2050 Programming Competition (CCPC)
Pro&Sol
链接: https://pan.baidu.com/s/17Tt3EPKEQivP2-3OHkYD2A 提取码: wbnu 复制这段内容后打开百度网盘手机App，操作更方便哦
6491时间间隔
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
using namespace std; #define ll long long const int maxn=1e5+;
const int inf=1e9;
const double eps=1e-; char s[maxn],str[]=""; int main()
{
int t,len,i;
scanf("%d",&t);
while (t--)
{
scanf("%s",s);
len=strlen(s);
for (i=;i<len;i++)
if (s[i]!=str[i%])
break;
if (i==len && len%==)
printf("Yes\n");
else
printf("No\n");
}
return ;
}
/* */
6491时间间隔
S距离2050年1月1日0点0时0分多少秒
2050_value-this_value 而不是差值绝对值模100 (不过感觉有点牵强)
负数的结果进行相应处理
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
using namespace std; #define ll long long const int maxn=1e4+;
const int inf=1e9;
const double eps=1e-; int main()
{
int t,a,b,c,d,e,f;
scanf("%d",&t);
while (t--)
{
scanf("%d-%d-%d %d:%d:%d",&a,&b,&c,&d,&e,&f);
printf("%d\n",((-e*-f)%+)%);
}
return ;
}
/*
7
2023-12-01 03:12:12
68
2024-05-01 03:12:12
68
3000-03-01 23:00:59
41
6000-03-01 00:00:00
0
6661-01-12 13:34:45
15
6661-01-12 13:34:46
14
2049-12-31 23:59:59
1
*/
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
using namespace std; #define ll long long const int maxn=1e4+;
const int inf=1e9;
const double eps=1e-; int mon[]={,,,,,,,,,,,,}; ll cal(int a,int b,int c,int d,int e,int f)
{
ll sum=;
int i;
for (i=;i<a;i++)
if (i%== && (i%== || i%!=))
sum+=1ll****;
else
sum+=1ll****;
if (i%== && (i%== || i%!=))
mon[]=;
for (i=;i<b;i++)
sum+=1ll****mon[i];
sum+=1ll****(c-)+d**+e*+f;
return sum;
} int main()
{
int t,a,b,c,d,e,f;
ll tot=;
tot=cal(,,,,,);
scanf("%d",&t);
while (t--)
{
scanf("%d-%d-%d %d:%d:%d",&a,&b,&c,&d,&e,&f);
printf("%lld\n",(tot-cal(a,b,c,d,e,f)%+)%);
}
return ;
}
/* */
import java.text.SimpleDateFormat;
import java.util.*; public class Main {
public static void main(String[] args) throws Exception {
//upper lower
SimpleDateFormat tf=new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
Date t1=tf.parse("2050-01-01 00:00:00");
Scanner in=new Scanner(System.in);
String s;
int t;
t=in.nextInt();
s=in.nextLine();
while (t-->0) {
s=in.nextLine();
Date t2=tf.parse(s);
//System.out.println(t2); //test
//System.out.println(t2.getTime()/1000);
//ms
System.out.println(( Math.abs(t1.getTime()-t2.getTime()) )/1000%100);
// System.out.println(dif);
}
in.close();
}
}
/*
7
2023-12-01 03:12:12
2024-05-01 03:12:12
2000-03-01 23:00:59
2049-03-01 00:00:00
2048-01-12 13:34:45
2023-01-12 13:34:46
2036-12-31 23:59:59 68
68
41
0
15
14
1
*/
6492分宿舍
贪心，参见下方注释
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
using namespace std; #define ll long long const int maxn=1e5+;
const int inf=1e9;
const double eps=1e-; ll n,m,k,a,b,c; ll cal(ll g)
{
/*
替代 2*3=3*2
两人间 <=2
三人间 <=1
*/
ll tot=1e18;
int i;
for (i=;i<=;i++)
tot=min(tot,i*a+(g-i*+)/*b);
for (i=;i<=;i++)
tot=min(tot,(g-i*+)/*a+i*b);
return tot;
} int main()
{
ll tot;
int t,i;
scanf("%d",&t);
while (t--)
{
scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&k,&a,&b,&c);
tot=1e18;
for (i=k;i>=;i--)
tot=min(tot,c*i+cal(n+k-i)+cal(m+k-i)); printf("%lld\n",tot);
}
return ;
}
/*
1
1000 1000 1000 1000000000 1000000000 1000000000 1
1 1 0 3 2 1
*/
预处理f[x]，x个人，需要花费的最小代码。对于每个k对应的n+k和m+k，O(1)得到答案。
\( f[i]=min(f[i-2]+a,f[i-3]+b) \\ f[j]=0 j<=0 \)
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
using namespace std; #define ll long long const int maxn=2e3+;
const int inf=1e9;
const double eps=1e-; ll n,m,k,a,b,c,f[maxn]; int main()
{
ll tot;
int t,i,lim;
scanf("%d",&t);
while (t--)
{
scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&k,&a,&b,&c);
memset(f,sizeof(0x3f),sizeof(f));
lim=max(n+k,m+k);
f[]=;
for (i=;i<=lim;i++)
if (i<)
f[i]=min(f[i],a);
else
f[i]=min(f[i],f[i-]+a);
for (i=;i<=lim;i++)
if (i<)
f[i]=min(f[i],b);
else
f[i]=min(f[i],f[i-]+b); tot=1e18;
for (i=k;i>=;i--)
tot=min(tot,c*i+f[n+k-i]+f[m+k-i]);
printf("%lld\n",tot);
}
return ;
}
/*
1
1000 1000 1000 1000000000 1000000000 1000000000 1
1 1 0 3 2 1
*/
暴力，不会超时
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
using namespace std; #define ll long long const int maxn=1e5+10;
const int inf=1e9;
const double eps=1e-8; int main()
{
int t;
ll n,m,k,a,b,c,i,j,tot,sum,x,y;
scanf("%d",&t);
while (t--)
{
scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&k,&a,&b,&c);
tot=1e18;
for (i=k;i>=0;i--)
{
sum=c*i; x=n+k-i;
y=1e18;
for (j=0;j<=(x+1)/2;j++)
y=min(y,j*a+(x-j*2+2)/3*b);
sum+=y; x=m+k-i;
y=1e18;
for (j=0;j<=(x+1)/2;j++)
y=min(y,j*a+(x-j*2+2)/3*b);
sum+=y; tot=min(tot,sum);
} printf("%lld\n",tot);
}
return 0;
}
/*
1
1000 1000 1000 1000000000 1000000000 1000000000 1
1 1 0 3 2 1
*/
以下方法是错误的
6493PASS
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cmath>
4 #include <cstring>
5 #include <string>
6 #include <algorithm>
7 #include <set>
8 #include <map>
9 #include <queue>
10 #include <iostream>
11 using namespace std;
12
13 #define ll long long
14
15 const int maxn=1e4+10;
16 const int inf=1e9;
17 const double eps=1e-8;
18
19 int g[maxn],a[maxn];
20
21 int main()
22 {
23 int t,n,m,k,i,tot;
24 scanf("%d",&t);
25 while (t--)
26 {
27 scanf("%d%d%d",&n,&m,&k);
28 memset(g,0,sizeof(g));
29 for (i=1;i<=n;i++)
30 {
31 scanf("%d",&a[i]);
32 g[a[i]]++;
33 }
34 for (i=1;i<=m;i++)
35 g[i]/=k;
36 tot=0;
37 for (i=1;i<=n/2;i++)
38 {
39 g[a[i]]--;
40 if (g[a[i]]>=0)
41 tot++;
42 }
43 printf("%d\n",tot);
44 }
45 return 0;
46 }
47 /*
48
49 */
1005球赛
一开始猜测贪心有可能是错的，所以选择使用dp
\( \begin{equation*}\begin{split}&condition \quad (a,b) \quad -> \quad condition (x,y) \quad + \quad c(+1/+0)\\&(10,k)/(k,10) \quad -> \quad (0,0) \quad [(11,k)/(k,11)] \quad + \quad 1 \quad k<=9\\&(10,11)/(11,10) \quad -> \quad (0,0) \quad [(10,12)/(12,10)] \quad +1\\&(10,11)/(11,10) \quad -> \quad (10,10) \quad [(11,11)]\\&f[x][y]=max(f[a][b]+c)\end{split}\end{equation*} \)
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
using namespace std; #define ll long long const int maxn=1e4+10;
const int inf=1e9;
const double eps=1e-8; char str[maxn];
int f[maxn][12][12]; int main()
{
int T,g,len,i,j,k,s,t;
scanf("%d",&T);
while (T--)
{ scanf("%s",str+1);
len=strlen(str+1);
for (i=0;i<=len;i++)
for (j=0;j<12;j++)
for (k=0;k<12;k++)
f[i][j][k]=-1;
f[0][0][0]=0;
for (i=1;i<=len;i++)
{
for (j=0;j<12;j++)
for (k=0;k<12;k++)
if (f[i-1][j][k]!=-1)
{
if (str[i]=='A' || str[i]=='?')
{
s=j+1;
t=k;
if (s==11 && t==11)
s=10,t=10;
if ((s>=11 && s-t>=2) || (t>=11 && t-s>=2))
f[i][0][0]=max(f[i][0][0],f[i-1][j][k]+1);
else
f[i][s][t]=max(f[i][s][t],f[i-1][j][k]);
}
if (str[i]=='B' || str[i]=='?')
{
s=j;
t=k+1;
if (s==11 && t==11)
s=10,t=10;
if ((s>=11 && s-t>=2) || (t>=11 && t-s>=2))
f[i][0][0]=max(f[i][0][0],f[i-1][j][k]+1);
else
f[i][s][t]=max(f[i][s][t],f[i-1][j][k]);
}
} }
g=0;
for (i=0;i<12;i++)
for (j=0;j<12;j++)
g=max(g,f[len][i][j]);
printf("%d\n",g);
}
return 0;
}
/*
1
AAAAAAAAAAA
??????????????????????
?????
*/
6495冰水挑战
无贪心方法等。应该往dp想。c不应该作为一维。应该很自然就能想到。。。
\( 前i个挑战中，选择j个挑战，剩余的最大体力\\ \begin{equation*}\begin{split}&f[i][j]=f[i-1][j]+c_{i}\\&f[i][j]=max(f[i][j],min(f[i-1][j-1],b_{i})-a_{i}+c_{i})) \quad if \, min(f[i-1][j-1],b_{i})-a_{i}>0 \, , \, i>0 \\\end{split}\end{equation*} \)
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
using namespace std; #define ll long long const int maxn=1e3+;
const ll inf=1e18;
const double eps=1e-; ll f[maxn][maxn]; int main()
{
int t,n,i,j;
ll m,a,b,c;
scanf("%d",&t);
while (t--)
{
scanf("%d%lld",&n,&m);
for (i=;i<=n;i++)
for (j=;j<=n;j++)
f[i][j]=-inf;
f[][]=m;
for (i=;i<=n;i++)
{
scanf("%lld%lld%lld",&a,&b,&c);
f[i][]=f[i-][]+c;
for (j=;j<=n;j++)
{
f[i][j]=f[i-][j]+c;
if (min(f[i-][j-],b)-a>)
f[i][j]=max(f[i][j],min(f[i-][j-],b)-a+c);
}
}
for (j=n;j>=;j--)
if (f[n][j]>)
break;
printf("%d\n",j);
}
return ;
}
/* */
6496大厦
1.新方法
用下面的方法求矩形
对于每个横坐标小块i，求纵坐标上下限[ d[i] , u[i] ]。
若对于某个纵坐标，横坐标小块为[x1,x2]，则往上，[x3,x4]，有x1<=x3 , x4<=x2。往下，同理。(里大，外小)
对于横坐标小块[x1,x2]，求[x1,x2]的纵坐标上下限([ max(d[x1],d[x1+1],...,d[x2]) , min(u[x1],u[x1+1],...,u[x2]) ])
st/线段树处理。
O(T*(n*m+n*n/2))
横坐标小块12；13；14；..；23；24；...；n-1 n
1*n 个数n+(n-1)+(n-2)+...+1 = n*(n+1)/2
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cmath>
4 #include <cstring>
5 #include <string>
6 #include <algorithm>
7 #include <set>
8 #include <map>
9 #include <queue>
10 #include <iostream>
11 using namespace std;
12
13 #define ll long long
14
15 const int maxn=1e3+10;
16 const int inf=1e9;
17 const double eps=1e-8;
18 const ll mod=1e9+7;
19
20 int c1[maxn],c2[maxn],d[maxn],u[maxn];
21
22 int main()
23 {
24 ll sum;
25 int t,w,h,n,m,i,j,down,up;
26 scanf("%d",&t);
27 while (t--)
28 {
29 scanf("%d%d%d%d",&w,&h,&n,&m);
30 for (i=1;i<=n;i++)
31 scanf("%d",&c1[i]);
32 for (i=1;i<=m;i++)
33 scanf("%d",&c2[i]);
34 memset(d,0x3f,sizeof(d));
35 memset(u,0,sizeof(u));
36
37 sort(c1+1,c1+n+1);
38 sort(c2+1,c2+m+1);
39 for (i=1;i<=n;i++)
40 ///其实用二分更快
41 for (j=1;j<=m;j++)
42 if (0<=c1[i]+c2[j] && c1[i]+c2[j]<=w*2
43 && 0<=c1[i]-c2[j] && c1[i]-c2[j]<=h*2) ///w*2 h*2 int range
44 d[i]=min(d[i],j),u[i]=max(u[i],j);
45
46 sum=0;
47 for (i=1;i<n;i++)
48 {
49 down=d[i],up=u[i];
50 for (j=i+1;j<=n;j++)
51 {
52 down=max(down,d[j]),up=min(up,u[j]);
53 if (up>=down)
54 sum=(sum+1ll*(up-down)*(up-down+1)/2)%mod;
55 }
56 }
57 printf("%lld\n",sum); ///ll
58 }
59 return 0;
60 }
61 /*
62
63 */
2.bitset(题解)
time(/32)
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <bitset>
using namespace std; #define ll long long const int maxn=1e3+;
const int maxm=1e3+;
const int inf=1e9;
const double eps=1e-;
const ll mod=1e9+; int c1[maxn],c2[maxn];
bitset<maxm> f[maxn]; int main()
{
ll sum;
int t,w,h,n,m,i,j,g;
scanf("%d",&t);
while (t--)
{
scanf("%d%d%d%d",&w,&h,&n,&m);
for (i=;i<=n;i++)
scanf("%d",&c1[i]);
for (i=;i<=m;i++)
scanf("%d",&c2[i]); for (i=;i<=n;i++)
f[i].reset(); sort(c1+,c1+n+);
sort(c2+,c2+m+);
for (i=;i<=n;i++)
for (j=;j<=m;j++)
if (<=c1[i]+c2[j] && c1[i]+c2[j]<=w*
&& <=c1[i]-c2[j] && c1[i]-c2[j]<=h*) ///w*2 h*2 int range
f[i].set(j,); sum=;
for (i=;i<n;i++)
for (j=i+;j<=n;j++)
{
g=(f[i]&f[j]).count();
sum=(sum+1ll*g*(g-)/)%mod;
}
printf("%lld\n",sum); ///ll
}
return ;
}
/* */
6497骑行
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <iostream>
using namespace std; #define ll long long const int maxn=1e3+;
const int inf=1e9;
const double eps=1e-; double v,t;
double w[maxn],s[maxn],a[maxn],lim[maxn]; void work(double v1,double v2,double v0,double len,double a)
{
double dist=(v0*v0-v1*v1)//a + (v0*v0-v2*v2)//a;
double dist1=(v2*v2-v1*v1)//a;
if (dist<=len)
{
t+=(v0-v1)/a + (v0-v2)/a + (len-dist)/v0;
v=v2;
}
else if (v2>v1 && dist1>len)
{
v=sqrt(*a*len + v1*v1);
t+=(v-v1)/a;
}
else
{
double v3=sqrt((*a*len + v1*v1 + v2*v2)/);
t+=(v3-v1)/a + (v3-v2)/a;
v=v2;
}
} int main()
{
double temp;
int T,n,i;
scanf("%d",&T);
while (T--)
{
scanf("%d",&n);
for (i=;i<=n;i++)
scanf("%lf%lf%lf",&w[i],&s[i],&a[i]);
lim[n]=s[n];
for (i=n-;i>=;i--)
{
temp=sqrt(*a[i+]*w[i+]+lim[i+]*lim[i+]);
lim[i]=min(min(s[i],s[i+]),temp);
} v=,t=;
for (i=;i<=n;i++)
work(v,lim[i],s[i],w[i],a[i]);
printf("%.10f\n",t);
}
return ;
}
/*
*/
6498跨洋飞行
一个点x到另外一个点y，到达点y时，要预留min(y to others)的油。参见题解。
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cmath>
4 #include <cstring>
5 #include <string>
6 #include <algorithm>
7 #include <set>
8 #include <map>
9 #include <queue>
10 #include <iostream>
11 using namespace std;
12
13 #define ll long long
14
15 const int maxn=1e3+10;
16 const int inf=1e9;
17 const double eps=1e-10;
18
19 double x[maxn],y[maxn],dist[maxn],road[maxn][maxn],mr[maxn];
20 bool vis[maxn];
21
22 int main()
23 {
24 double a,b,c,d,u;
25 int t,n,i,j,p;
26 scanf("%d",&t);
27 while (t--)
28 {
29 scanf("%d%lf%lf%lf%lf%lf",&n,&a,&b,&c,&d,&u);
30 a+=b;
31 for (i=1;i<=n;i++)
32 scanf("%lf%lf",&x[i],&y[i]);
33 for (i=1;i<=n;i++)
34 mr[i]=1.0e18;
35 for (i=1;i<n;i++)
36 for (j=i+1;j<=n;j++)
37 {
38 road[i][j]=road[j][i]=sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]) );
39 mr[i]=min(mr[i],road[i][j]);
40 mr[j]=min(mr[j],road[i][j]);
41 }
42 mr[1]=0; ///
43 for (i=0;i<=n;i++)
44 dist[i]=1.0e18;
45 dist[1]=0;
46 memset(vis,0,sizeof(vis));
47 for (i=1;i<n;i++)
48 {
49 p=0;
50 for (j=1;j<=n;j++)
51 if (!vis[j] && dist[p]>dist[j])
52 p=j;
53 if (p==n)
54 break;
55 vis[p]=1;
56 for (j=1;j<=n;j++)
57 if (!vis[j] && mr[j]+road[p][j]<u+eps && dist[j]>dist[p] +a + road[p][j]*d + (mr[j]+road[p][j]-mr[p])*c)
58 dist[j]=dist[p] +a + road[p][j]*d + (mr[j]+road[p][j]-mr[p])*c;
59 }
60 printf("%.10f\n",dist[n]);
61 }
62 return 0;
63 }
64 /*
65
66 */
错误想法：
下界提高，x到y，至少dist（x，y）+min y to others（z）。而miny to others是有用的，x到y，剩余z，而y到其它点，至少要使用z，不浪费。除了最后一个点，浪费min n to others。但倒数第二个点到点n，而倒数第二个点到其它点的最短距离有可能不是该点到点n的距离。
1 #include <cstdio>
2 #include <cstdlib>
3 #include <cmath>
4 #include <cstring>
5 #include <string>
6 #include <algorithm>
7 #include <set>
8 #include <map>
9 #include <queue>
10 #include <iostream>
11 using namespace std;
12
13 #define ll long long
14
15 const int maxn=1e3+10;
16 const int inf=1e9;
17 const double eps=1e-10;
18
19 double x[maxn],y[maxn],dist[maxn],road[maxn][maxn],mr[maxn];
20 bool vis[maxn];
21
22 int main()
23 {
24 double a,b,c,d,u;
25 int t,n,i,j,p;
26 scanf("%d",&t);
27 while (t--)
28 {
29 scanf("%d%lf%lf%lf%lf%lf",&n,&a,&b,&c,&d,&u);
30 a+=b;
31 for (i=1;i<=n;i++)
32 scanf("%lf%lf",&x[i],&y[i]);
33 for (i=1;i<=n;i++)
34 mr[i]=1.0e18;
35 for (i=1;i<n;i++)
36 for (j=i+1;j<=n;j++)
37 {
38 road[i][j]=road[j][i]=sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]) );
39 mr[i]=min(mr[i],road[i][j]);
40 mr[j]=min(mr[j],road[i][j]);
41 }
42 for (i=1;i<=n;i++)
43 for (j=1;j<=n;j++)
44 if (i!=j && road[i][j]+mr[i]>u+eps)
45 road[i][j]=-1;
46 else
47 road[i][j]=road[i][j]*(c+d)+a;
48 for (i=0;i<=n;i++)
49 dist[i]=1.0e18;
50 dist[1]=0;
51 memset(vis,0,sizeof(vis));
52 for (i=1;i<n;i++)
53 {
54 p=0;
55 for (j=1;j<=n;j++)
56 if (!vis[j] && dist[p]>dist[j])
57 p=j;
58 if (p==n)
59 break;
60 vis[p]=1;
61 for (j=1;j<=n;j++)
62 if (!vis[j] && road[p][j]!=-1 && dist[j]>dist[p]+road[p][j])
63 dist[j]=dist[p]+road[p][j];
64 }
65 printf("%.10f\n",dist[n]+mr[n]*c);
66 }
67 return 0;
68 }
69 /*
70
71 */
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