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?
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.
Print a single integer – the answer to the problem.
15 10 DZFDFZDFDDDDDDF
82
6 4 YJSNPI
4
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.
题意要拿k张卡片
每张卡片可以得到同种卡片总张数的钱
问这个收益的最大值
那么这个问题只需要简单的贪心就可以了
当卡片数目都小于k时优先拿卡片数量最多的即可
注意下数据范围
#include <iostream>
#include <stdio.h>
using namespace std;
const int M=1e5+5;
char s[M];
int ns[26];
int main()
{
int n,k;
long long mo=0;
scanf("%d%d%s",&n,&k,s);
for(int i=0;i<n;i++)
ns[s[i]-'A']++;
while(k)
{
int maxn=0;int p;
for(int i=0;i<26;i++)
{
if(ns[i]>k)
{
//cout<<k;
mo+=(long long)k*k;k=0;
break;
}
if(ns[i]>maxn)
{
p=i;
maxn=ns[i];
}
}
if(k)
{
k-=maxn;
ns[p]=0;
mo+=(long long)maxn*maxn;
}
}
printf("%I64d\n",mo);
return 0;
}