zoukankan      html  css  js  c++  java
  • PAT Advanced 1011 World Cup Betting (20分)

    With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excited as the best players from the best teams doing battles for the World Cup trophy in South Africa. Similarly, football betting fans were putting their money where their mouths were, by laying all manner of World Cup bets.

    Chinese Football Lottery provided a "Triple Winning" game. The rule of winning was simple: first select any three of the games. Then for each selected game, bet on one of the three possible results -- namely W for win, T for tie, and L for lose. There was an odd assigned to each result. The winner's odd would be the product of the three odds times 65%.

    For example, 3 games' odds are given as the following:

     W    T    L
    1.1  2.5  1.7
    1.2  3.1  1.6
    4.1  1.2  1.1
    
     

    To obtain the maximum profit, one must buy W for the 3rd game, T for the 2nd game, and T for the 1st game. If each bet takes 2 yuans, then the maximum profit would be ( yuans (accurate up to 2 decimal places).

    Input Specification:

    Each input file contains one test case. Each case contains the betting information of 3 games. Each game occupies a line with three distinct odds corresponding to WT and L.

    Output Specification:

    For each test case, print in one line the best bet of each game, and the maximum profit accurate up to 2 decimal places. The characters and the number must be separated by one space.

    Sample Input:

    1.1 2.5 1.7
    1.2 3.1 1.6
    4.1 1.2 1.1
    
     

    Sample Output:

    T T W 39.31

    这题考察,选择每组最大值,然后利用公式求值。属于简单题

    #include <iostream>
    #include <algorithm>
    using namespace std;
    int main(){
        double m_num[3];
        double w, t, l, m;
        char m_char[3];
        for(int i = 0; i < 3; i++){
            scanf("%lf%lf%lf", &w, &t, &l);
            m = max(max(w, t), l);
            if(m == w){
                m_char[i] = 'W';
                m_num[i] = m;
            }else if(m == t){
                m_char[i] = 'T';
                m_num[i] = m;
            }else {
                m_char[i] = 'L';
                m_num[i] = m;
            }
        }
        double res = (m_num[0] * m_num[1] * m_num[2] * 0.65 - 1) * 2;
        printf("%c %c %c %.2f", m_char[0], m_char[1], m_char[2], res);
        return 0;
    }
  • 相关阅读:
    XML文件操作指南
    数字判断和文本框提交事件,WEB SERVICE等
    【转】JS回车提交
    nchar,char,varchar与nvarchar区别
    .NET对象的序列化
    这是发布的第一篇随笔
    Eclipse!!!!!!!!!!!!!!!!!!
    关于JLink驱动升级后不能使用的解决方法
    Python基本问题笔记
    用Python写的学校图书馆视频下载工具
  • 原文地址:https://www.cnblogs.com/littlepage/p/12234671.html
Copyright © 2011-2022 走看看