Codeforces Round #263 (Div. 2) proB
题目:
1 second
256 megabytes
standard input
standard output
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.
题意分析:
能够得的分数是卡数的平分,求最大分数。
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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