PAT甲级(Advanced Level)练习题 World Cup Betting (20)

题目描述

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 3.0 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 (4.13.02.5*65%-1)*2 = 37.98 yuans (accurate up
to 2 decimal places).

输入描述:

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.

输出描述:

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.

输入例子:

1.1 2.5 1.7
1.2 3.0 1.6
4.1 1.2 1.1

输出例子:

T T W 37.98

题目分析

水题，唯一的困难是最后保留位数，在结果上加0.000001就非常灵性了。

代码实现（C++版本）

#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
    double input[3][3];
    double maxcol[3];
    for(int i =0;i<3;i++){
        for(int j =0;j<3;j++){
            cin >> input[i][j];
        }
    }

    for(int i =0;i<3;i++){
        double max1 = input[i][0];
        maxcol[i]=0;
        for(int j =0;j<3;j++){
            if(input[i][j]>max1){
                maxcol[i] = j;
                max1 = input[i][j];
            }
        }

    }
    double ans = 1.0;
    for(int i =0;i<3;i++){
        if(maxcol[i]==0){
            cout<< "W ";
            ans *= input[i][0];
        }
        else if(maxcol[i]==1){
            cout<<"T ";
            ans *= input[i][1];
        }
        else if(maxcol[i]==2){
            cout<<"L ";
            ans *= input[i][2];
        }
    }
    ans *= 0.65;
    ans -=1;
    ans *=2;
    ans+= 0.000001;
    cout<<setiosflags(ios::fixed)<<setprecision(2)<<ans<<endl;
    return 0;
}