Codeforces #219 (Div. 2) A. Collecting Beats is Fun

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本文探讨了在著名音乐游戏Kyubeat中，玩家Cucumberboy如何在限制时间内使用单手按下最多k个面板的问题。通过输入场景描述，判断是否能完美同步按下所有面板。

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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).

Sample test(s)

input
1
.135
1247
3468
5789

output
YES

input
5
..1.
1111
..1.
..1.

output
YES

input
1
....
12.1
.2..
.2..

output
NO

Note

In the third sample boy cannot press all panels in perfect timing. He can press all the panels in timing in time 1, but he cannot press the panels in time 2 in timing with his two hands.

一道水题，用数组统计1到9出现的次数，如果1到9出现的次数中有一个大于2*k，则输出NO，否则输出YES。

代码如下：

#include <stdio.h>
#define rep(n) for(int i=0; i<n; i++)
int main(void){

    int k,flag;
    char a[25];
    
    while(scanf("%d",&k)!=EOF){
     int count[25]={0};
     flag = 0;
     rep(20){
      scanf("%c",&a[i]);
      if(a[i] == '.')
       continue;
      count[a[i]-'1']++;
     }
     rep(9)
      if(count[i] > 2*k){
       flag = 1;
       break;
      }
     if(flag)
      printf("NO\n");
     else printf("YES\n");
    }
    return 0;
}