PTA甲级-1011 World Cup Betting c++

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

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一、题干大意

给出三场比赛的胜、平、负的概率，要求输出每场球应该买哪种输赢并把最后可以获得的金额按公式
$$(\Pi 每场球最大的概率\ast 0.65-1)\ast 2$$
输出。

二、题解要点

• 每次输入三个数，循环三次，找到每组数中最大的那个数及其下标，然后对按下标输出输赢结果，并把概率乘进结果中。
• 最后有一个四舍五入的处理不要忘记。

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三、具体实现

/**
*@Author：hhzheng
*@Date：2023/2/12  21:46
*@Filename：PAT甲级1011 World Cup Betting
*/

#include

using namespace std;

int main() {
    char WTL[3] = {'W', 'T', 'L'};
    float odds[3] = {0.0}, biggest = -1.0,theResult = 1.0;
    int biggestIndex = -1;
    /*每次输入三个数，循环三次，找到每组数中最大的那个数及其下标，然后对按下标输出输赢结果，并把概率乘进结果中*/
    for (int i = 0; i < 3; ++i) {
        for (int j = 0; j < 3; ++j) {
            cin >> odds[j];
            if (odds[j] > biggest){
                biggest = odds[j];
                biggestIndex = j;
            }
        }
        cout<< WTL[biggestIndex] << " ";
        theResult *= biggest;
        biggest = -1.0;
        biggestIndex = -1;
    }
    theResult = (theResult *0.65 -1) *2;
    printf("%.2f\n",theResult + 0.005); //要进行四舍五入的处理
    return 0;
}

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总结

题目难度不大，需要好好读题。