A. Generous Kefa
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 2aabb
Output
YES
Input
6 3aacaab
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?.
题意：每个人能不能分到均不相同的颜色气球……思路很简单，只要数目最多的颜色不超过人数就好了……（窝现在也就做做签到题了orz（○|￣|_ ） ）
感觉用map比较好处理……就懒懒の用了map……2333333
1 #include<iostream>
2 #include<algorithm>
3 #include<string>
4 #include<string.h>
5 #include<map>
6 using namespace std;
7 int n, k;
8 map<char, int>p;
9 char s;
10 int main()
11 {
12 while (cin >> n >> k)
13 {
14 p.clear();
15 for (int i = 0; i < n; i++)
16 {
17 cin >> s;
18 p[s]++;
19 }
20 map<char,int>::iterator ite;
21 int maxx = 0;
22 for (ite = p.begin(); ite != p.end(); ite++)
23 {
24 if (ite->second > maxx)
25 maxx = ite->second;
26 }
27 if (maxx > k) cout << "NO" << endl;
28 else cout << "YES" << endl;
29 }
30 return 0;
31 }