1011 World Cup Betting (20 point(s))

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本文介绍了一种在2010年世界杯期间，通过中国足彩提供的“TripleWinning”游戏进行投注的策略。该策略的目标是通过选择三场比赛并分别投注胜、平、负三种可能结果中的一种来最大化利润。每场比赛有三个不同的赔率，获胜者的赔率是三个选定赔率乘积的65%。示例显示了如何通过精确选择获得最大利润。

1011 World Cup Betting (20 point(s))

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：简单。

#include<iostream>
using namespace std;
int maxOfArray(double a[],int n){
	int index = 0;
	for(int i=0;i<n;i++){
		if(a[i]>a[index]) index=i;
	}
	return index;
} 
void printRes(int index){
	if(index==0) printf("W ");
	else if(index==1) printf("T ");
	else printf("L ");
}
int main(void){
	double res=1.0;
	double arr[3];
	for(int i=0;i<3;i++){
		for(int j=0;j<3;j++){
			scanf("%lf",&arr[j]);
		}
		int idx=maxOfArray(arr,3);
		printRes(idx);
		res=res*arr[idx];
	}
	printf("%.2f\n",(res*0.65-1)*2.0);
	return 0;
}