Codeforces Round #263 (Div. 2) proB

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本文介绍了一个涉及卡片选择的游戏策略问题，玩家需要从一系列带有字母的卡片中选择特定数量的卡片来获得最多的分数。文章提供了问题描述、输入输出格式、样例解析及实现代码，采用贪心算法对字符频率进行排序并计算最大得分。

题目：

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.

题意分析：

n张卡片，可以选取k张。可以得的分数是卡数的平分，求最大分数。

贪心吧，对字符进行记录，然后按个数排序。注意使用long long cha得残暴又是万恶的 long long

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char s[110000];
int j[500],cn[500];
void solve()
{
    int n,k;
    long long ans=0;
    scanf("%d%d",&n,&k);
    scanf("%s",s);
    int len=strlen(s);
    for(int i=0; i<len; i++)
    {
        j[s[i]]++;
    }
    while(k)
    {
        k--;
        int w=0,Max=0;
        for(int i='A'; i<='Z'; i++)
        {
            if(j[i]>Max&&cn[i]<j[i])
            {
                Max=j[i];
                w=i;
            }
        }
        cn[w]++;
    }
    for(int i='A'; i<='Z'; i++)
    {
        ans+=1LL*cn[i]*cn[i];
    }
    cout<<ans<<endl;
}
int main()
{
    solve();
    return 0;
}