Codefores Round #263 (div2) B. Appleman and Card Game_【codeforces #263 div2 b】appleman and card game 苹果人-优快云博客

本文探讨了一个涉及卡片选择的策略游戏，玩家需要从一系列带有字母的卡片中选出特定数量的卡片来获得最多的分数。通过分析卡片上的字母分布并采用适当的算法进行优化选择，文章详细解释了如何实现分数最大化的方法。

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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.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
int a[30];
char s[100100];
using namespace std;
bool cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int n,k;
    long long ans;
    while(~scanf("%d%d",&n,&k))
    {
        memset(a,0,sizeof(a));
        scanf("%s",s);
        for(int i=0;i<n;i++)
        {
           a[s[i]-'A']++;
        }
        sort(a,a+30,cmp);
        ans=0;
       int t=0;
        while(1)
        {
            if(k>=a[t])
            {
                ans+=(long long)a[t]*a[t];
                k-=a[t];
                t++;
            }
            else
            {
                ans+=(long long)k*k;
                break;
            }
            if(t>=30)
            break;
        }
        cout<<ans<<endl;
    }
    return 0;
}