【PAT甲级】World Cup Betting

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本文链接：https://blog.csdn.net/weixin_42449444/article/details/89714801

Problem Description：

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 (4.1×3.1×2.5×65%−1)×2=39.31 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 W, T 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

解题思路：

这是一道水题啊，甲级里面真的没有比这题更简单的啦。设最大利润为sum，利用max()来无脑找出胜负平的最高赔率ans，然后输出三局比赛最高赔率对应的彩果以及最大利润sum即可。

AC代码：

#include <bits/stdc++.h>
using namespace std;

int main()
{
    double sum = 1;   //最大利润sum
    for(int i = 0; i < 3; i++)
    {
        double W,T,L;   //胜W,平T,负L的赔率
        cin >> W >> T >> L;
        double ans = max(max(W,T),L);   //取最高赔率
        if(ans == W)
        {
            cout << "W ";
        }
        else if(ans == T)
        {
            cout << "T ";
        }
        else
        {
            cout << "L ";
        }
        sum *= ans;
    }
    sum = (sum*0.65-1)*2;
    printf("%.2f\n",sum);
    return 0;
}