[PAT-甲级]1011.World Cup Betting

1011. World Cup Betting (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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).

Input

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

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 3.0 1.6
4.1 1.2 1.1

Sample Output

T T W 37.98

解题思路：输入为一个3*3大小的矩阵，找到每一行中的最大值，然后代入公式(第一行最大值*第二行最大值*第三行最大值*0.65-1)*2，结果保留2位小数，注意：最终结果需要进行四舍五入

同时在输出的时候需要打印每一行最大值位于第几列，第一列则打印W，第二列则打印W，第三列则打印L

代码如下：

//1011. World Cup Betting
//备注：此题在ac的时候你发现答案是37.97，如果我们把注释那行不注释掉，要求你进行四舍五入
//得到37.98,但是你提交代码会发现是错误答案。如果注释那行的后提交代码是正确答案 
//正确答案给的是37.98。所采用的编译器是DEV C++ 

#include

int main()
{
	float buf[3][3];
	float sum = 1.0, tmp;
	for(int i = 0; i < 3; i ++)
		for(int j = 0; j < 3; j ++)
			scanf("%f", &buf[i][j]);
	for(int i = 0; i < 3; i ++)
	{
		int j = 0;
		if(buf[i][j] >= buf[i][j+1] && buf[i][j] >= buf[i][j+2])
		{
			printf("W ");
			tmp = buf[i][j];
		}
		else if(buf[i][j+1] >= buf[i][j] && buf[i][j+1] >= buf[i][j+2])
		{	
			printf("T ");
			tmp = buf[i][j+1];
		}
		else
		{
			printf("L ");
			tmp = buf[i][j+2];
		}
		
		sum = sum * tmp;		
	}
	sum = (sum*0.65-1) * 2;
	//sum = (sum * 100 + 0.5) / 100; 
	printf("%.2f\n", sum);
	return 0;
}