1011 World Cup Betting （20 分）

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本文介绍了一种基于2010年FIFA世界杯期间中国足球彩票TripleWinning游戏的策略。通过分析每场比赛的赔率，选择每场可能获得最高回报的结果，最终计算出最大利润。示例中，通过对三场比赛的赔率分析，选择了特定的比赛结果，最终计算出的最高利润为39.31元。

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：这道题的要求其实很简单，只要我们保证三场比赛每一场我们都压盈利最高的结果就好，比如：

第一场：押 W 可以获利 1.1，押 T 可以获利 2.5，押 L 可以获利 1.7，故我们选 2.5 即 T，此时直接输出；

第二场：押 W 可以获利 1.2，押 T 可以获利 3.1，押 L 可以获利 1.6，故我们选 3.1 即 T，此时直接输出；

第三场：押 W 可以获利 4.1，押 T 可以获利 1.2，押 L 可以获利 1.1，故我们选 4.1 即 W，此时直接输出；

之后按照题目要求进行结算：(4.1×3.1×2.5×65%−1)×2=39.31 得到结果。

代码如下：

#include<bits/stdc++.h>
using namespace std;
int main() 
{
	char S[3] = { 'W','T','L' };
	double Res[3][3],sum = 1;
	for (int i = 0; i < 3; i++)
	{
		double Max = 0;
		int temp;
		for (int j = 0; j < 3; j++)
		{
			cin >> Res[i][j];
			if (Max < Res[i][j])
			{
				Max = Res[i][j];
				temp = j;
			}
		}
		sum *= Max;
		cout << S[temp] << " ";
	}
	printf("%.2f", (sum * 0.65 - 1) * 2);
	return 0;
}