CodeForces 373A

Cucumber boy is fan of Kyubeat, a famous music game.

Kyubeat has 16 panels for playing arranged in 4 × 4 table. When a panel lights up, he has to press that panel.

Each panel has a timing to press (the preffered time when a player should press it), and Cucumber boy is able to press at most k panels in a time with
his one hand. Cucumber boy is trying to press all panels in perfect timing, that is he wants to press each panel exactly in its preffered time. If he cannot press the panels with his two hands in perfect timing, his challenge
to press all the panels in perfect timing will fail.

You are given one scene of Kyubeat's panel from the music Cucumber boy is trying. Tell him is he able to press all the panels in perfect timing.

Input

The first line contains a single integer k (1 ≤ k ≤ 5) — the number of panels Cucumber boy can press with his one hand.

Next 4 lines contain 4 characters each (digits from 1 to 9, or period) — table of panels. If a digit i was written on the panel, it means the boy has to press that panel in time i.
If period was written on the panel, he doesn't have to press that panel.

Output

Output "YES" (without quotes), if he is able to press all the panels in perfect timing. If not, output "NO" (without quotes).

题意是给你一个4*4的游戏台，当有灯亮起来时要用手去按，可以用两只手，需要输入一只手一秒可以按几个灯，和一个4*4的矩阵，点表示不用按，数字表示应该在第几秒的时候按，其实最终的意思就是统计有没有同一秒内需要按的灯数超过两只手可以按的总数，如果超过则输出NO。

#include <iostream>

#include<cstdio>

#include<cstring>

#define inf 1e9

using namespace std;

int main()

{

    int t;

    char a[4][4];

    int b[10];

    memset(b,0,sizeof(b));

    while(cin>>t)

    {
int y=t*2;

        for(int i=0;i<4;i++)

        {

            scanf("%s",a[i]);

            for(int j=0;j<4;j++)

            if(a[i][j]!='.')

            {

                int x=a[i][j]-'0';

                b[x]++;

            }

        }

        int flag=0;

        for(int i=1;i<10;i++)

        {

            if(b[i]>y)

            {

                 flag=1;

                 break;

            }

        }

        if(flag)

            printf("NO\n");

        else printf("YES\n");

    }

    return 0;

}