A. Generous Kefa（Round #429 (Div. 2)）

气球分配问题

最新推荐文章于 2019-07-05 10:16:00 发布

原创最新推荐文章于 2019-07-05 10:16:00 发布 · 367 阅读

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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一群朋友，确保没有两个朋友收到相同颜色的气球。文章通过一个示例程序详细解释了如何检查这种分配是否可行，并提供了代码实现。

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，如果有一种颜色的气球个数大于朋友数，则无法平均分。

#include <iostream>
#include <cstdio>

using namespace std;

int a[27];

int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)==2)
    {
        getchar();
        char ch;
        for(int i=0;i<n;i++)
        {
            scanf("%c",&ch);
            a[ch-'a']++;
        }
        int flag=1;
        for(int i=0;i<26;i++)
        {
            if(a[i]!=0)
            {
                if(a[i]>k)
                {
                    flag=0;
                    break;
                }
            }
        }
        if(flag)
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    return 0;
}