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:
    leetcode| Intersection of Two Arrays II
    Spring Boot起步依赖:定制starter
    SpringBoot自动配置的魔法是怎么实现的
    Dubbo中的IoC实现
    必须知道的String知识点
    Dubbo的SPI机制
    为什么要设置HTTP timeout?
    重构代码——简单工厂模式+模板方法模式
    计算机基础——位运算
    系统间HTTP调用代码封装
  • 原文地址:https://www.cnblogs.com/littlepage/p/12234671.html
Copyright © 2011-2022 走看看