PAT 1011 World Cup Betting

1011 World Cup Betting

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

 总结：一开始还以为这道题目很难，结果仔细的看懂了题目就发现还是很简单的，只需要按题目意思一步一步来就行了

代码：

#include 
using namespace std;
 
int main(){
    double sum=1,t;
    char c[3];
    char w[3]={'W','T','L'};
    
    for(int i=0;i<3;i++){
        double m=0;
        for(int j=0;j<3;j++){
            scanf("%lf",&t);
            if(m
                m=t;
                c[i]=w[j];
            }
        }
        sum=sum*m;
    }
    printf("%c %c %c",c[0],c[1],c[2]);
    printf(" %.2lf\n",(sum*0.65-1)*2);
    
    return 0;
}

依照惯例，看一眼大佬的代码

#include 
using namespace std;
int main() {
    char c[4] = {"WTL"};
    double ans = 1.0;
    for(int i = 0; i < 3; i++) {
        double maxvalue = 0.0;
        int maxchar = 0;
        for(int j = 0; j < 3; j++) {
            double temp;
            scanf("%lf", &temp);
            if(maxvalue <= temp) {
                maxvalue = temp;
                maxchar = j;
            }
        }
        ans *= maxvalue;
        printf("%c ", c[maxchar]);
    }
    printf("%.2f", (ans * 0.65 - 1) * 2);
    return 0;
}

好好学习，天天向上！

我要考研！