zoukankan      html  css  js  c++  java
  • NUC_TeamTEST_B(贪心)

    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
    ", 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
    ", 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;
    }
    

      

  • 相关阅读:
    【原创】构建高性能ASP.NET站点之一 剖析页面的处理过程(前端)
    .NET 并行(多核)编程系列之七 共享数据问题和解决概述
    架构设计解惑
    项目开发经验谈之:设计失败的挫败感
    项目开发经验谈之:忆第一次设计Framework
    盲目的项目开发
    扩展GridView之添加单选列
    日期转换格式
    动手完善个性化弹出提示框的过程及乐趣
    SQL开发中容易忽视的一些小地方(六)
  • 原文地址:https://www.cnblogs.com/ya-cpp/p/4052489.html
Copyright © 2011-2022 走看看