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 -- namelyWfor win,Tfor tie, andLfor 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 buyWfor the 3rd game,Tfor the 2nd game, andTfor the 1st game. If each bet takes 2 yuans, then the maximum profit would be(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 toW,TandL.

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：簡單的找最大值問題，注意題目中的長難句

#include<bits/stdc++.h>
using namespace std;
const int maxn=10010;
struct bet{
    double num[maxn];
};
int main(){
    bet B;
    double sum=1;
    string s="WTL";
    string st;
    for(int i=0;i<3;i++){
        double maxOdd=0;
        int ins=-1;
        for(int j=0;j<3;j++){
            scanf("%lf",&B.num[j]);
            if(B.num[j]>maxOdd){
                maxOdd=B.num[j];
                ins=j;
            }
        }
            sum*=maxOdd;
            st+=s[ins];
    }
    for(int i=0;i<st.size();i++){
        cout<<st[i]<<" ";
    }
    printf("%.2f\n",(sum*0.65-1)*2);
    return 0;
}