码迷,mamicode.com
首页 > 其他好文 > 详细

NUC_TeamTEST

时间:2014-10-26 19:37:44      阅读:154      评论:0      收藏:0      [点我收藏+]

标签:des   style   blog   http   io   os   ar   for   strong   

B - B
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
Submit Status

Description

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 Input

Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4

Hint

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.

 

 中间超int啊,不难的贪心。CodeForces,DIV2的A题
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int max_size = 100000+10;

int main()
{
    int n, k;
    char str[max_size];
    long long  arry[26];
    while(scanf("%d %d", &n, &k) == 2)
    {
        getchar();
        memset(arry, 0, sizeof(arry));
        scanf("%s", str);
        int len = strlen(str);
        int i;
        for(i = 0; i < len; ++i)
        {
            arry[str[i]-‘A‘]++;
        }
        i = 25;
        sort(arry, arry+26);
        //printf("%d\n", arry[25]);
        long long sum = 0;
        long long now = k;
        do{
            if(now > arry[i])
            {
                sum += arry[i] * arry[i];
                now -= arry[i];
                //printf("%d\n", sum);
                i--;
            }else if(now <= arry[i]){
                sum += now*now;          ///这个比较坑,没考虑到超int
                break;
            }
        }while(now != 0);
        cout << sum << endl;
    }
    return 0;
}

  

NUC_TeamTEST

标签:des   style   blog   http   io   os   ar   for   strong   

原文地址:http://www.cnblogs.com/ya-cpp/p/4052489.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!