1011. World Cup Betting (20)

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本文介绍了一个世界杯足球赛投注的策略，通过选择三场比赛并针对每场比赛的胜、平、负三种可能结果进行投注来获取最大利润。示例展示了如何计算最佳投注组合及预期利润。

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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

分析：这道题的pro如果定义为float，结果是37.98。读入4.1，在监视窗看到的是4.09999，计算出的结果也不一样，很奇怪，希望有大神能解释下。

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int Findmax(float* ratio)
{
	int i;
	float max = 0;
	int index;
	for (i = 0; i < 3; i++)
	{
		if (ratio[i] > max)
		{
			max = ratio[i];
			index = i;
		}
	}
	return index;
}

int main()
{
	
	int index;
	int i, j;
	float ratio[3];
	double pro = 1;
	for (i = 0; i < 3; i++)
	{
		for (j = 0; j < 3; j++)
			scanf( "%f", &ratio[j]);
		
		
		index = Findmax(ratio);
		pro *= ratio[index];
		if (index == 0)
			printf("W ");
		else if (index == 1)
			printf("T ");
		else
			printf("L ");
		
	}
	pro = (pro*0.65 - 1) * 2.0;
	pro = floor(pro * 100 + 0.51)/100.00;
	printf("%.2lf", pro);
	
	return 0;
}