zoukankan      html  css  js  c++  java
  • 1011 World Cup Betting

    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

    题意:

    三场比赛找出每场比赛中收益最大的name和profit,profit按照所给的公式计算。(好像是这样的^_^)

    Code:

    #include<iostream>
    #include<algorithm>
    #include<iomanip>
    
    using namespace std;
    
    int main() {
        double profit = 1.0;
        char arr[3];
        for (int i = 0; i < 3; ++i) {
            char c = 'W';
            double w, t, l, n;
            cin >> w >> t >> l;
            profit *= max(max(w, t), l);
            if (t > w && t > l) c = 'T';
            if (l > w && l > t) c = 'L';
            arr[i] = c;
        }
    
        cout << arr[0];
        for (int i = 1; i < 3; ++i) {
            cout << " " << arr[i];
        }
        cout << " " << fixed << setprecision(2) << (profit * 0.65 - 1) * 2;
    
        return 0;
    }
    

      

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    webpack4笔录
    在Salesforce中进行Report和Dashboard的配置
    在Salesforce中以PDF的格式显示对应的页面
    在Salesforce中处理Email的发送
    在Visualforce page中用自带的控件实现Ajax回调后台方法(并且可以用js去动态给parameters赋值)
    javascript settimeout and setinterval
    Convert XML to Object using LINQ
    Convert Object to XML using LINQ
    JS对Array进行自定制排序
    在Salesforce中通过 Debug Log 方式 跟踪逻辑流程
  • 原文地址:https://www.cnblogs.com/h-hkai/p/12587195.html
Copyright © 2011-2022 走看看