傳送門

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

題目

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 1.1
Sample Output
T T W 37.98

分析

方法比較傻，可以寫個循環搞定的，原諒我懶得改了，每次取最大賠率，然后將其所在的比賽狀況放入表示結果的字符串中，并且按公式計算
最大利潤，按照格式輸出結果即可。

源代碼

//C/C++實現
#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;
}