Codeforces Round #263 (Div. 2)B. Appleman and Card Game

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本文介绍了一个涉及卡片选择的游戏问题，目标是在限定条件下选择特定数量的卡片以获得最大收益。通过贪心算法解决此问题，并给出具体实现代码。

Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.

Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 105). The next line contains n uppercase letters without spaces — the i-th letter describes the i-th card of the Appleman.

Output

Print a single integer – the answer to the problem.

Sample test(s)

input
15 10
DZFDFZDFDDDDDDF

output
82

input
6 4
YJSNPI

output
4

Note

In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.

解题：贪心

代码如下：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#define LL long long
using namespace std;

int letter[27];
char str[100100];

bool cmp(const int &x,const int &y)
{
    return x > y;
}

int main()
{
    int n,k;
    LL ans;
    while(~scanf("%d %d",&n,&k))
    {
        memset(letter,0,sizeof(letter));
        scanf("%s",str);
        for(int i = 0; str[i]; i++) letter[str[i]-'A']++;
        sort(letter,letter+26,cmp);
        for(int i = ans = 0; k && i < 26; i++)
        {
            ans += 1LL*min(k,letter[i])*min(k,letter[i]);
            k -= min(k,letter[i]);
        }
        cout<<ans<<endl;
    }
    return 0;
}