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  • Codeforces Round #263 (Div. 2) proB

    题目:

    B. Appleman and Card Game
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    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?

    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,能够选取k张。

    能够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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  • 原文地址:https://www.cnblogs.com/yangykaifa/p/6801717.html
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