
if(a[i] > maxx)
{
maxx = a[i];
last = i;
}
}
if(first != last) break;
now++;
}
swap(a[first],a[last]);
cout << a << endl;
}
return ;
}
1125.分成三个三角形，由面积判断。
#include<bits/stdc++.h>
using namespace std; double f(double x1,double y1,double x2,double y2,double x3,double y3)
{
double a,b,c,p;
a = sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
b = sqrt((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3));
c = sqrt((x3-x2)*(x3-x2)+(y3-y2)*(y3-y2));
p = (a+b+c)/;
return sqrt(p*(p-a)*(p-b)*(p-c));
} int main()
{
double x1,x2,x3,y1,y2,y3,x,y;
while(cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3 >> x >> y)
{
if(abs(f(x1,y1,x2,y2,x3,y3)-f(x,y,x2,y2,x3,y3)-f(x1,y1,x,y,x3,y3)-f(x1,y1,x2,y2,x,y)) < 1e-) cout << "Orz" << endl;
else cout << "stO" << endl;
}
return ;
}
*1128.待补。
1129.记录交换后的映射。
#include<bits/stdc++.h>
using namespace std; int a[][],x[],y[],n,m,k; int main()
{
while(~scanf("%d%d%d",&n,&m,&k))
{
for(int i = ;i <= n;i++) x[i] = i;
for(int i = ;i <= m;i++) y[i] = i;
for(int i = ;i <= n;i++)
{
for(int j = ;j <= m;j++) scanf("%d",&a[i][j]);
}
int o,xx,yy;
while(k--)
{
scanf("%d%d%d",&o,&xx,&yy);
if(o == ) swap(x[xx],x[yy]);
else swap(y[xx],y[yy]);
}
for(int i = ;i <= n;i++)
{
printf("%d",a[x[i]][y[]]);
for(int j = ;j <= m;j++) printf(" %d",a[x[i]][y[j]]);
printf("\n");
}
}
return ;
}
1130.C(n,0)+C(n,1)+...+C(n,n) =2^n，0*C(n,0)+1*C(n,1)+2*C(n,2)+...n*C(n,n) = n*2^(n-1)。
#include<bits/stdc++.h>
#define LL long long
#define MOD 542
using namespace std; int p,d,n; LL qmod(LL a, LL b, LL c)
{
LL ans = ;
a = a%c;
while(b)
{
if(b&) ans = ans*a%c;
b >>= ;
a = a*a%c;
}
return ans;
} int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&p,&d,&n);
LL ans = qmod(,n,MOD)*p%MOD;
ans = (qmod(,n-,MOD)*d*n+ans)%MOD;
printf("%lld\n",ans);
}
return ;
}
1131.dp[i][j]表示i,j两个下面为首的最序列的个数，从后往前更新，用map记录是否出现过某个值。
#include<bits/stdc++.h>
using namespace std; int n,dp[][];
long long a[]; int main()
{
//freopen("1.txt","r",stdin);
while(~scanf("%d",&n))
{
memset(dp,,sizeof(dp));
for(int i = ;i <= n;i++) scanf("%lld",&a[i]);
map<long long,int> mp;
long long ans1 = 2e18,ans2 = 2e18,maxx = ;
for(int j = n;j >= ;j--)
{
for(int i = j-;i >= ;i--)
{
if(mp.count(a[i]+a[j])) dp[i][j] = max(dp[i][j],dp[j][mp[a[i]+a[j]]]+);
dp[i][j] = max(dp[i][j],);
if(dp[i][j] > maxx || dp[i][j] == maxx && (a[i] < ans1 || a[i] == ans1 && a[j] < ans2))
{
maxx = dp[i][j];
ans1 = a[i];
ans2 = a[j];
}
}
mp[a[j]] = j;
}
printf("%d\n",maxx);
int now1 = mp[ans1],now2 = mp[ans2];
printf("%lld %lld",ans1,ans2);
for(int i = ;i <= maxx;i++)
{
long long t = a[now1]+a[now2];
now1 = now2;
while(a[now2] != t) now2++;
printf(" %lld",a[now2]);
}
printf("\n");
}
return ;
}
*1132.待补。
1138.a+b。
#include<bits/stdc++.h>
using namespace std; char s;
int a,b,c,d; int main()
{
ios::sync_with_stdio();
while(cin >> a >> s >> b >> s >> c >> s >> d >> s) cout << a+c << "+" << b+d << "i" << endl;
return ;
}
1139.打表二分，注意前面几个特例。
#include<bits/stdc++.h>
using namespace std; long long n,f[],sum[]; int main()
{
ios::sync_with_stdio();
f[] = ;
f[] = ;
sum[] = ;
for(int i = ;i <= ;i++)
{
f[i] = f[i-]+f[i-];
sum[i] = sum[i-]+f[i];
}
while(cin >> n)
{
if(n == || n == || n == ) cout << "1 1" << endl;
else
{
int x = lower_bound(sum,sum+,n)-sum;
long long y = n-sum[x-];
if(y > f[x-]) cout << x << " " << y << endl;
else cout << x- << " " << f[x-] << endl;
}
}
return ;
}
1140.kmp，c中的strstr比kmp快10来倍。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
while(~scanf("%d",&n))
{
getchar();
int s = ;
char a[],b[] = " ";
while(n--)
{
gets(a);
strcat(b,a);
char *p = b,*o;
while(o = strstr(p,"wanshen"))
{
p = o+;
s++;
}
strcpy(b,p);
}
printf("%d\n",s);
}
return ;
}
1141.取余后模拟。
#include<bits/stdc++.h>
using namespace std; long long x,y,a; int main()
{
while(cin >> x >> y >> a)
{
a %= x+y;
if(a == )
{
cout << "light" << endl;
continue;
}
while(a)
{
a -= x;
if(a <= )
{
cout << "wanshen" << endl;
break;
}
a -= y;
if(a <= )
{
cout << "light" << endl;
break;
}
}
}
return ;
}
1142.记录字符串中出现的字符，然后删除。
#include<bits/stdc++.h>
using namespace std; char a[],b[];
int s[]; int main()
{
while(cin >> a >> b)
{
memset(s,,sizeof(s));
int flag=,lena = strlen(a),lenb = strlen(b);
for(int i = ;i < lenb;i++) s[b[i]] = ;
for(int i = ;i < lena;i++)
{
if(!s[a[i]])
{
putchar(a[i]);
flag = ;
}
}
if(!flag) cout << "EMPTY";
cout << endl;
}
return ;
}
1143.暴力dfs，注意过程中还原。
#include<bits/stdc++.h>
using namespace std; int a[],vis[],num[],ans; bool ok(int x,int z)
{
if(x == ) return ;
if(x > && __gcd(z,num[x-]) != ) return ;
if(x% != && __gcd(z,num[x-]) != ) return ;
return ;
} void dfs(int x)
{
if(x == )
{
ans++;
return;
}
for(int i = ;i <= ;i++)
{
if(vis[i]) continue;
if(ok(x+,a[i]))
{
vis[i] = ;
num[x+] = a[i];
dfs(x+);
vis[i] = ;
}
}
}
int main()
{
while(cin >> a[])
{
for(int i = ;i <= ;i++) cin >> a[i];
ans = ;
dfs();
cout << ans << endl;
}
return ;
}
1144.合并小的，不一定每一次都是合并k个，第一次先合并不足k的次数，后面都合并k个。
#include<bits/stdc++.h>
using namespace std; priority_queue< long long,vector<long long>,greater<long long> > q;
int n,k; int main()
{
ios::sync_with_stdio();
while(cin >> n >> k)
{
while(!q.empty()) q.pop();
long long x,ans = ;
for(int i = ;i <= n;i++)
{
cin >> x;
q.push(x);
}
while((n-)%(k-))
{
q.push();
n++;
}
while(q.size() != )
{
long long t = ,sum = ;
while(t++ < k)
{
sum += q.top();
q.pop();
}
ans += sum;
q.push(sum);
}
cout << ans << endl;
}
return ;
}
1145.矩阵快速幂。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; long long l,r,anss[] = {,,,,};
struct matrix
{
long long m[][];
};
matrix ans = { ,,,,
,,,,
,,,,
,,,,};
matrix base = { ,,,,
,,,,
,,,,
,,,,}; matrix mul(matrix a, matrix b)
{
matrix tmp;
for(int i = ; i < ;i++)
{
for(int j = ; j < ;j++)
{
tmp.m[i][j] = ;
for(int k = ; k < ;k++) tmp.m[i][j] = (tmp.m[i][j]+a.m[i][k]*b.m[k][j])%MOD;
}
}
return tmp;
} long long fast_mod(long long n)
{
matrix x = ans,y = base;
while(n)
{
if(n & )
{
x = mul(x,y);
}
y = mul(y,y);
n >>= ;
}
return (x.m[][]+x.m[][]+x.m[][]+*x.m[][])%MOD;
} int main()
{
ios::sync_with_stdio();
while(cin >> l >> r)
{
long long x,y;
if(l == ) x = ;
else x = l < ?anss[l-]:fast_mod(l-);
y = r < ?anss[r]:fast_mod(r-);
cout << (y-x+MOD)%MOD << endl;
}
return ;
}
1146.01背包的变形，达成每种V所需的最小体力。
#include<bits/stdc++.h>
using namespace std; int dp[],v[],w[],n,W; int main()
{
ios::sync_with_stdio();
while(cin >> n >> W)
{
memset(dp, 0x3f, sizeof(dp));
dp[] = ;
int sum = ;
for(int i = ;i <= n;i++)
{
cin >> w[i] >> v[i];
sum += v[i];
}
for(int i = ;i <= n;i++)
{
for(int j = sum;j >= v[i];j--) dp[j] = min(dp[j-v[i]]+w[i],dp[j]);
}
for(int i = sum;i >= ;i--)
{
if(dp[i] <= W)
{
cout << i << endl;
break;
}
}
}
return ;
}
1147.先判断和在不在最大和最小的区间内。若在，从后往前贪心。
#include<bits/stdc++.h>
using namespace std; int n,l[],r[],ans[],s; int main()
{
while(~scanf("%d%d",&n,&s))
{
int sum1 = ,sum2 = ;
for(int i = ;i <= n;i++)
{
scanf("%d%d",&l[i],&r[i]);
sum1 += l[i];
sum2 += r[i];
}
if(s < sum1 || s > sum2)
{
printf("Xue Beng\n");
continue;
}
int x = s-sum1;
for(int i = n;i >= ;i--)
{
ans[i] = x >= r[i]-l[i]?r[i]:l[i]+x;
x -= ans[i]-l[i];
}
printf("%d",ans[]);
for(int i = ;i <= n;i++) printf(" %d",ans[i]);
printf("\n");
}
return ;
}
1148.b ≥ phi[c]时，a^b%c = a^(b%phi[c]+phi[c])%c，否则直接算。
#include<bits/stdc++.h>
#define LL long long
using namespace std; LL a,b,c,y; LL gcd(LL a,LL b)
{
return b?gcd(b,a%b):a;
} LL PowerMod(LL a, LL b, LL c)
{
LL ans = ;
a = a % c;
while(b)
{
if(b&) ans = (ans*a)%c;
b = b>>;
a = (a*a)%c;
}
return ans;
} LL euler(LL n)
{
LL res = n,a = n;
for(LL i = ;i*i <= a;i++)
{
if(a%i == )
{
res = res/i*(i-);
while(a%i == ) a /= i;
}
}
if(a > ) res = res/a*(a-);
return res;
} int main()
{
while(~scanf("%lld%lld%lld%lld",&a,&b,&c,&y))
{
LL t = euler(y);
if(log(b) < /c)
{
printf("%lld\n",PowerMod(a,PowerMod(b,c,1e9+),y));
}
else printf("%lld\n",PowerMod(a,PowerMod(b,c,t)+t,y));
}
return ;
}
1149.打表预处理，容斥。
#include<bits/stdc++.h>
#define MOD 1000000007
#define LL long long
using namespace std; LL fac[],inv[],n,m,k; LL PowerMod(LL a, LL b, LL c)
{
LL ans = ;
a = a % c;
while(b)
{
if(b&) ans = (ans*a)%c;
b = b>>;
a = (a*a)%c;
}
return ans;
} inline LL c(LL n,LL m)
{
if(m > n) return ;
return fac[n]*inv[m]%MOD*inv[n-m]%MOD;
} int main()
{
ios::sync_with_stdio();
fac[] = ;
for(int i = ;i <= ;i++) fac[i] = fac[i-]*i%MOD;
inv[] = PowerMod(fac[],MOD-,MOD);
for(int i = ;i >= ;i--) inv[i] = inv[i+]*(i+)%MOD;
while(cin >> n >> m >> k)
{
LL ans = c(n+m-,m),x = -;
for(int i = ;i <= n;i++,x = -x)
{
LL t = m-i*(k+);
if(t < ) break;
ans = (ans+x*(c(n,i)*c(n+t-,t)%MOD)+MOD)%MOD;
}
cout << ans << endl;
}
return ;
}
1150.减法。
#include<bits/stdc++.h>
using namespace std; int a; int main()
{
while(cin >> a) cout << a- << endl;
return ;
}
1151.分情况讨论。
#include<bits/stdc++.h>
using namespace std; long long a,b,c; int main()
{
while(cin >> a >> b >> c)
{
if(a==&&b==&&c==)
{
cout << "inf" << endl;
continue;
}
if(a==&&b!=)
{
cout << << endl;
continue;
}
if(a==&&b==&&c!=)
{
cout << << endl;
continue;
}
long long x=b*b-*a*c;
if(x==) cout << << endl;
else if(x>) cout << << endl;
else cout << << endl;
}
return ;
}
1152.map记录。
#include<bits/stdc++.h>
using namespace std; struct st
{
string name,course;
};
map<int,st> mp;
map<int,st>::iterator it; int main()
{
int n,m;
while(cin >> n >> m)
{
mp.clear();
string a,c,d;
int b;
cin >> a >> c;
for(int i = ;i <= n;i++)
{
cin >> a >> b;
mp[b].name = a;
mp[b].course = "NULL";
}
cin >> a >> c >> d;
for(int i = ;i <= m;i++)
{
cin >> a >> b >> c;
mp[b].course = c;
}
cout << "Name StuNum CourseName" << endl;
for(it = mp.begin();it != mp.end();it++) cout << it->second.name << ' ' << it->first << ' ' << it->second.course << endl;
}
return ;
}
1153.比较斜率排序。
#include<bits/stdc++.h>
#define LL long long
using namespace std; struct line
{
LL x,y;
}a[];
int n; bool cmp(line X,line Y)
{
return X.y*Y.x < X.x*Y.y;
} int main()
{
while(~scanf("%d",&n))
{
int cnt = ;
LL x,y,xx,yy,temp = ;
for(int i = ;i <= n;i++)
{
scanf("%lld%lld%lld%lld",&x,&y,&xx,&yy);
if(x == xx) temp++;
else if(x < xx)
{
a[cnt].x = xx-x;
a[cnt].y = yy-y;
cnt++;
}
else
{
a[cnt].x = x-xx;
a[cnt].y = y-yy;
cnt++;
}
}
sort(a,a+cnt,cmp);
LL ans = temp*(temp-)/;
LL num = ;
for(int i = ;i < cnt;i++)
{
if(a[i].x*a[i-].y == a[i].y*a[i-].x) num++;
else
{
ans += num*(num-)/;
num = ;
}
}
ans += num*(num-)/;
printf("%lld\n",ans);
}
return ;
}
1154.统计连续0串的个数。
#include<bits/stdc++.h>
using namespace std; string s; int main()
{
ios::sync_with_stdio();
while(cin >> s)
{
int len = s.length();
int ans = ,flag = ;
for(int now = ;now < len;now++)
{
if(flag)
{
while(now < len)
{
if(s[now] == '')
{
ans++;
flag = ;
break;
}
else now++;
}
}
else if(s[now] == '') flag = ;
}
if(flag) ans++;
cout << ans << endl;
}
return ;
}
1155.先确定在往里数的第几个圈上，然后根据位置确定数字。
#include<bits/stdc++.h>
using namespace std; long long k,x,y,ans; int main()
{
ios::sync_with_stdio();
while(cin >> k >> x >> y)
{
long long p = min(x,y);
p = min(p,k-x+);
p = min(p,k-y+);
long long dis = x-p+y-p;
if(x <= y) ans = *(p-)*(k-p+)++dis;
else ans = *p*(k-p)+-dis;
cout << ans << endl;
}
return ;
}
1156.递减的单调队列，保存人的编号和入队耐心-入队时间。
#include<bits/stdc++.h>
using namespace std; struct st
{
int num,t;
}s;
deque<st> q;
int n,x; int main()
{
while(~scanf("%d",&n))
{
while(!q.empty()) q.pop_back();
int out = ,cnt = ;
for(int i = ;i <= n;i++)
{
scanf("%d",&x);
switch(x)
{
case :
int tt;
scanf("%d",&tt);
s.t = tt-i;
s.num = cnt++;
while(!q.empty() && q.back().t <= s.t) q.pop_back();
q.push_back(s);
break;
case :
if(q.front().num == out++) q.pop_front();
break;
case :
printf("%d\n",q.front().t+i);
}
}
}
return ;
}
1157.一条一条加边，求max和从小边往大边加，求min和从大边往小边加，用并查集计点数。
#include<bits/stdc++.h>
using namespace std; int n,pre[],point[];
struct line
{
int v,u,w;
friend bool operator<(line x,line y)
{
return x.w < y.w;
}
}l[]; int findd(int x)
{
return x == pre[x]?x:pre[x] = findd(pre[x]);
} int main()
{
while(~scanf("%d",&n))
{
for(int i = ;i < n;i++) scanf("%d%d%d",&l[i].u,&l[i].v,&l[i].w);
sort(l+,l+n);
long long ans = ;
for(int i = ;i <= n;i++)
{
pre[i] = i;
point[i] = ;
}
for(int i = ;i < n;i++)
{
int x = findd(l[i].u),y = findd(l[i].v);
ans += (long long)point[x]*point[y]*l[i].w;
pre[x] = y;
point[y] += point[x];
}
for(int i = ;i <= n;i++)
{
pre[i] = i;
point[i] = ;
}
for(int i = n-;i >= ;i--)
{
int x = findd(l[i].u),y = findd(l[i].v);
ans -= (long long)point[x]*point[y]*l[i].w;
pre[x] = y;
point[y] += point[x];
}
printf("%lld\n",ans);
}
return ;
}
1158.规律C(x+k,x+y)，打表预处理。
#include<bits/stdc++.h>
#define LL long long
#define MOD (long long)(1e9+7)
using namespace std; int x,y,k;
LL fac[]; LL PowerMod(LL a, LL b, LL c)
{
LL ans = ;
a = a % c;
while(b)
{
if(b&) ans = (ans*a)%c;
b = b>>;
a = (a*a)%c;
}
return ans;
} LL inv(LL x)
{
return PowerMod(x,MOD-,MOD);
} int main()
{
fac[] = ;
for(int i = ;i <= ;i++) fac[i] = fac[i-]*i%MOD;
while(~scanf("%d%d%d",&x,&y,&k))
{
if(y == k) printf("1\n");
else if(y < k) printf("0\n");
else printf("%lld\n",fac[x+y]*inv(fac[x+k])%MOD*inv(fac[y-k])%MOD);
}
return ;
}
1159.floyd动态更新，首要要考虑重边和环，然后开放一个点就要更新相关最短路，而且更新的顺序有要求。
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std; int n,m,t,a[][],ok[]; int main()
{
while(~scanf("%d%d%d",&n,&m,&t))
{
memset(a,0x3f,sizeof(a));
memset(ok,,sizeof(ok));
for(int i = ;i < n;i++) a[i][i] = ;
while(m--)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
if(u == v) continue;
a[u][v] = min(a[u][v],w);
}
while(t--)
{
int o;
scanf("%d",&o);
if(o == )
{
int x;
scanf("%d",&x);
if(ok[x]) printf("lab %d has been repaired!\n",x);
else
{
ok[x] = ;
for(int i = ;i < n;i++)
{
for(int j = ;j < n;j++)
{
if(!ok[i] || !ok[j]) continue;
a[j][x] = min(a[j][x],a[j][i]+a[i][x]);
}
}
for(int i = ;i < n;i++)
{
for(int j = ;j < n;j++)
{
if(!ok[i] || !ok[j]) continue;
a[x][j] = min(a[x][j],a[x][i]+a[i][j]);
}
}
for(int i = ;i < n;i++)
{
for(int j = ;j < n;j++)
{
if(!ok[i] || !ok[j]) continue;
a[i][j] = min(a[i][j],a[i][x]+a[x][j]);
}
}
}
}
else
{
int u,v;
scanf("%d%d",&u,&v);
if(!ok[u] || !ok[v]) printf("help %d %d\n",u,v);
else if(a[u][v] == INF) printf("no path\n");
else printf("%d\n",a[u][v]);
}
}
}
return ;
}
1160.set模拟，貌似代码有点问题，直接过了，好像是询问都是按时间顺序来的。
#include<bits/stdc++.h>
using namespace std; int n,m;
struct ss
{
int num,t,o;
}a[];
set<int> s; bool cmp(ss x,ss y)
{
return x.t < y.t;
} int main()
{
scanf("%d%d",&n,&m);
for(int i = ;i <= n;i++) scanf("%d%d%d",&a[i].o,&a[i].t,&a[i].num);
sort(a+,a++n,cmp);
int cnt = ;
for(int i = ;i <= m;i++)
{
int p,q;
scanf("%d%d",&p,&q);
while(cnt <= n&& a[cnt].t < p)
{
if(a[cnt].o == ) s.insert(a[cnt].num);
else s.erase(a[cnt].num);
cnt++;
}
if(s.count(q)) printf("YES\n");
else printf("NO\n");
}
return ;
}
1161.规律，列方程推一下，其中某一个村庄假设不动，动其他n-1个，最后转化成求绝对值最小问题。
#include<bits/stdc++.h>
#define LL long long
using namespace std; LL a[] = {};
int n; int main()
{
while(~scanf("%d",&n) && n > )
{
LL sum1 = ,sum2 = ;
for(int i = ;i <= n;i++)
{
scanf("%lld",&a[i]);
sum1 += a[i];
}
for(int i = ;i <= n;i++)
{
LL t;
scanf("%lld",&t);
sum2 += t;
a[i] = a[i-]+t-a[i];
}
if(sum1 != sum2)
{
printf("No way\n");
continue;
}
sort(a,a+n);
LL t = a[n/],ans = ;
for(int i = ;i < n;i++) ans += abs(t-a[i]);
printf("%lld\n",ans);
}
return ;
}
1162.二分补课次数，树形dp判断。
#include<bits/stdc++.h>
using namespace std; int n,m,k,ok,a[],b[],minn[],cnt[];
bool vis[];
vector<int> v[]; void dfs(int now)
{
minn[now] = a[now];
for(int i = ;i < v[now].size();i++)
{
int t = v[now][i];
if(!vis[t])
{
vis[t] = ;
dfs(t);
}
minn[now] = min(minn[now],minn[t]);
}
minn[now] += k*cnt[now];
} bool pan(int x)
{
memset(cnt,,sizeof(cnt));
memset(vis,,sizeof(vis));
memset(minn,,sizeof(minn));
for(int i = ;i <= x;i++) cnt[b[i]]++;
ok = ;
for(int i = ;i <= n;i++)
{
if(vis[i]) continue;
dfs(i);
if(minn[i] < ) return ;
}
ok = ;
return ;
} int main()
{
ios::sync_with_stdio();
scanf("%d%d%d",&n,&m,&k);
for(int i = ;i < n;i++)
{
int x,y;
scanf("%d%d",&x,&y);
v[x].push_back(y);
}
for(int i = ;i <= n;i++) scanf("%d",&a[i]);
for(int i = ;i <= m;i++) scanf("%d",&b[i]);
int l = ,r = m;
while(l < r)
{
int mid = (l+r)/;
if(pan(mid)) r = mid;
else l = mid+;
}
pan(l);
if(ok) printf("%d\n",l);
else printf("mdzz\n");
return ;
}
1163.扫一边，动态维护每个质因数对应的最大最小值，树状数组位数累和。
#include<bits/stdc++.h>
using namespace std; int n,a[],c[],minn[],maxx[] = {};
long long tree[] = {},ans[]; inline int lowbit(int x)
{
return x&-x;
} void add(int pos,int x)
{
while(pos <= )
{
tree[pos] += x;
pos += lowbit(pos);
}
} long long getsum(int pos)
{
long long sum = ;
while(pos > )
{
sum += tree[pos];
pos -= lowbit(pos);
}
return sum;
} int main()
{
while(~scanf("%d",&n) && n)
{
for(int i = ;i <= n;i++) scanf("%d",&a[i]);
for(int i = ;i <= n;i++) scanf("%d",&c[i]);
memset(maxx,,sizeof(maxx));
memset(minn,0x3f,sizeof(minn));
memset(tree,,sizeof(tree));
for(int i = ;i <= n;i++)
{
int x = a[i];
int mi = 0x3f3f3f3f,ma = ;
for(int j = ;j*j <= x;j++)
{
if(x%j == )
{
while(x%j == ) x /= j;
mi = min(mi,minn[j]);
ma = max(ma,maxx[j]);
minn[j] = min(minn[j],a[i]);
maxx[j] = max(maxx[j],a[i]);
}
}
if(x > )
{
mi = min(mi,minn[x]);
ma = max(ma,maxx[x]);
minn[x] = min(minn[x],a[i]);
maxx[x] = max(maxx[x],a[i]);
}
if(mi == 0x3f3f3f3f) mi = ;
ans[i] = getsum(ma)-getsum(mi-);
add(a[i],c[i]);
}
printf("%lld",ans[]);
for(int i = ;i <= n;i++) printf(" %lld",ans[i]);
printf("\n");
}
return ;
}
1164.因为存孩子会爆，只能用孩子存父节点，然后逆向的dfs，期间需要一个数组保存累加值。
#include<bits/stdc++.h>
using namespace std; int a[],pre[] = {},sum[] = {},n,u,v;
bool vis[] = {}; long long dfs(int pos)
{
if(vis[pos]) return ;
vis[pos] = ;
if(!pre[pos])
{
sum[pos] -= a[pos];
return abs(sum[pos]);
}
long long t = dfs(pre[pos]);
a[pos] += sum[pre[pos]];
t += abs(a[pos]);
sum[pos] += sum[pre[pos]]-a[pos];
return t;
} int main()
{
scanf("%d",&n);
for(int i = ;i < n;i++)
{
scanf("%d%d",&u,&v);
pre[v] = u;
}
for(int i = ;i <= n;i++) scanf("%d",&a[i]);
long long ans = ;
for(int i = ;i <= n;i++) ans += dfs(i);
printf("%lld\n",ans);
return ;
}
1165.题目https://pan.baidu.com/s/1Ro3vR-L_7mnDfbIKjRj2sg
旋转坐标，求二维前缀和。
#include<bits/stdc++.h>
using namespace std; int n,m,k,a[][],b[][],mp[][],sum[][],cnt[][]; int main()
{
while(~scanf("%d%d%d",&m,&n,&k))
{
memset(a,,sizeof(a));
memset(b,,sizeof(b));
int maxx = max(n,m),nn = *maxx;
for(int i = ;i <= n;i++)
{
for(int j = ;j <= m;j++)
{
int ii = i-j+maxx,jj = i+j;
scanf("%d",&a[ii][jj]);
b[ii][jj] = ;
mp[i][j] = a[ii][jj];
}
}
for(int i = ;i <= nn;i++)
{
for(int j = ;j <= nn;j++)
{
a[i][j] += a[i][j-]+a[i-][j]-a[i-][j-];
b[i][j] += b[i][j-]+b[i-][j]-b[i-][j-];
}
}
for(int i = ;i <= n;i++)
{
for(int j = ;j <= m;j++)
{
int ii = i-j+maxx,jj = i+j;
int x1 = max(,ii-k),y1 = max(,jj-k);
int x2 = min(nn,ii+k),y2 = min(nn,jj+k);
sum[i][j] = a[x2][y2]-a[x1-][y2]-a[x2][y1-]+a[x1-][y1-];
cnt[i][j] = b[x2][y2]-b[x1-][y2]-b[x2][y1-]+b[x1-][y1-];
}
}
for(int i = ;i <= n;i++)
{
printf("%.0f",1.0*(sum[i][]-mp[i][])/(cnt[i][]-)+1e-);
for(int j = ;j <= m;j++) printf(" %.0f",1.0*(sum[i][j]-mp[i][j])/(cnt[i][j]-)+1e-);
printf("\n");
}
}
return ;
}
1166.题目https://pan.baidu.com/s/1Ro3vR-L_7mnDfbIKjRj2sg
将一个串扩充至两倍，另一个串变化几种形态在里面找是否存在相同字串，strstr方便又快速。
#include<bits/stdc++.h>
using namespace std; char a[],b[];
int n; void change()
{
for(int i = ;i < n;i++)
{
if(b[i] == '') b[i] = '';
else b[i] += ;
}
} int main()
{
while(~scanf("%d",&n))
{
getchar();
gets(a);
gets(b);
for(int i = ;i < n;i++) a[i+n] = a[i];
a[*n] = ;
char *p = NULL;
for(int i = ;i < ;i++)
{
p = strstr(a,b);
if(p) break;
if(i != ) change();
}
if(p) printf("yes\n");
else printf("no\n");
}
return ;
}
*1167.待补。
1168.都取出现次数最少的那个字母就可以了。
#include<bits/stdc++.h>
using namespace std; char a[];
int x[],y[]; int main()
{
int T;
scanf("%d",&T);
for(int c = ;c <= T;c++)
{
memset(x,,sizeof(x));
memset(y,,sizeof(y));
scanf("%s",a);
int len = strlen(a);
for(int i = ;i < len;i++)
{
if('a' <= a[i] && a[i] <= 'z') x[a[i]-'a']++;
else y[a[i]-'A']++;
}
sort(x,x+);
sort(y,y+);
printf("Case %d: %d\n",c,min(x[],y[]));
}
return ;
}
1169.不断除以2，直到奇数，操作次数的奇偶性。
#include<bits/stdc++.h>
using namespace std; int f(long long x)
{
if(x&) return ;
else return f(x/)^;
} int main()
{
int T;
scanf("%d",&T);
while(T--)
{
long long x;
scanf("%d",&x);
if(f(x) == ) printf("First Box\n");
else printf("Second Box\n");
}
}
1170.优化一下暴力的过程，将数量一样的放一起算。
#include<bits/stdc++.h>
#define LL long long
using namespace std; LL n,m; LL gcd(LL x,LL y)
{
return y?gcd(y,x%y):x;
} LL c(LL x,LL y)
{
return (y*(y+)*(*y+))/-((x-)*x*(*x-))/;
} int main()
{
while(~scanf("%lld%lld",&n,&m) && n && m)
{
LL ans = ;
int now = ;
while(now <= n)
{
int t = n/now,endd = min(m,n/t);
ans += t*c(now,endd);
if(endd == m) break;
now = endd +;
}
LL x = n*m;
LL g = gcd(ans,x);
printf("%lld/%lld\n",ans/g,x/g);
}
}
1171.dp[i][j][k]表示前i行里有j列1个，k列两个的方案数，自底向上dp可以省略一维空间，记忆化应该快一点。
#include<bits/stdc++.h>
#define MOD 14020130063
using namespace std; int n,m;
long long dp[][][]; long long dfs(int x,int y,int z)
{
if(x < || y < || z < || y+z > m) return ;
if(dp[x][y][z] != -) return dp[x][y][z];
dp[x][y][z] = (dfs(x-,y,z)+(m-(y-)-z)*dfs(x-,y-,z)+(y+)*dfs(x-,y+,z-)+(m-(y-)-z)*(m-(y-)-z-)/*dfs(x-,y-,z)+(y+)*(y+-)/*dfs(x-,y+,z-)+y*(m-y-(z-))*dfs(x-,y,z-))%MOD;
return dp[x][y][z];
} int main()
{
ios::sync_with_stdio();
while(cin >> n >> m)
{
if(m > n) swap(n,m);
memset(dp,-,sizeof(dp));
dp[][][] = ;
dp[][][] = m;
dp[][][] = m*(m-)/;
long long ans = ;
for(int i = ;i <= m;i++)
{
for(int j = ;i+j <= m;j++) ans = (ans+dfs(n,i,j))%MOD;
}
cout << ans << endl;
}
return ;
}
1172.容量为sum/2的01背包。
#include<bits/stdc++.h>
using namespace std; int n,w[],dp[]; int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
memset(dp,,sizeof(dp));
int sum = ;
for(int i = ;i <= n;i++)
{
cin >> w[i];
sum += w[i];
}
int endd = sum/;
for(int i = ;i <= n;i++)
{
for(int j = endd;j >= w[i];j--)
{
dp[j] = max(dp[j],dp[j-w[i]]+w[i]);
}
}
cout << sum-dp[endd]* << endl;
}
return ;
}
1173.是否为3的倍数。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
if(n%) cout << "Yes" << endl;
else cout << "No" << endl;
}
return ;
}
*1174.待补。
1175.拿k长度的线段扫一遍，map维护数量。
#include<bits/stdc++.h>
using namespace std; int n,k,a[] = {}; int main()
{
while(~scanf("%d%d",&n,&k))
{
for(int i = ;i <= n;i++) scanf("%d",&a[i]);
map<int,int> mp;
for(int i = ;i < k;i++) mp[a[i]]++;
int ans = ;
for(int i = k;i <= n;i++)
{
mp[a[i]]++;
if(mp.size() == k) ans++;
if(--mp[a[i-k+]] == ) mp.erase(a[i-k+]);
}
printf("%d\n",ans);
}
return ;
}
1176.渡河dp，每次分两批送走最大的两个人。
#include<bits/stdc++.h>
using namespace std; int a[],n; int main()
{
while(cin >> n)
{
for(int i = ;i <= n;i++) cin >> a[i];
sort(a+,a++n);
long long ans = ;
while(n)
{
if(n == )
{
ans += a[];
break;
}
if(n == )
{
ans += a[];
break;
}
if(n == )
{
ans += a[]+a[]+a[];
break;
}
else
{
ans += min(*a[]+a[n]+a[n-],a[]+*a[]+a[n]);
n -= ;
}
}
cout << ans << endl;
}
return ;
}
1177.排序后树状数组维护。
#include<bits/stdc++.h>
using namespace std; int tree[],n,m,cnt = ,ans[]; struct star
{
int num,x,y,o;
}a[]; bool cmp(star X,star Y)
{
if(X.x == Y.x) return X.y < Y.y;
return X.x < Y.x;
} inline int lowbit(int x)
{
return x & (-x);
} void update(int pos,int x)
{
while(pos <= )
{
tree[pos] += x;
pos += lowbit(pos);
}
} int getsum(int pos)
{
int sum = ;
while(pos > )
{
sum += tree[pos];
pos -= lowbit(pos);
}
return sum;
} int main()
{
while(~scanf("%d%d",&n,&m))
{
printf("Case #%d:\n",cnt++);
memset(tree,,sizeof(tree));
for(int i = ;i <= n;i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
a[i].o = ;
}
for(int i = n+;i <= n+m;i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
a[i].num = i-n;
a[i].o = ;
}
sort(a+,a++n+m,cmp);
for(int i = ;i <= n+m;i++)
{
if(a[i].o) update(a[i].y+,);
else ans[a[i].num] = getsum(a[i].y+);
}
for(int i = ;i <= m;i++) printf("%d\n",ans[i]);
}
return ;
}
1179.先打表单个硬币能凑成的数的最小硬币个数，枚举每种硬币及个数，更新答案。
#include<bits/stdc++.h>
using namespace std; int a[],ans[] = {},n,s,q;
int main()
{
memset(ans,0x3f,sizeof(ans));
cin >> n >> s;
for(int i = ;i <= n;i++)
{
cin >> a[i];
for(int j = ;j <= s;j++) ans[a[i]*j] = min(ans[a[i]*j],j);
} cin >> q;
while(q--)
{
int t,x = 2e9;
cin >> t;
for(int i = ;i <= n;i++)
{
int endd = min(s,t/a[i]);
for(int j = ;j <= endd;j++)
{
int left = t-j*a[i];
if(left == ) x = min(x,j);
else if(ans[left] && ans[left]+j <= s) x = min(x,ans[left]+j);
}
}
if(x == 2e9) cout << - << endl;
else cout << x << endl;
}
return ;
}
1180.dp[i][j][k]表示取第i个珠子颜色为j且末尾有k个连续的珠子的总数。
k = 1时，dp[i][j][k] = sumx≠j且1≤y≤k(dp[i-1][x][y])
k ≠ 1时，dp[i][j][k] = dp[i-1][j][k-1]
由于对称性，j不影响dp[i][j][k]结果，则忽略第二维，dp[i][1] = sum1≤y≤k(dp[i-1][y])*(m-1)
利用矩阵快速幂，计算sum1≤x≤k-1(dp[n][x])，再乘以m种颜色即为结果。
#include<bits/stdc++.h>
#define MOD 23333
using namespace std; long long n,m,k;
struct matrix
{
long long m[][];
}; matrix one = {},base = {}; matrix mul(matrix a, matrix b)
{
matrix tmp;
for(int i = ; i < k-;i++)
{
for(int j = ; j < k-;j++)
{
tmp.m[i][j] = ;
for(int K = ;K < k-;K++) tmp.m[i][j] = (tmp.m[i][j]+a.m[i][K]*b.m[K][j])%MOD;
}
}
return tmp;
} long long fast_mod(long long n)
{
matrix ans = one,y = base;
while(n)
{
if(n&) ans = mul(ans,y);
y = mul(y,y);
n /= ;
}
long long res = ;
for(int i = ;i < k-;i++) res = (res+ans.m[i][])%MOD;
return res*m%MOD;
} int main()
{
ios::sync_with_stdio(false);
for(int i = ;i < ;i++) one.m[i][i] = ;
for(int i = ;i < ;i++) base.m[i][i-] = ;
while(cin >> n >> m >> k)
{
for(int i = ;i < k-;i++) base.m[][i] = m-;
cout << fast_mod(n-) << endl;
}
return ;
}
1181.因为只有一组数据，直接开vector存每一次的状态，把树状数组改一下就可以了。
#include<bits/stdc++.h>
using namespace std; int a[] = {},n,q,o,t = ,c = ;
vector< pair<int,int> > v[];
vector< pair<int,int> >::iterator it; inline int lowbit(int x)
{
return x&(-x);
} void update(int pos,int x)
{
while(pos <= n)
{
int temp = ;
if(!v[pos].empty()) temp = v[pos].back().second;
v[pos].push_back(make_pair(t,temp+x));
pos += lowbit(pos);
}
} int getsum(int pos,int tt)
{
int sum = ;
while(pos > )
{
it = upper_bound(v[pos].begin(),v[pos].end(),make_pair(tt,));
if(it != v[pos].end() && it->first == tt) sum += it->second;
else if(it != v[pos].begin())
{
it--;
sum += it->second;
}
pos -= lowbit(pos);
}
return sum;
} int main()
{
ios::sync_with_stdio(false);
cin >> n >> q;
while(q--)
{
cin >> o;
if(o == )
{
t++;
int i,x;
cin >> i >> x;
c = i^c;
update(c,x);
a[c] += x;
c = a[c];
}
else
{
int l,r,tt;
cin >> l >> r >> tt;
c = getsum(r^c,tt)-getsum((l^c)-,tt);
cout << c << endl;
}
}
return ;
}
1182.快速乘。
#include<bits/stdc++.h>
#define LL long long
#define MOD 23333333333
using namespace std; LL n,m; LL qmul(LL a, LL b, LL c)
{
LL ans = ;
a = a%c;
while(b)
{
if(b&) ans = (ans+a)%c;
b >>= ;
a = (a+a)%c;
}
return ans;
} int main()
{
while(cin >> n >> m)
{
LL x = m/,y = m%,ans = ;
if(x%) ans = qmul(x*,(*n+x+)/,MOD);
else ans = qmul(x*/,*n+x+,MOD);
for(int i = ;i < y;i++) ans = (ans+n+x+i)%MOD;
cout << ans << endl;
}
return ;
}
1183.划分问题，dp[i][j] = dp[i-1][j-1]+dp[i-j][j]。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; long long dp[][] = {};
int n,m; int main()
{
for(int i = ;i <= ;i++) dp[i][] = ;
for(int i = ;i <= ;i++)
{
for(int j = ;j <= i;j++) dp[i][j] = (dp[i-][j-]+dp[i-j][j])%MOD;
}
for(int i = ;i <= ;i++)
{
for(int j = ;j <= ;j++) dp[i][j] = (dp[i][j-]+dp[i][j])%MOD;
}
while(scanf("%d%d",&n,&m) && n && m) printf("%lld\n",dp[n][m]);
return ;
}
1184.Sprague-Grundy定理。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
while(~scanf("%d",&n))
{
int ans = ,t;
for(int i = ;i <= n;i++)
{
scanf("%d",&t);
ans ^= t%;
}
if(ans) printf("Yes\n");
else printf("No\n");
}
return ;
}
1185.二分时间。
#include<bits/stdc++.h>
using namespace std; int m,k,t[]; bool ok(int x)
{
int cnt = ,sum = ;
for(int i = ;i <= m;i++)
{
if(sum+t[i] < x) sum += t[i];
else if(sum+t[i] == x)
{
cnt++;
sum = ;
}
else
{
cnt++;
sum = t[i];
}
}
if(sum) cnt++;
if(cnt > k) return ;
return ;
} int main()
{
ios::sync_with_stdio(false);
while(cin >> m >> k)
{
for(int i = ;i <= m;i++) cin >> t[i];
int l = ,r = ;
while(l < r)
{
int mid = (l+r)/;
if(ok(mid)) l = mid+;
else r = mid;
}
cout << l << endl;
}
return ;
}
1186.dfs。
#include<bits/stdc++.h>
using namespace std; int n = ,m,len[],vis[][] = {},sum,ok;
int dx[] = {-,,,};
int dy[] = {,-,,};
string a[]; void dfs(int x,int y)
{
vis[x][y] = ;
sum++;
for(int i = ;i < ;i++)
{
int xx = x+dx[i],yy = y+dy[i];
if(xx < || xx > n || yy < || yy > len[x])
{
ok = ;
continue;
}
if(vis[xx][yy]) continue;
if(a[xx][yy] == '#') continue;
dfs(xx,yy);
}
} int main()
{
ios::sync_with_stdio(false);
while(cin >> a[++n])
{
len[n] = a[n].length();
a[n] = ' '+a[n];
}
int ans = ;
for(int i = ;i <= n;i++)
{
for(int j = ;j <= len[i];j++)
{
if(vis[i][j] || a[i][j] == '#') continue;
sum = ;
ok = ;
dfs(i,j);
if(!ok) ans += sum;
}
}
cout << ans << endl;
return ;
}
1187.每个点按度排序，从度小的点开始，把每个点邻接表中在该点前面的点都删掉。然后对于每一条边，统计公共点的个数和。
#include<bits/stdc++.h>
using namespace std; int n,m,vis[],x[],y[];
vector<int> v[];
struct xx
{
int cnt,id;
friend bool operator<(xx a,xx b)
{
return a.cnt < b.cnt;
}
}a[]; int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
memset(vis,,sizeof(vis));
for(int i = ;i <= n;i++) v[i].clear();
for(int i = ;i <= n;i++) a[i].id = i,a[i].cnt = ;
for(int i = ;i <= m;i++)
{
scanf("%d%d",&x[i],&y[i]);
v[x[i]].push_back(y[i]);
v[y[i]].push_back(x[i]);
a[x[i]].cnt++;
a[y[i]].cnt++;
}
sort(a+,a++n);
for(int i = ;i <= n;i++)
{
int tt = a[i].id;
for(int j = ;j < v[tt].size();j++)
{
int t = v[tt][j];
if(vis[t]) v[tt].erase(v[tt].begin()+j),j--;
}
sort(v[tt].begin(),v[tt].end());
vis[tt] = ;
}
long long ans = ;
for(int i = ;i <= m;i++)
{
int t = x[i],tt = y[i];
int now1 = ,now2 = ;
while(now1 < v[t].size() && now2 < v[tt].size())
{
if(v[t][now1] < v[tt][now2]) now1++;
else if(v[t][now1] > v[tt][now2]) now2++;
else
{
ans++;
now1++;
now2++;
}
}
}
printf("%lld\n",ans);
}
return ;
}
1188.递推，找公式。
#include<bits/stdc++.h>
using namespace std; long long n; int main()
{
ios::sync_with_stdio(false);
while(cin >> n) cout << +n*(n+)/ << endl;
return ;
}
1189.输出注意空格。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
ios::sync_with_stdio(false);
while(cin >> n)
{
for(int i = ;i <= n;i++)
{
int t = -i;
while(t--) cout << " ";
for(int j = ;j < i;j++) cout << (char)('A'+j);
for(int j = i-;j >= ;j--) cout << (char)('A'+j);
cout << endl;
}
}
return ;
}
1190.连续平方数求和，公式，逆元。
#include<bits/stdc++.h>
#define LL long long
using namespace std; long long n,p; LL qmod(LL a, LL b, LL c)
{
LL ans = ;
a = a%c;
while(b)
{
if(b&) ans = ans*a%c;
b >>= ;
a = a*a%c;
}
return ans;
} LL inv(LL a,LL b)
{
return qmod(a,b-,b);
} int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> p)
{
cout << n%p*(n+)%p*(*n+)%p*inv(,p)%p << endl;
} return ;
}
1191.给每一类记录一个标准。
#include<bits/stdc++.h>
#define LL long long
using namespace std; long long n;
vector<string> v[]; int main()
{
ios::sync_with_stdio(false);
while(cin >> n)
{
int cnt= ;
for(int i = ;i <= n;i++) v[i].clear();
for(int i = ;i <= n;i++)
{
string s;
cin >> s;
string ss = s;
sort(ss.begin(),ss.end());
int p = ,flag = ;
for(;p <= cnt;p++)
{ if(v[p][] == ss)
{
v[p].push_back(s);
flag = ;
break;
}
}
if(!flag)
{
v[++cnt].push_back(ss);
v[cnt].push_back(s);
}
}
for(int i = ;i <= cnt;i++)
{
cout << v[i][];
for(int j = ;j <v[i].size();j++) cout << " " << v[i][j];
cout << endl;
}
} return ;
}
1192.线段扫一遍比较每一段。
#include<bits/stdc++.h>
#define LL long long
using namespace std; int n,k,a[]; int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> k)
{
long long ans = ,now = ;
for(int i = ;i <= n;i++) cin >> a[i];
for(int i = ;i <= k;i++) now += a[i];
ans = now;
for(int i = k+;i <= n;i++)
{
now = now+a[i]-a[i-k];
ans = min(now,ans);
}
cout << ans << endl;
}
return ;
}
1193.Sprague-Grundy定理。
#include<bits/stdc++.h>
#define LL long long
using namespace std; int a,b,c,d; int main()
{
ios::sync_with_stdio(false);
while(cin >> a >> b >> c >> d)
{
int t = a%(c+),tt = b%(d+);
if(t^tt) cout << "NUO!" << endl;
else cout << "NO!" << endl;
}
return ;
}
1194.求解模方程a^x=b(mod n)，n为素数。大步小步模版。
#include<bits/stdc++.h>
#define LL long long
using namespace std; LL pow_mod(LL a,LL b,LL n)
{
LL s=;
while(b)
{
if(b&)
s=(s*a)%n;
a=(a*a)%n;
b=b>>;
}
return s;
} LL log_mod (LL a,LL b,LL n)
{
LL m,v,e=,i;
m=ceil(sqrt(n+0.5)); //x=i*m+j
//v=inv(pow_mod(a,m,n),n); //a^m*v=1(mod n)
v=pow_mod(a,n-m-,n);
map<LL,LL>x;
x[]=m;
for(i=;i<m;i++)
{
e=(e*a)%n;
if(!x[e])x[e]=i;
}
for(i=;i<m;i++)
{
if(x[b])
{
LL num=x[b];
x.clear();
return i*m+(num==m?:num);
}
b=(b*v)%n; //b=b/(a^m)
} return -;
} int main()
{
LL a,b;
while(scanf("%lld%lld",&a,&b)!=EOF)
{
LL ans1=log_mod(,a,),ans2 = log_mod(,b,); if(ans1==- || ans2 == -) printf("No Solution\n");
else printf("%lld\n",pow_mod(,ans1*ans2,));
}
return ;
}
1195.最短路模版。
#include<bits/stdc++.h>
using namespace std; int n,m,a[][],vis[],dis[]; int main()
{
ios::sync_with_stdio(false);
int T;
cin >> T;
while(T--)
{
cin >> m;
n = ;
memset(a,0x3f,sizeof(a));
while(m--)
{
int x,y,z;
cin >> x >> y >> z;
a[x][y] = a[y][x] = min(a[x][y],z);
n = max(n,x);
n = max(n,y);
}
int s,d;
cin >> s >> d;
memset(vis,,sizeof(vis));
memset(dis,0x3f,sizeof(dis));
dis[s] = ;
for(int i = ;i <= n;i++)
{
int k = -,minn = 0x3f3f3f3f;
for(int j = ;j <= n;j++)
{
if(vis[j]) continue;
if(minn > dis[j])
{
minn = dis[j];
k = j;
}
}
if(k == -) break;
vis[k] = ;
for(int j = ;j <= n;j++)
{
if(vis[j]) continue;
if(dis[j] > dis[k]+a[j][k]) dis[j] = dis[k]+a[j][k];
}
}
cout << *dis[d] << endl;
}
return ;
}
1196.直接判断。
#include <bits/stdc++.h>
using namespace std; int n; int main()
{
ios::sync_with_stdio(false);
while(cin >> n)
{
if(n <= ) cout << "XiaoZhen" << endl;
else if(n < ) cout << "Zh0ngshen" << endl;
else cout << "DaNuo" << endl;
}
return ;
}
1197.直接判断。
#include <bits/stdc++.h>
using namespace std; string s; int main()
{
ios::sync_with_stdio(false);
while(cin >> s)
{
if(s == "v8") cout << "SingleDog&YangRouHuoGuo" << endl;
else if(s == "qsqx") cout << "Couple&Program" << endl;
else cout << "SingleDog&GoodGoodStud" << endl;
}
return ;
}
1198.递推，找公式。
#include <bits/stdc++.h>
using namespace std; long long a[];
int n; int main()
{
ios::sync_with_stdio(false);
a[] = ;
a[] = ;
for(int i = ;i <= ;i++) a[i] = a[i-]+*(i-);
while(cin >> n) cout << a[n] << endl;
return ;
}
1199.圆排列，除数取模不好处理，只要用在乘的时候抵消就可以了。
#include <bits/stdc++.h>
using namespace std; long long n,m,p; int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> m >> p)
{
long long ans = ;
int flag = ;
for(int i = n-m+;i <= n;i++)
{
if(!flag && i%m == )
{
ans = ans*i/m%p;
flag = ;
}
else ans = ans*i%p;
}
cout << ans << endl;
}
return ;
}
1200.递推，找公式。
#include <bits/stdc++.h>
using namespace std; long long n; long long qpower(long long a,long long b,long long c)
{
a %= c;
long long ans = ;
while(b)
{
if(b%) ans = ans*a%c;
a = a*a%c;
b /= ;
}
return ans;
} int main()
{
ios::sync_with_stdio(false);
while(cin >> n) cout << ((qpower(,n,)-)*+)% << endl;
return ;
}
1201.素数筛。
#include <bits/stdc++.h>
using namespace std; int vis[] = {},prime[],num[] = {};
long long n; int main()
{
int cnt = ;
for(int i = ;i < ;i++)
{
if(!vis[i]) prime[++cnt] = i;
for(int j = ;i*prime[j] < && j <= cnt;j++)
{
vis[prime[j]*i] = ;
if(!(i%prime[j])) break;
}
}
for(int i = ;i <= ;i++)
{
for(int j = ;j <= i;j++) num[i] += (j+)*(i-j+);
}
while(~scanf("%lld",&n))
{
long long ans = ;
for(int i = ;(long long)prime[i]*prime[i] <= n;i++)
{
if(n%prime[i] == )
{
int t = ;
while(n%prime[i] == )
{
t++;
n /= prime[i];
}
ans *= num[t];
}
}
if(n != ) ans *= ;
printf("%lld\n",ans);
}
return ;
}
1202.最短路构造虚点。
#include<bits/stdc++.h>
using namespace std; struct xx
{
int to;
long long w;
xx(int a,long long b):to(a),w(b){};
friend bool operator <(xx X,xx Y)
{
return X.w > Y.w;
}
}; vector<xx> v[];
int n,m,r,h[][];
long long dis[];
bool vis[],has[]; void add(int a,int b,int x)
{
v[a].push_back(xx(b,x));
v[b].push_back(xx(a,x));
} void dij(int beg)
{
priority_queue<xx> q;
memset(dis,0x3f,sizeof(dis));
memset(vis,,sizeof(vis));
dis[beg] = ;
q.push(xx(beg,));
while(!q.empty())
{
int now = q.top().to;
q.pop();
if(vis[now]) continue;
vis[now] = ;
for(int i = ;i < v[now].size();i++)
{
int tt = v[now][i].to,ww = v[now][i].w;
if(!vis[tt] && dis[now]+ww < dis[tt])
{
dis[tt] = dis[now]+ww;
q.push(xx(tt,dis[tt]));
}
}
}
} int main()
{
ios::sync_with_stdio(false);
int T;
cin >> T;
while(T--)
{
for(int i = ;i <= ;i++) v[i].clear();
cin >> n >> m >> r;
for(int i = ;i <= n;i++)
{
for(int j = ;j <= m;j++) cin >> h[i][j];
}
for(int i = ;i <= n;i++)
{
for(int j = ;j <= m;j++)
{
if(i+ <= n) add(*i+j,*i+j+,*(h[i][j]+h[i+][j]));
if(j+ <= m) add(*i+j,*i+j+,*(h[i][j]+h[i][j+]));
}
}
for(int k = ;k <= r;k++)
{
memset(has,,sizeof(has));
int ax,ay,bx,by,t,kk;
cin >> ax >> ay >> bx >> by >> t >> kk;
for(int i = ax;i <= bx;i++)
{
for(int j = ay;j <= by;j++)
{
add(+*k+h[i][j],*i+j,t);
v[+*k+h[i][j]].push_back(xx(*i+j,));
v[*i+j].push_back(xx(+*k+h[i][j],));
has[h[i][j]] = ;
}
}
for(int i = ;i <= ;i++)
{
if(!has[i]) continue;
for(int j = i+;j <= min(i+kk,);j++)
{
if(!has[j]) continue;
add(+*k+i,+*k+j,*t);
add(+*k+i,+*k+j,*t);
}
}
}
int sx,sy,ox,oy;
cin >> sx >> sy >> ox >> oy;
dij(*sx+sy);
cout << dis[*ox+oy]/ << endl;
}
}
1203.状压dp。
#include<bits/stdc++.h>
using namespace std;
#define MOD 1000000007 int n,m,k,cnt,sta[<<],a[][<<]; void dfs(int l,int now,int pre1,int pre2)
{
if(l == m)
{
sta[now] = ;
return;
}
for(int i = ;i < k;i++)
{
if(i == pre1 && i == pre2) continue;
dfs(l+,(now<<)|i,pre2,i);
dfs(l+,(now<<)|(i+),pre2,i);
}
} bool ok(int x)
{
int t = x%;
x >>= ;
int tt = x%;
if(t != tt) return ;
x >>= ;
int ttt = x%;
if(t != ttt) return ;
return ;
} int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> m >> k)
{
if(n < m) swap(n,m);
memset(sta,,sizeof(sta));
memset(a,,sizeof(a));
cnt = ;
dfs(,,-,-);
int endd = <<(*m);
for(int i = ;i < endd;i++)
{
if(!sta[i]) continue;
int flag = ;
for(int j = ;j < m;j++)
{
if(i&(<<(j*))) flag = ;
}
if(!flag) a[m][i] = ;
}
for(int i = ;i < n;i++)
{
for(int j = ;j < m;j++)
{
int now = i*m+j+;
for(int l = ;l < endd;l++)
{
if(a[now-][l] == ) continue;
if((l&(<<(j*))) == && (j < || ok(l>>(*j-)))) a[now][l|(<<(j*))] = (a[now][l|(<<(j*))]+a[now-][l])%MOD;
for(int t = ;t < k;t++)
{
if(t == (l>>(j*))%) continue;
int tt = (l-(l&(<<(j*))))|(t<<(j*));
if(j > && !ok(tt>>(*j-))) continue;
a[now][tt] = (a[now][tt]+a[now-][l])%MOD;
}
}
}
}
int t = n*m;
int ans = ;
for(int i = ;i < endd;i++) ans = (ans+a[t][i])%MOD;
cout << ans << endl;
}
return ;
}
1204.大模拟的恐惧。
#include<bits/stdc++.h>
using namespace std; int n,x1,x2,a[][] = {},ok[],vis[]; double calcF()
{
double kin = ,kout = ;
for(int i = ;i <= n;i++)
{
for(int j = i+;j <= n;j++)
{
if(ok[i] && ok[j]) kin += *a[i][j];
else if(ok[i] || ok[j]) kout += a[i][j];
}
}
return kin/(kin+kout);
} double calcf(int x)
{
int t = ok[x];
ok[x] = ;
double t1 = calcF();
ok[x] = ;
double t2 = calcF();
ok[x] = t;
return t1-t2;
} void op1()
{
double maxx = -1e18;
int pos = -;
for(int i = ;i <= n;i++)
{
if(ok[i]) continue;
int flag = ;
for(int j = ;j <= n;j++)
{
if(ok[j] && a[i][j]) flag = ;
}
if(!flag) continue;
double t = calcf(i);
if(t-maxx > 1e- || abs(t-maxx) <= 1e-)
{
maxx = t;
pos = max(pos,i);
}
}
ok[pos] = ;
} void op2()
{
double minn = 1e18;
int pos = -;
for(int i = ;i <= n;i++)
{
if(!ok[i]) continue;
double t = calcf(i);
if(minn-t > 1e- || abs(minn-t) <= 1e-)
{
minn = t;
pos = max(pos,i);
}
}
if(minn < ) ok[pos] = ;
} bool isok()
{
for(int i = ;i <= n;i++)
{
if(ok[i] && calcf(i) <= ) return ;
if(ok[i]) continue;
int flag = ;
for(int j = ;j <= n;j++)
{
if(ok[j] && a[i][j]) flag = ;
}
if(flag && calcf(i) >= ) return ;
}
return ;
} void op(int x)
{
ok[x] = ;
while()
{
op1();
op2();
if(isok()) return;
}
} int main()
{
//freopen("1.txt","r",stdin);
ios::sync_with_stdio();
cin >> x1 >> x2;
int x,y,z;
while(cin >> x >> y >> z)
{
n = max(n,x);
n = max(n,y);
a[x][y] += z;
a[y][x] += z;
}
memset(ok,,sizeof(ok));
op(x1);
if(ok[x1] && ok[x2])
{
cout << "YES" << endl;
return ;
}
memset(ok,,sizeof(ok));
op(x2);
if(ok[x1] && ok[x2])
{
cout << "YES" << endl;
return ;
}
cout << "NO" << endl;
return ;
}
1205.构造x为10。
#include<bits/stdc++.h>
using namespace std; string s;
stack<char> ss; int main()
{
ios::sync_with_stdio(false);
while(cin >> s)
{
if(s.length() == )
{
cout << "1 10" << endl;
cout << << " " << s[] << endl;
continue;
}
for(int i = ;i < s.length();i++) ss.push(s[i]);
cout << s.length()- << " " << << endl;
cout << ss.top();
ss.pop();
while(!ss.empty())
{
cout << " " << ss.top();;
ss.pop();
}
cout << endl;
}
return ;
}
1206.等比求和推。
#include<bits/stdc++.h>
#define LL long long
using namespace std; string s; int main()
{
ios::sync_with_stdio(false);
while(cin >> s)
{
LL t = ,x = ,y = ;
int i = ,len = s.length();
int a = ,b = ;
for(;i < len && s[i] != '.';i++) t = t*+s[i]-'';
i++;
for(;i < len && s[i] != '_';i++)
{
x = x*+s[i]-'';
a++;
}
i++;
y = x;
for(;i < len && s[i] != '.';i++)
{
y = y*+s[i]-'';
b++;
}
LL sub,mom,z = ;
if(a == && b != )
{
while(b--) z = z*+;
LL g = __gcd(y,z);
sub = y/g,mom = z/g;
}
else if(a == && b == )
{
sub = ;
mom = ;
}
else if(b == )
{
z = ;
while(a--) z *= ;
LL g = __gcd(x,z);
sub = x/g,mom = z/g;
}
else
{
y -= x;
while(b--) z = z*+;
while(a--) z *= ;
LL g = __gcd(y,z);
sub = y/g,mom = z/g;
}
sub += t*mom;
cout << sub << "/" << mom << endl;
}
return ;
}
1207.dp，过程中有个累和优化。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; int n,l,r,sum[],t[]; int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> l >> r)
{
memset(sum,,sizeof(sum));
memset(t,,sizeof(t));
t[] = ;
sum[] = ;
while(n--)
{
int c,a;
cin >> c >> a;
for(int i = ;i+c <= r;i++)
{
t[i+c] = (t[i]+t[i+c])%MOD;
int tt = i+(a+)*c;
if(tt <= r) t[tt] = (t[tt]-sum[i]+MOD)%MOD;
}
memcpy(sum,t,(r+)*sizeof(int));
}
for(int i = l+;i <= r;i++) sum[i] = (sum[i-]+sum[i])%MOD;
cout << sum[r] << endl;
}
return ;
}
1208.每次少k-1个瓶盖，推公式。
#include<bits/stdc++.h>
#define MOD 1000000007
#define LL long long
using namespace std; int a,b,k; LL qmod(LL a, LL b, LL c)
{
LL ans = ;
a = a%c;
while(b)
{
if(b&) ans = ans*a%c;
b >>= ;
a = a*a%c;
}
return ans;
} int main()
{
ios::sync_with_stdio(false);
while(cin >> a >> b >> k)
{
LL x = qmod(a,b,k-);
if(x == ) x = k-;
cout << (qmod(a,b,MOD)-x+MOD)%MOD*qmod(k-,MOD-,MOD)%MOD << endl;
}
return ;
}
1209.dp，f[i] = sum[i-1]*2/i+i。
#include<bits/stdc++.h>
using namespace std; int n;
double f[],sum[]; int main()
{
ios::sync_with_stdio();
f[] = ;
sum[] = ;
for(int i = ;i <= ;i++)
{
f[i] = *sum[i-]/i+i;
sum[i] = sum[i-]+f[i];
}
int T;
cin >> T;
while(T--)
{
cin >> n;
cout << f[n] << endl;
}
return ;
}
1210.
#include<bits/stdc++.h>
#define MOD 19260817
using namespace std; int n,q,a[];
long long euler(long long n)
{
long long ans = ;
for(int i = ;(long long)i*i <= n;i++)
{
if(n%i == )
{
n /= i;
ans *= i-;
while(n%i == )
{
n /= i;
ans *= i;
}
}
}
if(n > ) ans *= n-;
return ans;
} int main()
{
while(~scanf("%d",&n))
{
for(int i = ;i <= n;i++)
{
int x;
scanf("%d",&x);
a[i] = euler(x);
}
scanf("%d",&q);
while(q--)
{
long long ans = ;
int x;
scanf("%d",&x);
for(int i = ;i <= n;i++) ans = ans*(__gcd(x-,a[i])+)%MOD;
printf("%lld\n",ans);
}
}
return ;
}
1211.首项系数为1，根肯定为整数，且为an因子，找个大素数取模防止溢出。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; int n,a[]; long long qpower(long long a, long long b, long long c)
{
long long ans = ;
a = a%c;
while(b)
{
if(b%) ans = ans*a%c;
a = a*a%c;
b /= ;
}
return ans;
} bool ok(long long x)
{
long long t = ;
for(int i = ;i <= n;i++) t = (t+a[i]*qpower(x,n-i,MOD)+MOD)%MOD;
return t == ;
} int main()
{
ios::sync_with_stdio();
a[] = ;
while(cin >> n)
{
for(int i = ;i <= n;i++) cin >> a[i];
long long x = abs(a[n]),ans = ;
for(long long i = ;i*i <= x;i++)
{
if(x%i) continue;
if(ok(i)) ans += i;
if(ok(-i)) ans -= i;
if(i*i == x) continue;
if(ok(x/i)) ans += x/i;
if(ok(-x/i)) ans -= x/i;
}
cout << fixed << setprecision() << (double)ans << endl;
}
return ;
}
1213.对m排序，双指针维护m符合的序列，树状数组维护w。
#include<bits/stdc++.h>
using namespace std; int n,X,Y,tree[]; inline int lowbitt(int x)
{
return x&-x;
} void add(int pos,int x)
{
while(pos <= )
{
tree[pos] += x;
pos += lowbitt(pos);
}
} int getsum(int pos)
{
int ans = ;
while(pos > )
{
ans += tree[pos];
pos -= lowbitt(pos);
}
return ans;
} struct xx
{
int x,y;
friend bool operator<(xx a,xx b)
{
return a.x < b.x;
}
}a[]; int main()
{
ios::sync_with_stdio(false);
while(~scanf("%d%d%d",&n,&X,&Y))
{
memset(tree,,sizeof(tree));
for(int i = ;i <= n;i++) scanf("%d",&a[i].x);
for(int i = ;i <= n;i++) scanf("%d",&a[i].y);
sort(a+,a++n);
int now = ,r = ;
long long ans = ;
for(int i = n;i >= ;i--)
{
while(a[r].x <= X-a[i].x) r++;
while(now > i)
{
--now;
add(a[now].y,-);
}
while(now < i && now < r)
{
add(a[now].y,);
now++;
}
ans += getsum(Y-a[i].y);
}
printf("%lld\n",ans);
}
return ;
}
1216.dfs序+划分树，或dfs序+主席树。
#include<bits/stdc++.h>
#define N 100005
using namespace std; int n,m,cnt,l[N],r[N],pos[N];
int a[N];
vector<int> v[N]; void dfs(int now,int pre)
{
l[now] = ++cnt;
pos[cnt] = now;
for(int i = ;i < v[now].size();i++)
{
int t = v[now][i];
if(t == pre) continue;
dfs(t,now);
}
r[now] = cnt;
} int tree[][N],sorted[N],toleft[][N]; void build(int l,int r,int dep)
{
if(l == r)return;
int mid = (l+r)/,same = mid-l+;
for(int i = l;i <= r;i++)
{
if(tree[dep][i] < sorted[mid]) same--;
}
int lpos = l,rpos = mid+;
for(int i = l;i <= r;i++)
{
if(tree[dep][i] < sorted[mid]) tree[dep+][lpos++] = tree[dep][i];
else if(tree[dep][i] == sorted[mid] && same > )
{
tree[dep+][lpos++] = tree[dep][i];
same--;
}
else tree[dep+][rpos++] = tree[dep][i];
toleft[dep][i] = toleft[dep][l-]+lpos-l;
}
build(l,mid,dep+);
build(mid+,r,dep+);
} int query(int l,int r,int ql,int qr,int dep,int k)
{
if(ql == qr) return tree[dep][ql];
int mid = (l+r)/,cnt = toleft[dep][qr]-toleft[dep][ql-];
if(cnt >= k)
{
int ll = l+toleft[dep][ql-]-toleft[dep][l-],rr = ll+cnt-;
return query(l,mid,ll,rr,dep+,k);
}
else
{
int rr = qr+toleft[dep][r]-toleft[dep][qr],ll = rr-(qr-ql-cnt);
return query(mid+,r,ll,rr,dep+,k-cnt);
}
} int main()
{
while(~scanf("%d",&n))
{
for(int i = ;i <= n;i++) scanf("%d",&a[i]);
for(int i = ;i <= n;i++) v[i].clear();
cnt = ;
for(int i = ;i < n;i++)
{
int x,y;
scanf("%d%d",&x,&y);
v[x].push_back(y);
v[y].push_back(x);
}
dfs(,-);
for(int i = ;i <= n;i++) sorted[i] = tree[][i] = a[pos[i]];
sort(sorted+,sorted+n+);
build(,n,);
scanf("%d",&m);
while(m--)
{
int x,y;
scanf("%d%d",&x,&y);
printf("%d\n",query(,n,l[x],r[x],,y));
}
}
return ;
}
1226.待补。
1127.lucas+中国剩余。
#include<bits/stdc++.h>
using namespace std; int n;
long long a[]; long long qmod(long long a,long long b,long long c)
{
long long ans = ;
a = a%c;
while(b)
{
if(b%) ans = (ans*a)%c;
a = (a*a)%c;
b /= ;
}
return ans;
} long long c(long long m,long long n,long long MOD)
{
if(m < n) return ;
if(m == n) return ;
if(n > m-n) n = m-n;
long long mm = ,nn = ;
for(long long i = ;i < n;i++)
{
mm = mm*(m-i)%MOD;
nn = nn*(n-i)%MOD;
}
return mm*qmod(nn,MOD-,MOD)%MOD;
} long long lucas(long long m,long long n,long long MOD)
{
long long ans = ;
while(m && n && ans)
{
ans = ans%MOD*c(m%MOD,n%MOD,MOD)%MOD;
n /= MOD;
m /= MOD;
}
return ans;
} long long e_gcd(long long a,long long b,long long &x,long long &y)
{
if(!b)
{
x = ;
y = ;
return a;
}
long long d = e_gcd(b,a%b,y,x);
y -= a/b*x;
return d;
} long long solve(long long *m,long long *a,long long n)
{
long long M = m[],A = a[],x,y;
for(int i = ;i <= n;i++)
{
long long d = e_gcd(M,m[i],x,y);
if((a[i]-A)%d) return -;
x = (a[i]-A)/d*x%(m[i]/d);
A += x*M;
M = M/d*m[i];
A %= M;
}
if(A < ) A += M;
return A;
} int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i = ;i <= n;i++) scanf("%lld",&a[i]);
long long t1 = ,t2 = ,t3 = ;
for(int i = ;i <= n;i++)
{
t1 = (t1+lucas(n-,i-,)*a[i])%;
t2 = (t2+lucas(n-,i-,)*a[i])%;
t3 = (t3+lucas(n-,i-,)*a[i])%;
}
long long a[],m[];
a[] = t1;
a[] = t2;
a[] = t3;
m[] = ;
m[] = ;
m[] = ;
printf("%lld\n",solve(m,a,));
}
return ;
}
1129.len-串和反串的LCS。
#include<bits/stdc++.h>
using namespace std; int dp[][];
char a[],b[]; int main()
{
ios::sync_with_stdio(false);
int T;
scanf("%d",&T);
while(T--)
{
scanf("%s",&a);
int len = strlen(a);
for(int i = ;i < len;i++) b[i] = a[len-i-];
memset(dp,,sizeof(dp));
for(int i = ;i <= len;i++)
{
for(int j = ;j <= len;j++)
{
if(a[i-] == b[j-]) dp[i][j] = dp[i-][j-]+;
else dp[i][j] = max(dp[i-][j],dp[i][j-]);
}
}
printf("%d\n",len-dp[len][len]);
}
return ;
}
1233.转化为LIS，nlogn。
#include<bits/stdc++.h>
using namespace std; int n,a[],b[],mp[]; int main()
{
ios::sync_with_stdio();
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i = ;i <= n;i++) scanf("%d",&a[i]);
for(int i = ;i <= n;i++) scanf("%d",&b[i]);
for(int i = ;i <= n;i++) mp[b[i]] = i;
for(int i = ;i <= n;i++) a[i] = mp[a[i]];
b[] = a[];
int len = ;
for(int i = ;i <= n;i++)
{
int t = lower_bound(b+,b+len+,a[i])-b;
b[t] = a[i];
if(t == len+) len++;
}
printf("%d\n",len); }
return ;
}
1234.打表会发现规律。
#include<bits/stdc++.h>
using namespace std; int n;
double a[],b[],R; int main()
{
ios::sync_with_stdio();
cin >> n;
if(n == )
{
cout << "0.00" << endl;
return ;
}
for(int i = ;i <= n;i++) cin >> a[i];
cin >> R;
sort(a+,a++n);
reverse(a+,a++n);
int l = ,r = n,cnt = ;
while(l <= r)
{
if(cnt%) b[r--] = a[cnt++];
else b[l++] = a[cnt++];
}
double ans = b[]+b[n];
for(int i = ;i < n;i++) ans += *sqrt(R*(b[i]+b[i+]-R));
cout << fixed << setprecision() << ans << endl;
return ;
}
1237.二分轮数。
#include<bits/stdc++.h>
using namespace std; int n,maxx = ,cnt[] = {}; bool ok(int x)
{
if(x < maxx) return ;
if(x-maxx > ) return ;
int now = ;
for(int i = ;i <= x;i++)
{
if(now > 1e6) break;
if(cnt[x-i] > now) return ;
now -= cnt[x-i];
now *= ;
}
return ;
} int main()
{
scanf("%d",&n);
for(int i = ;i <= n;i++)
{
int x;
scanf("%d",&x);
cnt[x]++;
maxx = max(maxx,x);
}
int l = ,r = 1e6+;
while(l < r)
{
int mid = (l+r)/;
if(ok(mid)) r = mid;
else l = mid+;
}
printf("%d\n",l);
return ;
}
1239.从高位往低位贪心，尽可能将1和并成偶数，维护每个可以划分的位置。
#include<bits/stdc++.h>
using namespace std; int n,k,a[][],ok[][]; int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
for(int i = ;i <= n;i++)
{
int x;
scanf("%d",&x);
for(int j = ;j <= ;j++)
{
a[i][j] = x%;
x /= ;
}
}
for(int i = ;i <= n;i++) ok[i][] = ;
int ans = ;
for(int j = ;j >= ;j--)
{
int cnt = ,now = ;
for(int i = ;i <= n;i++)
{
now ^= a[i][j];
if(now == && ok[i][j+])
{
ok[i][j] = ;
cnt++;
}
else ok[i][j] = ;
}
if(cnt < k || now == )
{
ans += (<<(j-));
for(int i = ;i <= n;i++) ok[i][j] = ok[i][j+];
}
}
cout << ans << endl;
}
return ;
}
1240.贪心找大于等于且最小的。
#include<bits/stdc++.h>
using namespace std; int n,a[]; int main()
{
while(~scanf("%d",&n))
{
multiset<int> s;
for(int i = ;i <= n;i++)
{
int x;
scanf("%d",&x);
s.insert(x);
}
for(int i = ;i <= n;i++) scanf("%d",&a[i]);
int ans = ;
for(int i = ;i <= n;i++)
{
auto it = s.lower_bound(a[i]);
if(it != s.begin())
{
it--;
ans++;
s.erase(it);
}
}
if(ans) printf("%d\n",ans);
else printf("Godv too strong\n");
}
return ;
}
1241.dp[i][j]表示前i个时刻，j个芯片最大的rp。
#include<bits/stdc++.h>
using namespace std; int n,m,a[],dp[][],x[][]; int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
cin >> n >> m;
for(int i = ;i <= n;i++) cin >> a[i];
for(int i = ;i <= n;i++)
{
int minn = ;
for(int j = i;j <= n;j++)
{
minn = min(minn,a[j]);
x[i][j] = minn;
}
}
memset(dp,,sizeof(dp));
for(int j = ;j <= m;j++)
{
for(int i = n;i >= ;i--)
{
for(int k = i;k >= ;k--) dp[i][j] = max(dp[i][j],dp[k-][j-]+(i-k+)*x[k][i]);
}
for(int i = ;i <= n;i++) dp[i][j] = max(dp[i][j],dp[i-][j]);
}
cout << dp[n][m] << endl;
}
return ;
}
*1242.待补。
1243.移项，开根号。
#include<bits/stdc++.h>
using namespace std; long long z,ans; void ok(long long x)
{
long long endd = sqrt(x/);
for(long long i = ;i <= endd;i++)
{
long long a = i*i,b = x-a;
long long t = sqrt(b);
if(t*t == b && __gcd(a,b) == && a != b) ans++;
}
}
int main()
{
int T;
cin >> T;
while(T--)
{
cin >> z;
ans = ;
long long endd = sqrt(*z);
for(long long i = ;i <= endd;i++)
{
if(*z%i) continue;
ok(i);
ok(*z/i);
}
cout << ans* << endl;
}
return ;
}
*1244.待补。
1245.数位dp+kmp。
#include<bits/stdc++.h>
using namespace std; int a[],b[],x[],cnt1,cnt2;
long long n,m,dp[][]; long long dfs(int now1,int now2,int limit)
{
if(now2 == ) return ;
if(now1 == ) return ;
if(!limit && dp[now1][now2] != -) return dp[now1][now2];
int endd = limit?a[now1]:;
long long ans = ;
for(int i = endd;i >= ;i--)
{
int t = now2;
while(b[t] != i && t != cnt2) t = x[t];
if(b[t] == i) t--;
ans += dfs(now1-,t,limit && i == endd);
}
if(!limit) dp[now1][now2] = ans;
return ans;
} int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
cin >> n >> m;
cnt1 = ;
cnt2 = ;
while(n)
{
a[++cnt1] = n%;
n /= ;
}
while(m)
{
b[++cnt2] = m%;
m /= ;
}
int i = cnt2,j = cnt2+;
x[cnt2] = cnt2+;
while(i >= )
{
if(j == cnt2+ || b[i] == b[j]) x[--i] = --j;
else j = x[j];
}
x[cnt2] = cnt2;
memset(dp,-,sizeof(dp));
cout << dfs(cnt1,cnt2,)- << endl;
}
return ;
}
*1247.待补。
*1248.待补。
*1250.待补。
1254.高精度。
#include<bits/stdc++.h>
using namespace std; string s1,s2,s3,s4; int compare(string str1,string str2)
{
if(str1.length() > str2.length()) return ;
else if(str1.length() < str2.length()) return -;
else return str1.compare(str2);
} string sub(string str1,string str2)
{
string str;
int flag = ;
if(compare(str1,str2) < )
{
flag = ;
swap(str1,str2);
}
int tmp = str1.length()-str2.length(),cf = ;
for(int i = str2.length()-;i >= ;i--)
{
if(str1[tmp+i] < str2[i]+cf)
{
str = char(str1[tmp+i]-str2[i]-cf+''+)+str;
cf = ;
}
else
{
str = char(str1[tmp+i]-str2[i]-cf+'')+str;
cf = ;
}
}
for(int i = tmp-;i >= ;i--)
{
if(str1[i]-cf >= '')
{
str = char(str1[i]-cf)+str;
cf = ;
}
else
{
str = char(str1[i]-cf+)+str;
cf = ;
}
}
str.erase(,str.find_first_not_of(''));
if(str.empty()) str = "";
if(flag) str = "-"+str;
return str;
} int main()
{
cin >> s1 >> s2 >> s3 >> s4;
string t = sub(s2,s1),y = sub(s4,s3);
int x = compare(t,y);
if(x > ) cout << "Ting" << endl;
else if(x < ) cout << "Yu" << endl;
else cout << "Excellent" << endl;
return ;
}
1255.进制转化。
#include<bits/stdc++.h>
using namespace std; string s; int main()
{
while(getline(cin,s))
{
stringstream ss(s);
char c;
int x,y,z;
ss >> c >> x >> c >> y >> c >> z;
cout << "#" << hex << uppercase << setw() << setfill('') << x << setw() << setfill('') << y << setw() << setfill('') << z << endl;
}
return ;
}
1256.处理每一位。
#include<bits/stdc++.h>
using namespace std; int n;
string yi[] = {"","I","X","C","M"};
string wu[] = {"","V","L","D"}; int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
string s = "";
int now = ;
while(n)
{
int t = n%;
n /= ;
if(t <= )
{
while(t--) s = yi[now]+s;
}
else if(t <= )
{
s = wu[now]+s;
if(t == ) s = yi[now]+s;
}
else if(t <= )
{
t -= ;
while(t--) s = yi[now]+s;
s = wu[now]+s;
}
else s = yi[now]+yi[now+]+s;
now++;
}
cout << s << endl;
}
return ;
}
1257.两个绝对值的最大值可转化为非绝对值的形式，维护两个最小值。
#include<bits/stdc++.h>
using namespace std; int n;
struct xx
{
int x,y;
friend bool operator<(xx a,xx b)
{
return a.x < b.x;
}
}a[]; int main()
{
ios::sync_with_stdio();
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i = ;i <= n;i++) scanf("%d%d",&a[i].x,&a[i].y);
sort(a+,a++n);
long long ans = ,minn1 = 1e18,minn2 = 1e18;
for(int i = ;i <= n;i++)
{
ans = max(ans,max(a[i].x+a[i].y-minn1,a[i].x-a[i].y-minn2));
minn1 = min(minn1,(long long)a[i].x+a[i].y);
minn2 = min(minn2,(long long)a[i].x-a[i].y);
}
cout << ans << endl;
}
return ;
}
1259.n-1。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
cin >> n;
cout << n- << endl;
}
return ;
}
1260.加权平均值。
#include<bits/stdc++.h>
using namespace std; int n;
map<string,double> mp1;
map<string,int> mp2; int main()
{
ios::sync_with_stdio();
cin >> n;
for(int i = ;i <= n;i++)
{
string s;
double x;
int y;
cin >> s >> x >> y;
if(mp2.count(s))
{
mp2[s] = max(mp2[s],y);
}
else
{
mp1[s] = x;
mp2[s] = y;
}
}
double sum1 = ,sum2 = ,sum3;
for(auto it = mp1.begin();it != mp1.end();it++)
{
int t = mp2[it->first];
sum1 += t*it->second;
if(t < ) sum2 += ;
else if(t < ) sum2 += *it->second;
else if(t < ) sum2 += 1.5*it->second;
else if(t < ) sum2 += 1.7*it->second;
else if(t < ) sum2 += *it->second;
else if(t < ) sum2 += 2.3*it->second;
else if(t < ) sum2 += 2.7*it->second;
else if(t < ) sum2 += *it->second;
else if(t < ) sum2 += 3.3*it->second;
else if(t < ) sum2 += 3.7*it->second;
else sum2 += *it->second;
sum3 += it->second;
}
cout << fixed << setprecision() << sum1/sum3 << endl;
cout << fixed << setprecision() << sum2/sum3 << endl;
return ;
}
1261.带属性的并查集。
#include<bits/stdc++.h>
using namespace std; int n,m,a[],pre[]; int findd(int x)
{
if(x == pre[x]) return x;
int t = findd(pre[x]);
a[x] ^= a[pre[x]];
pre[x] = t;
return pre[x];
} void join(int x,int y)
{
int xx = findd(x),yy = findd(y);
if(xx != yy)
{
a[xx] = a[x]^a[y]^;
pre[xx] = yy;
}
} int main()
{
ios::sync_with_stdio();
while(~scanf("%d%d",&n,&m))
{
memset(a,,sizeof(a));
for(int i = ;i <= n;i++) pre[i] = i;
for(int i = ;i <= m;i++)
{
int t,x,y;
scanf("%d%d%d",&t,&x,&y);
if(t)
{
int xx = findd(x),yy = findd(y);
if(xx != yy) printf("Not sure yet.\n");
else if(a[x] == a[y]) printf("In the same category.\n");
else printf("In different category.\n");
}
else join(x,y);
}
}
return ;
}
1262.先求Q的前缀和后缀和。
#include<bits/stdc++.h>
using namespace std; string s;
long long l[] = {},r[] = {}; int main()
{
ios::sync_with_stdio();
cin >> s;
int n = s.length();
s = " "+s;
for(int i = ;i <= n;i++) l[i] = l[i-]+(s[i] == 'Q');
for(int i = n;i >= ;i--) r[i] = r[i+]+(s[i] == 'Q');
long long ans = ;
for(int i = ;i <= n;i++)
{
if(s[i] == 'A') ans += l[i-]*r[i+];
}
cout << ans << endl;
return ;
}
1263.用r来分隔g和b，注意r的个数限制，以及r只有两个时g和b的限制，以及n=1的特例。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
cin >> n;
int r = ,g = ,b = ;
for(int i = ;i <= ;i++)
{
string s;
cin >> s;
for(int j = ;j < n;j++)
{
if(s[j] == 'R') r++;
else if(s[j] == 'G') g++;
else b++;
}
}
if(n == && r != && (!g || !b) || n > && r <= n && (!g || !b || r == && g% == && b% == || r > )) cout << "YES" << endl;
else cout << "NO" << endl;
}
return ;
}
1264.二进制输出。
#include<bits/stdc++.h>
using namespace std; int a[];
long long n; int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
memset(a,,sizeof(a));
int now = ;
while(n)
{
a[now--] = n%;
n /= ;
}
for(int i = ;i <= ;i++) cout << char(a[i]?'R':'G');
cout << endl;
}
return ;
}
*1266.待补。
1269.bfs。
#include<bits/stdc++.h>
using namespace std; string s[];
bool vis[][][] = {};
int dis[][][];
int dx[] = {-,,,};
int dy[] = {,,-,};
struct xxx
{
int x,y,b;
xxx(){};
xxx(int xx,int yy,int bb = ):x(xx),y(yy),b(bb){};
}; int main()
{
ios::sync_with_stdio();
memset(dis,0x3f,sizeof(dis));
for(int i = ;i <= ;i++)
{
cin >> s[i];
s[i] = " "+s[i];
}
queue<xxx> q;
int endx,endy;
for(int i = ;i <= ;i++)
{
for(int j = ;j <= ;j++)
{
if(s[i][j] == 'S')
{
q.push(xxx(i,j,));
vis[i][j][] = ;
dis[i][j][] = ;
}
else if(s[i][j] == 'E')
{
endx = i;
endy = j;
}
}
}
while(!q.empty())
{
int x = q.front().x,y = q.front().y,b = q.front().b;
q.pop();
if(x == endx && y == endy) break;
for(int i = ;i < ;i++)
{
int xx = x+dx[i],yy = y+dy[i],bb = b;
if(xx < || xx > || yy < || yy > ) continue;
if(s[xx][yy] == 'r' && (bb == || bb == )) continue;
if(s[xx][yy] != 'r' && !vis[xx][yy][bb])
{
if(bb == && s[xx][yy] == 'B') bb = ;
vis[xx][yy][bb] = ;
dis[xx][yy][bb] = min(dis[xx][yy][bb],dis[x][y][b]+);
q.push(xxx(xx,yy,bb));
}
else if(s[xx][yy] == 'r' && bb == && !vis[xx][yy][])
{
vis[xx][yy][] = ;
dis[xx][yy][] = min(dis[xx][yy][],dis[x][y][b]+);
q.push(xxx(xx,yy,));
}
}
}
int ans = min(min(dis[endx][endy][],dis[endx][endy][]),dis[endx][endy][]);
if(ans != 0x3f3f3f3f) cout << ans << endl;
else cout << - << endl;
return ;
}
1272.排序和去重。
#include<bits/stdc++.h>
using namespace std; int n,a[],b[],c[] = {}; int main()
{
ios::sync_with_stdio();
cin >> n;
for(int i = ;i <= n;i++) cin >> a[i];
for(int i = ;i <= n;i++) b[i] = a[i];
sort(b+,b++n);
cout << b[];
for(int i = ;i <= n;i++) cout << " " << b[i];
cout << endl;
int maxx = b[n-],minn = b[];
for(int i = ;i <= n;i++) b[i] = a[i];
vector<int> v;
map<int,int> mp;
for(int i = ;i <= n;i++)
{
if(mp.count(b[i])) continue;
v.push_back(b[i]);
mp[b[i]] = ;
}
cout << v[];
for(int i = ;i < v.size();i++) cout << " " << v[i];
cout << endl;
cout << maxx << " " << minn << endl;
for(int i = ;i <= n;i++) c[a[i]]++;
maxx = ;
for(int i = ;i <= ;i++) maxx = max(maxx,c[i]);
v.clear();
for(int i = ;i <= ;i++)
{
if(c[i] == maxx) v.push_back(i);
}
cout << v[];
for(int i = ;i < v.size();i++) cout << " " << v[i];
cout << endl;
return ;
}
1273.不断交换到对应下标位置。
#include<bits/stdc++.h>
using namespace std; int n,b[]; struct xx
{
int x,pos;
friend bool operator<(xx a,xx b)
{
return a.x < b.x;
}
}a[]; int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
cin >> n;
for(int i = ;i <= n;i++)
{
cin >> a[i].x;
a[i].pos = i;
}
sort(a+,a++n);
for(int i = ;i <= n;i++) b[a[i].pos] = i;
int ans = ;
for(int i = ;i <= n;i++)
{
while(b[i] != i)
{
ans++;
swap(b[i],b[b[i]]);
}
}
cout << ans << endl;
}
return ;
}
1274.推公式。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
cin >> n;
cout << (n-)*(n-)/ << endl;
}
return ;
}
1275.枚举左下，右上两个点。
#include<bits/stdc++.h>
using namespace std; int n,a[][]; int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
memset(a,,sizeof(a));
cin >> n;
for(int i = ;i <= n;i++)
{
int x,y;
cin >> x >> y;
a[x][y] = ;
}
int ans = ;
for(int i = ;i <= ;i++)
{
for(int j = ;j <= ;j++)
{
if(!a[i][j]) continue;
int endd = min(-i,-j);
for(int k = ;k <= endd;k++)
{
if(a[i+k][j] && a[i][j+k] && a[i+k][j+k]) ans++;
}
}
}
cout << ans << endl;
}
return ;
}
1276.维护个前缀和累计值。
#include<bits/stdc++.h>
using namespace std; int n,m,a[] = {}; int main()
{
ios::sync_with_stdio();
while(cin >> n >> m)
{
memset(a,,sizeof(a));
for(int i = ;i <= m;i++)
{
int x;
cin >> x;
a[]++;
a[x+]--;
}
for(int i = ;i <= n;i++) a[i] += a[i-];
for(int i = ;i <= n;i++) cout << a[i] << endl;
}
return ;
}
1277.判断旋转的9种情况是否可行。
#include<bits/stdc++.h>
using namespace std; int a[],b[];
int pan[][] = {{,,,},{,,,},{,,,},{,,,},{,,,},{,,,}};
int zhuan[][] = {{,,,,,,,},{,,,,,,,},{,,,,,,,}};
bool ok()
{
for(int i = ;i < ;i++)
{
if(!(b[pan[i][]] == b[pan[i][]] && b[pan[i][]] == b[pan[i][]] && b[pan[i][]] == b[pan[i][]])) return ;
}
return ;
} int main()
{
ios::sync_with_stdio();
while(cin >> a[])
{
int flag = ;
for(int i = ;i <= ;i++) cin >> a[i];
for(int i = ;i < ;i++)
{
for(int k = ;k <= ;k++)
{
for(int j = ;j <= ;j++) b[j] = a[j];
for(int j = ;j < ;j++) b[zhuan[i][j]] = a[zhuan[i][(j+*k)%]];
if(ok()) flag = ;
}
}
if(flag) cout << "YES" << endl;
else cout << "NO" << endl;
}
return ;
}
1278.n小于等于100时，必为91。
#include<bits/stdc++.h>
using namespace std; long long n; int main()
{
ios::sync_with_stdio();
int T;
scanf("%d",&T);
while(T--)
{
scanf("%lld",&n);
if(n > ) printf("%lld\n",n-);
else printf("91\n");
}
return ;
}
1281.f(n) = f(n-1)+f(n-2)+f(n-4)，矩阵快速幂。
#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std; long long n; struct matrix
{
long long m[][];
}; matrix one = { ,,,,
,,,,
,,,,
,,, }; matrix base = { ,,,,
,,,,
,,,,
,,, };
matrix mul(matrix a, matrix b)
{
matrix tmp;
for(int i = ; i < ;i++)
{
for(int j = ; j < ;j++)
{
tmp.m[i][j] = ;
for(int k = ; k < ;k++) tmp.m[i][j] = (tmp.m[i][j]+a.m[i][k]*b.m[k][j])%MOD;
}
}
return tmp;
} matrix maqower(matrix a,long long b)
{
matrix ans = one;
while(b)
{
if(b%) ans = mul(ans,a);
a = mul(a,a);
b /= ;
}
return ans;
} int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
if(n == ) cout << << endl;
else if(n == ) cout << << endl;
else if(n == ) cout << << endl;
else if(n == ) cout << << endl;
else
{
matrix ans = maqower(base,n-);
cout << (*ans.m[][]+*ans.m[][]+*ans.m[][]+ans.m[][])%MOD << endl;
}
}
return ;
}
1282.推公式。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
cin >> n;
cout << n*(n-)/ << endl;
}
return ;
}
1283.从中位数开始向上发，使这n/2+1个人最小值最大。
#include<bits/stdc++.h>
using namespace std; int n,m;
long long a[]; int main()
{
ios::sync_with_stdio();
while(cin >> n >> m)
{
for(int i = ;i <= n;i++) cin >> a[i];
sort(a+,a++n);
reverse(a+,a++n);
int k = (n+)/,now = k;
long long sum = ;
while(now > && (a[now-]-a[now])*(k-now+)+sum <= m)
{
sum += (a[now-]-a[now])*(k-now+);
now--;
}
m -= sum;
cout << a[now]+m/(k-now+) << endl;
}
return ;
}
1284.最大公约数。
#include<bits/stdc++.h>
using namespace std; int a,b; int main()
{
ios::sync_with_stdio();
while(cin >> a >> b) cout << __gcd(a,b) << endl;
return ;
}
1285.简单推一下公式，记忆化dp。
#include<bits/stdc++.h>
using namespace std; int n;
double dp[]; double dfs(int x)
{
if(dp[x] > -1e-) return dp[x];
dp[x] = ;
for(int i = ;i <= x;i++)
{
if(x%i == ) dp[x] += 1.0/x;
else dp[x] += (dfs(x%i)+)/x;
}
double t = 1.0*x/n;
dp[x] += (-t)/t;
return dp[x];
} int main()
{
ios::sync_with_stdio();
while(cin >> n)
{
for(int i = ;i <= n;i++) dp[i] = -;
double ans = ;
for(int i = ;i <= n;i++) ans += dfs(i)/n;
cout << fixed << setprecision() << ans << endl;
}
return ;
}
1286.dp或者规律。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
int T;
cin >> T;
while(T--)
{
cin >> n;
int endd = sqrt(n+0.1);
long long ans = ;
for(int i = ;i <= endd;i++)
{
while(n%i == )
{
ans += i-;
n /= i;
}
}
if(n > ) ans += n-;
cout << ans <<endl;
}
return ;
}
1284.二分，三人或以上相同=1-无人相同-两人相同，计算时注意精度问题。
#include<bits/stdc++.h>
using namespace std; int n,ans[] = {,,,,,,,,,,};
double c[][]; void get_C(int maxn)
{
c[][] = ;
for(int i = ;i <= maxn;i++)
{
c[i][] = ;
for(int j = ;j <= i;j++) c[i][j] = c[i-][j]+c[i-][j-];
}
} bool ok(int x)
{
if(x > *n) return ;
double t1 = ,t2 = ,tt = ;
if(x > n) t1 = ;
else
{
for(int i = n-x+;i <= n;i++)
{
t1 *= 1.0*i/n;
}
}
for(int k = ;k <= x/;k++)
{
tt *= 1.0*c[*k][];
tt /= k*(n-x+k);
t2 += tt*c[x][*k];
}
for(int i = n-x+;i <= n;i++) t2 *= 1.0*i/n;
double t = -t1-t2;
return t > 0.5+1e-;
} int main()
{
ios::sync_with_stdio();
get_C();
while(cin >> n)
{
if(n <= )
{
cout << ans[n] << endl;
continue;
}
int l = ,r = min(,n);
while(l < r)
{
int mid = (l+r)/;
if(ok(mid)) r = mid;
else l = mid+;
}
cout << l << endl;
}
return ;
}
1288.根据逆序数判断是否有解，dfs，康拓展开储存状态。
#include<bits/stdc++.h>
using namespace std; struct Node
{
int a[],num,Hash;
}start; int HASH[] = {,,,,,,,,};
int vis[],dis[] = {-,-,,},step[];
int a[][]; int get_Hash(Node x)
{
int ans = ;
for(int i = ;i < ;i++)
{
int k = ;
for(int j =;j < i;j++)
{
if(x.a[j] > x.a[i]) k++;
}
ans += HASH[i]*k;
}
return ans;
} int main()
{
ios::sync_with_stdio();
int T;
cin >> T;
while(T--)
{
memset(vis,,sizeof(vis));
for(int i = ;i < ;i++)
{
cin >> start.a[i];
if(start.a[i] == ) start.num = i;
}
int num = ;
for(int i = ;i < ;i++)
{
for(int j = i+;j <;j++)
{
if(start.a[i] && start.a[j] && start.a[i]>start.a[j]) num++;
}
}
if(num%)
{
cout << "cannot" << endl;
continue;
}
start.Hash = get_Hash(start);
step[start.Hash] = ;
vis[start.Hash] = ;
queue<Node> q;
q.push(start);
while(!q.empty())
{
Node temp = q.front();
q.pop();
if(temp.Hash == ) break;
int xx = temp.num/,yy = temp.num%;
for(int i = ;i < ;i++)
{
Node x = temp;
x.num += dis[i];
if(yy == && i == ) continue;
if(yy == && i == ) continue;
if(x.num < || x.num > ) continue;
swap(x.a[x.num],x.a[temp.num]);
x.Hash = get_Hash(x);
if(vis[x.Hash]) continue;
step[x.Hash] = step[temp.Hash]+;
q.push(x);
vis[x.Hash] = ;
}
}
cout << step[] << endl;
}
return ;
}
1289.最优解必定为递增的相邻或递减的相邻，对于每种情况，找大于等于当前的最小值或小于等于当前的最大值。
#include<bits/stdc++.h>
using namespace std; int n,a[],l[]; int main()
{
while(~scanf("%d",&n))
{
for(int i = ;i <= n;i++) scanf("%d",&a[i]);
double ans = 1e9;
l[] = ;
for(int i = ;i <= n;i++)
{
l[i] = i-;
while(l[i] >= && a[i] > a[l[i]]) l[i] = l[l[i]];
if(l[i] != ) ans = min(ans,1.0*(a[l[i]]-a[i])/(i-l[i]));
}
for(int i = ;i <= n;i++) a[i] = -a[i];
l[] = ;
for(int i = ;i <= n;i++)
{
l[i] = i-;
while(l[i] >= && a[i] > a[l[i]]) l[i] = l[l[i]];
if(l[i] != ) ans = min(ans,1.0*(a[l[i]]-a[i])/(i-l[i]));
}
printf("%.2f\n",ans);
}
return ;
}
1290.dp或者规律。
#include<bits/stdc++.h>
using namespace std; int n; int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
int endd = sqrt(n+0.1);
long long ans = ;
for(int i = ;i <= endd;i++)
{
while(n%i == )
{
ans += i-;
n /= i;
}
}
if(n > ) ans += n-;
printf("%d\n",ans);
}
return ;
}
1291.转化一下。
#include<bits/stdc++.h>
using namespace std; map<string,string> mp; int main()
{
mp["zero"] = "ling";
mp["one"] = "yi";
mp["two"] = "er";
mp["three"] = "san";
mp["four"] = "si";
mp["five"] = "wu";
mp["six"] = "liu";
mp["seven"] = "qi";
mp["eight"] = "ba";
mp["nine"] = "jiu";
mp["ten"] = "shi";
int T;
cin >> T;
while(T--)
{
string s;
cin >> s;
cout << mp[s] << endl;
}
return ;
}
1292.记录一下出现过的字母。
#include<bits/stdc++.h>
using namespace std; int n;
string s; int main()
{
ios::sync_with_stdio();
while(cin >> s)
{
set<char> st;
for(int i = ;i < s.length();i++) st.insert(s[i]);
cin >> n;
int maxx = ;
while(n--)
{
cin >> s;
int ok = ;
for(int i = ;i < s.length();i++)
{
if(!st.count(s[i])) ok = ;
}
if(ok) maxx = max(maxx,(int)s.length());
}
cout << maxx << endl;
}
return ;
}
1293.二分答案。
#include<bits/stdc++.h>
using namespace std; int n,k;
long long a[]; bool ok(int x)
{
int cnt = ;
for(int i = ;i <= n;i++)
{
for(int j = i+;j <= n;j++)
{
if(min(a[i],a[j]) <= x) cnt++;
}
}
return cnt >= n*(n-)/-k+;
} int main()
{
ios::sync_with_stdio();
int t;
cin >> t;
while(t--)
{
cin >> n >> k;
for(int i = ;i <= n;i++) cin >> a[i];
int cnt = ;
sort(a+,a++n);
int l = ,r = n;
while(l < r)
{
int mid = (l+r)/;
if(ok(a[mid])) r = mid;
else l = mid+;
}
cout << a[l] << endl;
}
return ;
}
1294.求均值向下取整。
#include<bits/stdc++.h>
using namespace std; int n,k;
long long a[]; int main()
{
ios::sync_with_stdio();
int t;
cin >> t;
while(t--)
{
cin >> n;
int sum = ;
for(int i = ;i <= n;i++)
{
cin >> a[i];
sum += a[i];
}
cout << sum/n <<endl;
}
return ;
}
1295.dp[i][j]表示前i位起，余数为j的个数。
#include<bits/stdc++.h>
using namespace std; string s;
long long dp[][]; int main()
{
ios::sync_with_stdio();
while(cin >> s)
{
memset(dp,,sizeof(dp));
if(s[] == '') dp[][] = ;
else dp[][] = ;
for(int i = ;i < s.length();i++)
{
if(s[i] == '')
{
dp[i][] += dp[i-][]+;
dp[i][] += dp[i-][];
dp[i][] += dp[i-][];
}
else
{
dp[i][] += dp[i-][]+;
dp[i][] += dp[i-][];
dp[i][] += dp[i-][];
}
}
long long ans = ;
for(int i = ;i < s.length();i++) ans += dp[i][];
cout << ans << endl;
}
return ;
}
*1296.待补。
*1297.待补。
*1298.待补。
XDOJ的更多相关文章
- xdoj 1241--余神的rp机(区间dp)
xdoj 1241---余神的rp机 核
- xdoj新生现场赛1269——带有限制条件的bfs 寻找最短路径
bfss是解决最短路径的强大武器 (尝试dfs寻找最短路径 -(7*7)就会爆炸) 例题1 ccf 201604-4 游戏 问题描述 小明在玩一个电脑游戏,游戏在一个n×m的方格图上进行,小明控制 ...
- XDOJ 1046 - 高精度模板综合测试 - [高精度模板]
题目链接:http://acm.xidian.edu.cn/problem.php?id=1046 题目描述 请输出两个数的和,差,积,商,取余.注意不要有前导零. 输入 多组数据,每组数据是两个整数 ...
- [xdoj]1303jlz的刷题黑科技
先分析复杂度,给的数据是1e5的,那么我们至少需要一个nlogn的算法才可以.由于答案是一个数字,首先想到是二分法(一般答案是一个数字都可以通过二分法来完成) 下面是思路: 1.可以完成题目的条件是, ...
- [xdoj]1299&1300朱神的烦恼 朱神的序列
http://acm.xidian.edu.cn/problem.php?id=1299 1.第一道题简单的很,数据范围最多只有1e4,对于数组中的每一个元素进行两个for循环,i=0;i<n; ...
- [xdoj] 1310 DSKer的卡牌游戏
http://acm.xidian.edu.cn/problem.php?id=1310 1. 这道题可以类比括号匹配,YY和yy是两组可以匹配的信号,当然要注意逻辑是否正确,一开始进行括号匹配算法的 ...
- [xdoj] 1301&1302 数字计数 数字计数的复仇
1.首先需要掌握二进制数的一种特性,00=0,01=1,10=2,11=3.每一个二进制的值代表他前面的二进数的个数,比如11=3,他的前面就有三个二进制的数字,不过在本题中,题目数据是1-n,故把0 ...
- xdoj 1330---异或（找规律）
我是打表找的规律 233 样例什么作用都没有 只会迷惑作用... 1330: 天才琪露诺的完美算数教室 时间限制: 1 Sec 内存限制: 128 MB Special Judge提交: 37 ...
- xdoj 1067组合数学+动态规划 (一个题断断续续想了半年 233）
题目分析 : (8 4) 可以由(7 4),(6,4),( 4,4) 基础上转化 意味着一个新加入的元素可以按照它加入的方式分类,从而实现动态规划 核心:加入方式 新加入的元素构成转换环的元素个数(n ...
- xdoj 1237 (贪心+逆向思维）
提示: 当有的元素分裂的同时,其他元素也可以+1 分析: 逆向思维,把当前数列变成一个0: 相应得操作相反: 每个元素减1 相同得两个元素可以合并 设数列中最大的数是max,则一共需要减max次才可 ...
随机推荐
- 设置本地上网IP
在局域网中,我们经常需要根据网络连接环境来对本地连接的IP地址进行手动设置,那么如何对IP地址进行设置呢?下面小编就把教程介绍给大家. 1. 右击桌面“网上邻居”->选择“属性”,打开“网络共享 ...
- 【转】出现“ValueError : numpy.ufunc has the wrong size, try recompiling" 解决方法
出现这个问题的原因是:numpy版本和scikit-learn版本不搭配. 解决方法: 升级numpy即可: pip install -U numpy
- nor flash之写保护
背景 没有电池的嵌入式设备,很容易发生随机掉电.因此要让产品可靠稳定,就必须保证各种场景下的掉电安全. 例如系统更新过程随机掉电,不能导致系统无法启动.例如正常读写flash过程中掉电,最多正在传输的 ...
- 【Java基础总结】网络编程
网络编程 InetAddress tcp udp
- vue根据选择的月份动态展示当前月份的每一天并展示每一天所对应的星期几
我们在开发过程中所遇到的所有的奇奇怪怪的交互美其名曰用(奇)户(葩)体(需)验(求),而产品所谓的良好的交互效果,在我等开发人员眼里不值一提.不屑一顾.讨厌至极! 对于这样的需求,我通常都是: 但胳膊 ...
- SQL Server2012高可用之日志传送测试
(一)日志传送架构 (1.1)相关服务器 主服务器 :用于生产的服务器,上面运行这生产SQL Server数据库: 辅助服务器:用于存放主服务器上数据库的一个“镜像”数据库: 监控服务器:用来监控 ...
- C#实现把查询出的Table作为参数更新到数据库
1.ImportData主方法 把传入为object数组类型,按照下标取出对应的参数,此处为Table和Username public object[] ImportData(object[] Par ...
- 《C# 爬虫 破境之道》：第二境 爬虫应用 — 第三节：处理压缩数据
续上一节内容,本节主要讲解一下Web压缩数据的处理方法. 在HTTP协议中指出,可以通过对内容压缩来减少网络流量,从而提高网络传输的性能. 那么问题来了,在HTTP中,采用的是什么样的压缩格式和机制呢 ...
- 20190925Java课堂记录（二）
1. testrandom public class test2{ public static void main(String[] args) { int[] n=new int [1001]; n ...
- Qt Installer Framework翻译(5-1)
创建离线安装程序 脱机安装程序在安装过程中根本不会尝试连接在线存储库.但是,元数据配置(config.xml)使用户可以在线添加和更新组件. 在公司防火墙不允许用户连接到Web服务器的情况下,脱机安装 ...