#include<bits/stdc++.h>

using namespace std;

int f(int n,int k)

{

int s=1;

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

 s*=10;

return  n%s;

}

int main()

{

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

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

int n,q;

cin>>n>>q;

long long int a[1001],b[1001][3];

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

   cin>>a[i];

  for(int i=1;i<=q;i++)

&

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该题解等待再次审核

........................................................................(剩余 615 个中英字符)

#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)