PAT (Advanced Level) Practice 1011

1011 World Cup Betting (20分)

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

									   题意分析

原文大致意思，每行输入有三个数字，找出每行的最大数，并输出其对应的字符编号，并通过每行的最大数计算最大盈利。最大盈利计算公式为**profits = [(max1max2max3)0.65-1]2.
很多人可能觉得这题采用数组保存中间结果再进行操作比较直观，实际上对于PTA这种单组测试的OJ，我们可以取巧采用如下做法。

#include 
int main()
{
    char game[3] = {'W','T','L'};
    double bet,max,res = 1.0;
    int pos;
    
    for(int i = 0;i < 3;i++)
    {
        max = 0.0;
        for(int j = 0;j < 3;j++)
        {
            scanf("%lf", &bet);
            if(bet > max)
            {
                max = bet;
                pos = j;
            }
        }
        printf("%c ", game[pos]);
        res *= max;
    }
    printf("%.2f", (res*0.65-1)*2);
    
    return 0;
}

思路即是每输入一组数据就输出结果max profits最后输出。