Codeforces Round #262 (Div. 2)

A

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 100000
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 int n,m;
 int main()
 {
     int n,m,i,j;
     cin>>n>>m;
     if(m>n)
     cout<<n<<endl;
     else
     {
         int o = ;
         while(n)
         {
             n--;
             o++;
             if(o%m==) n++;
         }
         cout<<o<<endl;
     }
     return ;
 }

B

枚举和值

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 100000
 #define LL long long
 #define INF 0xfffffff
 #define M 1000000000
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 int o[N];
 int judge(LL x)
 {
     int ans = ;
     while(x)
     {
         ans+=x%;
         x/=;
     }
     return ans;
 }
 int main()
 {
     int a,b,c,i,j;
     cin>>a>>b>>c;
     int g = ;
     for(i = ; i <=  ; i++)
     {
         LL k = (LL)b*pow(i*1.0,a)+c;
         if(judge(k)==i&&k>&&k<M)
         o[++g] = k;
     }
     cout<<g<<endl;
     if(g)
     {
 
     sort(o+,o+g+);
     for(i = ; i < g ; i ++)
     cout<<o[i]<<" ";
     cout<<o[i]<<endl;
     }
     return ;
 }

C

二分高度，之后用高度减去原有高度，就可以知道每株花被浇了多少水，用线段树维护一下就可以得到最少浇水的次数。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 100010
 #define LL long long
 #define INF 0xfffffff
 #define M 1000200000
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 int a[N],o[N];
 LL s[N<<];
 int n,m,w;
 void build(int l,int r,int w)
 {
     s[w] = ;
     if(l==r)
     {
         s[w] = o[l];
         return ;
     }
     int m = (l+r)>>;
     build(l,m,w<<);
     build(m+,r,w<<|);
 }
 void down(int w,int m)
 {
     if(s[w])
     {
         s[w<<] += s[w];
         s[w<<|] += s[w];
         s[w] = ;
     }
 }
 void update(int a,int b,int d,int l,int r,int w)
 {
     if(a<=l&&b>=r)
     {
         s[w]+=d;
         return ;
     }
     down(w,r-l+);
     int m = (l+r)>>;
     if(a<=m) update(a,b,d,l,m,w<<);
     if(b>m) update(a,b,d,m+,r,w<<|);
 }
 LL query(int p,int l,int r,int w)
 {
     if(l==r)
         return s[w];
     down(w,r-l+);
     int m = (l+r)>>;
     if(p<=m) return query(p,l,m,w<<);
     else return query(p,m+,r,w<<|);
 }
 int cal(int k)
 {
     int i,j;
     for(i = ; i <= n; i++)
         if(k>a[i]) o[i] = k-a[i];
         else o[i] = ;
     build(,n,);
     LL num = ;
     for(i = ; i <= n; i++)
     {
         int pp = query(i,,n,);
 
         if(pp>) {update(i,min(n,i+w-),-pp,,n,);num+=pp;} if(num>m) return ;
     }
     if(num<=m) return ;
     return ;
 }
 int main()
 {
     int i,j;
     cin>>n>>m>>w;
     for(i = ; i <= n; i++)
         scanf("%d",&a[i]);
     int low = ,high = M,mid;
     int ans = ;
     while(low<=high)
     {
         mid = (low+high)>>;
         if(cal(mid)==)
             high = mid-;
         else
         {
             low = mid+;
             ans = max(ans,mid);
         }
     }
     cout<<ans<<endl;
     return ;
 }

D

对于

k=1 ,ans = L。

连续偶奇异或是为1的 比如10 11    12 13等 ，而奇偶则是不一定的

k= 2, 如果l%2==0，ans = L^(L+1)=1 ,否则要根据R-L+1的取值决定

k= 4 ，如果L%2=0，ans = L^(L+1)^(L+2)^(L+3)  = 0，否则如果R-L+1>4 也是为偶奇偶奇=0的，若R-L+1=4 就可转化为3的时候做。

k=3 ，有可能为1也有可能为0 ，假设L-R范围内的三个数x,y,z异或为0，x,y,z不同，就设x>y>z，那么x,y已确定，那么就是尽可能让z大，

比如

x  100110001111   那么y值的第一位一定为1，不然z就为1就不符合假设y>z了，因为y<x，所以之后遇到x为0的位，y，z也是为0的，一旦再遇到一位1，就可以把y置为0，z置为1，再之后z就可以一直为1，而y的变化可以依据x,z而定。

x  100110001111...1

y  100001110000...0

z  000111111111...1

这样z将获得<R的最大值，再与L相比较就可以知道结果了。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 100000
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 LL a[];
 int di[];
 void judge(LL l,LL r)
 {
     LL x = r;
     LL i;
     int g = ;
     while(x)
     {
         di[g++] = x%;
         x/=;
     }
     LL y = ;
     LL z = ;
     int flag = ;
     y+=((LL)<<(g-));
     for(i = g- ; i >=  ; i--)
     {
         if(di[i])
         {
             flag = ;
             z += ((LL)<<i);
         }
         else if(flag)
         {
             z+=((LL)<<i);
             y+=((LL)<<i);
         }
     }
     if(z>=l)
     {
         cout<<"0\n";
         cout<<"3\n";
         cout<<r<<" "<<y<<" "<<z<<endl;
     }
     else
     {
         cout<<"1\n";
         cout<<"2\n";
         if(l%)
         cout<<l+<<" "<<l+<<endl;
         else
         cout<<l<<" "<<l+<<endl;
     }
 }
 int main()
 {
     LL l,r,i,k;
     cin>>l>>r>>k;
     if(r==l)
     {
         cout<<l<<endl;
         cout<<"1\n";
         cout<<l<<endl;
         return ;
     }
     if(l%==)
     {
         if(r-l+==||k<=)
         {
             if(k==)
             {
                 cout<<l<<endl;
                 cout<<"1\n";
                 cout<<l<<endl;
             }
             else
             {
                 cout<<"1\n";
                 cout<<"2\n";
                 cout<<l<<" "<<l+<<endl;
             }
         }
         else if(r-l+==||k==)
         {
             judge(l,r);
         }
         else
         {
             cout<<"0\n";
             cout<<"4\n";
             for(i = l  ; i < l +  ; i++)
                 cout<<i<<" ";
             puts("");
         }
     }
     else
     {
         if(r-l+==||k<=)
         {
             if(k==||(r-l+==&&(l^(l+))>l))
             {
                 cout<<l<<endl;
                 cout<<"1\n";
                 cout<<l<<endl;
             }
             else
             {
                 if(r-l+==)
                 {
                     cout<<(l^(l+))<<endl;
                     cout<<"2\n";
                     cout<<l<<" "<<l+<<endl;
                 }
                 else
                 {
                     cout<<"1\n";
                     cout<<"2\n";
                     cout<<l+<<" "<<l+<<endl;
                 }
             }
         }
         else if(r-l+==||k==)
         {
             judge(l,r);
         }
         else if(r-l+==)
         {
             int flag = ;
             judge(l,r);
         }
         else
         {
             cout<<"0\n";
             cout<<"4\n";
             for(i = l+ ; i < l+ ; i++)
                 cout<<i<<" ";
             puts("");
         }
 
     }
     return ;
 }