PAT Advanced 1011 World Cup Betting (20 分)

PAT Advanced 1011 World Cup Betting (20 分)
强烈推介IDEA2021.1.3破解激活,IntelliJ IDEA 注册码,2021.1.3IDEA 激活码  

大家好,我是架构君,一个会写代码吟诗的架构师。今天说一说PAT Advanced 1011 World Cup Betting (20 分),希望能够帮助大家进步!!!

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 ( 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 WT 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


#include <iostream>
using namespace std;
int main(){
    int a=3;double res=1.0,W,T,L;
    while(a--){
        cin>>W>>T>>L;
        if(W>T&&W>L) {
            cout<<"W ";
            res*=W;
        }
        if(T>W&&T>L) {
            cout<<"T ";
            res*=T;
        }
        if(L>W&&L>T) {
            cout<<"L ";
            res*=L;
        }
    }
    res=(res*0.65-1)*2;
    printf("%.2f",res);
    system("pause");
    return 0;
}

 

转载于:https://www.cnblogs.com/littlepage/p/11286354.html

本文来源weixin_30595035,由架构君转载发布,观点不代表Java架构师必看的立场,转载请标明来源出处:https://javajgs.com/archives/29561

发表评论