Code Forces 462B Appleman and Card Game

D - D
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
SubmitStatusPracticeCodeForces
462B

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.

贪心水题。

题目大概意思是说，给一个大写字母的序列，每个字母的得分是其出现的次数。现在要在这n个字符里找出k个字符，使得得分最大。求最大值。

要找最大，就要贪心。排序后从最大值开始找。

注意要用long long.

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 100005
using namespace std;
int n;
long long k;
char s
;
long long letter[26];
int main()
{
memset(letter,0,sizeof(letter));
scanf("%d%I64d",&n,&k);
scanf("%s",s);
for(int i=0;i<n;i++)
letter[s[i]-'A']++;
sort(letter,letter+26);
long long ans=0;
for(int i=25;i>=0;i--)
{
if(letter[i]<=k)
{
ans+=letter[i]*letter[i];
k-=letter[i];
}
else
{
ans=ans+k*k;
break;
}
}
printf("%I64d\n",ans);
return 0;
}