#include<bits/stdc++.h>

using namespace std;

int main()

{

freopen("poker.in","r",stdin);

freopen("poker.out","w",stdout);

int n,b[5][53]={0},h,di,s=0;

cin>>n;

char a[2];

for(int i=1;i<=n;i++)

{

cin>>a[0]>>a[1];

if(a[0]=='D')  h=1;

if(a[0]=='C')  h=2;

if(a[0]=='H')  h=3;

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该题解等待再次审核

........................................................................(剩余 493 个中英字符)

#include<bits/stdc++.h>

using namespace std;

int main()

{

freopen("apple.in","r",stdin);

freopen("apple.out","w",stdout);

long long int n,date=0,ans[10000],i=0;

cin>>n;

while(n>0)

{

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该题解等待再次审核

........................................................................(剩余 322 个中英字符)

#include<bits/stdc++.h>

using namespace std;

int main()

{

int l,r,sum;

cin>>l>>r;

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该题解等待再次审核

........................................................................(剩余 175 个中英字符)

#include <iostream>

using namespace std;

int n,m,num=0;

int bz[100][100];

int dx[4]={-1,0,1,0},

   dy[4]={0,1,0,-1};

void doit(int p,int q){

num++;

int h[1000][3];

int x,y,t,w;

t=0;w=1;h[1][1]=p;h[1][2]=q;

do{

t++;

for(int i=0;i<4;i++){

x=h[t][1]+dx[i];y=h[t][2]+dy[i];

if(x>=0&&x<m&&y>=0&&y<n&&bz[x][y])

{

w++;h[w][1]=x;h[w][2]=y;

bz[x][y]=0;

}

}

}while(t<w);

}

int main(){

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该题解等待再次审核

........................................................................(剩余 723 个中英字符)

#include<bits/stdc++.h>

using namespace std;

int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};

int sum=0,m,n,xb[100][100];

void doit(int p,int q)

{

int x,y,t,w,i;

int h[1000][3];

sum++;xb[p][q]=0;

t=0;w=1;h[1][1]=p;h[1][2]=q;

do{

t++;

for(i=0;i<=3;i++){

x=h[t][1]+dx[i];

y=h[t][2]+dy[i];

if((x>=0)&&(x<m)&&(y>=0)&&(y<n)&&(xb[x][y])){

w++;

h[w][1]=x;

h[w][2]=y;

xb[x][y]=0;

}

}

}while(t<w);

}

int main()

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该题解等待再次审核

........................................................................(剩余 748 个中英字符)

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#define x first

#define y second

using namespace std;

typedef pair<char,char> PCC;

const int N = 3010;

PCC pcc[N];

int main()

{

 int n, len;

 cin >> n >> len;

 if(n == 1){

   puts("1");

   return 0;

 }

 for(int i = 1; i <= n; i ++)

 {

   string s;

   cin >> s;

   char minc = s[0], maxc = s[0];

   for(int j = 1; j < len; j ++ )

   {

     if(s[j] < minc) minc = s[j

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该题解等待再次审核

........................................................................(剩余 955 个中英字符)