PAT Advanced 1011. World Cup Betting (20) (C语言实现)

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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  3.1  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\times 3.1\times 2.5\times 65\%-1)\times 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

思路

比较简单的题,写了个不用数组的代码。

代码

最新代码@github,欢迎交流

#include <stdio.h>

int main()
{
    char c;
    float odd, maxodd, maxprofit = 1;

    for(int i = 0; i < 3; i++)
    {
        maxodd = 0;
        for(int j = 0; j < 3; j++)
        {
            scanf("%f", &odd);
            if(maxodd < odd)
            {
                maxodd = odd;
                c = "WTL"[j];
            }
        }
        printf("%c ", c);
        maxprofit *= maxodd;
    }
    printf("%.2f", (maxprofit * 0.65 - 1) * 2);

    return 0;
}
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