codeforces 841A Generous Kefa

最新推荐文章于 2021-08-03 18:05:50 发布

风吹过的烟花

最新推荐文章于 2021-08-03 18:05:50 发布

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分类专栏： ACM算法_水题文章标签：水题

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本文介绍了一个简单的编程问题：如何将不同颜色的气球公平地分配给朋友们，确保没有朋友收到两个相同颜色的气球。通过使用C++实现的示例代码，展示了如何统计每种颜色气球的数量并判断是否可以成功分配。

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One day Kefa found n baloons. For convenience, we denote color of i-th baloon as si — lowercase letter of the Latin alphabet. Also Kefa has k friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 100) — the number of baloons and friends.

Next line contains string s — colors of baloons.

Output

Answer to the task — «YES» or «NO» in a single line.

You can choose the case (lower or upper) for each letter arbitrary.

        Examples
      
          input
        
4 2
aabb

          output
        
YES

          input
        
6 3
aacaab

          output
        
NO

Note

In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.

In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».

水题。因为给朋友气球不能重的，所以直接看相同颜色气球的最大数量，如果不大于k就YES，否则就NO

#include<cstdio>
#include<cstring>
#include<map>
using namespace std;
int main()
{
    map<char,int>ff;
    int n,k,i;
    char str[110],x;
    scanf("%d%d",&n,&k);
    scanf("%s",str);
    for(i=0; i<n; i++)
    {
        x=str[i];
        ff[x]++;
        if(ff[x]>k)
        {
            printf("NO\n");
            return 0;
        }
    }
    printf("YES\n");
    return 0;
}