PAT(甲) 1011 World Cup Betting (20)（20 分）

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本文介绍了一个关于2010年世界杯足球彩票的投注策略案例，通过选择三场比赛并针对每场赛事的胜、平、负进行投注，旨在获取最大的收益。文章提供了具体的例子和实现代码。

1011 World Cup Betting (20)（20 分）

题目描述：

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	3.0	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*3.0*2.5*65%-1)*2 = 37.98 yuans (accurate up to 2 decimal places).

输入格式：
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.
输出格式：
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.

解题方法：
这道题目用常规方法求解即可。其中只需要注意最终的收益为（MAX1*MAX2*MAX3*0.65-1）*2即可。

程序：

#include <stdio.h>
#include <stdlib.h>

int main()
{
    double sum = 1, bet[3][3], Max[3] = {0, 0, 0};
    int MaxIdx[3] = {0, 0, 0};
    char flag[3] = {'W', 'T', 'L'};
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            scanf("%lf", &bet[i][j]);
            if (bet[i][j] > Max[i])
            {
                Max[i] = bet[i][j];     /* 保存最大赔率下标 */  
                MaxIdx[i] = j;
            }
        }
        sum *= Max[i];
        if (i != 0)
            printf(" ");
        printf("%c", flag[MaxIdx[i]]);
    }
    printf(" %.2lf\n", (sum*0.65-1)*2);
}