1011 World Cup Betting (20)（20 分）

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本文介绍了一个针对2010年南非世界杯足球比赛的投注策略案例，通过选择每场比赛可能的结果（胜、平、负）来实现最大化的收益。以一个具体的例子展示了如何计算投注回报，并提供了一段C++代码实现最佳投注方案的选择。

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

这题就是三行三列的输入，在输入的时候就可以就算乘的值了，顺便把从“WTL”中要输出的一个字母也输出了

这题的样例有问题应该是37.97

#include<bits/stdc++.h>
using namespace std;
double ans=1.0;
char s[5]="WTL";
int main()
{
    ios::sync_with_stdio(false);
    double x,maxn;
    int d;
    for(int i=0;i<3;i++){
        maxn=0.0;
        for(int j=0;j<3;j++){
            cin>>x;
            if(maxn<x){
                d=j;
                maxn=x;
            }
        }
        ans*=maxn;
        cout<<s[d]<<" ";

    }
    cout<<fixed;
    cout<<setprecision(2);
    cout<<(ans*0.65-1)*2<<endl;
    return 0;
}