PAT甲级—— 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 (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

 思路：

这道题要求的其实就是三注彩票中能得到最大收益的方案。每输入一组数据就找出其中的最大值，然后输出对应的字符就行了。下面代码，要注意求完每组的max值之后要重新初始化为0，否则后面组的数据的最大值如果没有超过max就会出错。

代码：

#include 
using namespace std;

int main()
{
	double data[3] = { 0 }, result[3] = { 0 }; char ch[3] = { 'W','T','L' };
	int index[3] = { 0 };
	double ans = 0.0, max = 0.0;
	for (int i = 0; i < 3; i++)
	{
		for (int j = 0; j < 3; j++)
		{
			scanf("%lf", &data[j]);
			if (data[j] > max)
			{
				max = data[j];
				index[i] = j;
			}
		}
		result[i] = max;
        max=0;//这里一定要重置为0.否则后面的结果就不对了，因为max会保持最大的值
		
	}
	for (int i = 0; i < 3; i++)
	{
		printf("%c ", ch[index[i]]);
		//printf_s("%f ", result[i]);
	}
	ans = (result[0] * result[1] * result[2] * 0.65 - 1) * 2;
	printf("%.2f", ans);
	return 0;

}