# PAT-A 1011. World Cup Betting (20)

# 传送门

https://www.patest.cn/contests/pat-a-practise/1011

# 题目

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 1.1Sample Output

T T W 37.98

# 分析

方法比较傻，可以写个循环搞定的，原谅我懒得改了，每次取最大赔率，然后将其所在的比赛状况放入表示结果的字符串中，并且按公式计算

最大利润，按照格式输出结果即可。

# 源代码

```
//C/C++实现
#include <iostream>
#include <string>
using namespace std;
int main(){
string result = "";
float a, b, c;
float first, second, third;
scanf("%f %f %f", &a, &b, &c);
if(a > b && a > c){
result += "W ";
first = a;
}
else if(b > a && b > c){
result += "T ";
first = b;
}
else if(c > a && c > b){
result += "L ";
first = c;
}
scanf("%f %f %f", &a, &b, &c);
if(a > b && a > c){
result += "W ";
second = a;
}
else if(b > a && b > c){
result += "T ";
second = b;
}
else if(c > a && c > b){
result += "L ";
second = c;
}
scanf("%f %f %f", &a, &b, &c);
if(a > b && a > c){
result += "W ";
third = a;
}
else if(b > a && b > c){
result += "T ";
third = b;
}
else if(c > a && c > b){
result += "L ";
third = c;
}
cout << result;
printf("%.2f", (first * second * third * 0.65 - 1) * 2);
return 0;
}
```