zoukankan      html  css  js  c++  java
  • hdu-1338 game predictions(贪心题)

    Suppose there are M people, including you, playing a special card game. At the beginning, each player receives N cards. The pip of a card is a positive integer which is at most N*M. And there are no two cards with the same pip. During a round, each player chooses one card to compare with others. The player whose card with the biggest pip wins the round, and then the next round begins. After N rounds, when all the cards of each player have been chosen, the player who has won the most rounds is the winner of the game. 
    Given your cards received at the beginning, write a program to tell the maximal number of rounds that you may at least win during the whole game. 

    InputThe input consists of several test cases. The first line of each case contains two integers m (2 <= m <= 20) and n (1 <= n <= 50), representing the number of players and the number of cards each player receives at the beginning of the game, respectively. This followed by a line with n positive integers, representing the pips of cards you received at the beginning. Then a blank line follows to separate the cases. 

    The input is terminated by a line with two zeros. 
    OutputFor each test case, output a line consisting of the test case number followed by the number of rounds you will at least win during the game. 
    Sample Input

    2 5
    1 7 2 10 9
    
    6 11
    62 63 54 66 65 61 57 56 50 53 48
    
    0 0

    Sample Output

    Case 1: 2
    Case 2: 4

    题意:模拟打牌,m个人,每人n张牌。点数1-n*m,出得最大的那个人赢一局,问你最多赢几局

    题解:贪心,我出大的,对面出最小的,我出小的,对面最小但比我大的
    
    
    
    
    
    
    #include<iostream>
    #include<algorithm>
    #include<cstring>
    #include<sstream>
    #include<cmath>
    #include<cstdlib>
    #include<queue>
    using namespace std;
    #define PI 3.14159265358979323846264338327950
    bool a[1050];
    int main()
    {
        int m=0,n=0,count=0,t;
        while(scanf("%d %d",&m,&n) && (m||n))
        {
            count++;
            int i=0;
            int win=0,big=0;
            memset(a,0,sizeof(a));
            for(i=0;i<n;i++)
            {
                scanf("%d",&t);
                a[t]=1;
            }
            for(i=n*m;i>0;i--)
            {
                if (a[i])
                {
                    if (big==0)
                        ++win;
                    else
                        big--;
                }
                else
                    big++;
            }
            printf("Case %d: %d
    ",count,win);
        }
        return 0;
    }
  • 相关阅读:
    django -- 信号
    django缓存设置
    django-debug-toolbar 插件的使用
    scrapy基本操作流程
    scrapy框架持久化存储
    scrapy基础
    phantomJS,谷歌无头浏览器, 模拟登陆qq空间
    python爬虫--selenium
    pytorch掉坑记录:model.eval的作用
    numpy常用函数
  • 原文地址:https://www.cnblogs.com/smallhester/p/9498878.html
Copyright © 2011-2022 走看看