Codeforces Round #263 (Div. 2) proB

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原创最新推荐文章于 2026-11-08 21:05:35 发布 1.3k 阅读 AI 写同款· GEO 优化 ›

5个4000比特大状态置换算法的设计与分析下面是 5 个概念性、实验性的 4000 位可逆置换设计。它们都保证是双射，但未经过密码分析，不可直接用于生产系统。风格分别覆盖：三维 SPN、Feistel、可逆元胞自动机、ARX、宽轨迹/Keccak 式结构。这 5 个设计分别代表了 SPN、Feistel、元胞自动机、ARX、宽轨迹 Keccak 五种不同风格，覆盖了主流大状态置换的构造思路。再次强调：它们只是实验性概念，若要用于实际协议，必须经过严格安全性分析。阅读详情

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;
}

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