[PAT A1011]World Cup Betting

[PAT A1011]World Cup Betting

题目描述

1011 World Cup Betting （20 分）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).

输入格式

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.

输出格式

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.

输入样例

1.1 2.5 1.7
1.2 3.1 1.6
4.1 1.2 1.1

输出样例

T T W 39.31

解析

1. 就是一道简单题，使用的方法是简单模拟
2. 题目解读：意思是给出三场比赛的赔率，开始输入三行数，每行代表一场比赛，每一行的三个数字分别代表了赢，平和输，然后我们按照公式（第一行大的数x第二行最大的数x第三行最大的数x0.65-1）x2得到最终的结果

#include
#include
using namespace std;
int main()
{
 double sum = 0, temp;
 char bet[3];   //存放的是赌的胜负
 double odd[3];  //存放的是每一行最大的数
 double w, t, l;
 for (int i = 0; i < 3; i++)
 {
  cin >> w >> t >> l;
  bet[i] = (w > t) ? (t > l ? 'W' : (w > l ? 'W' : 'L')) : (w > l ? 'T' : (t > l ? 'T' : 'L'));    //计算出bet[i]
  odd[i] = max(max(w, l), t);
 }
 sum = (odd[0] * odd[1] * odd[2] * 0.65 - 1) * 2;
 cout << bet[0] << " " << bet[1] << " " << bet[2] << " ";
 printf("%.2f", sum); //注意格式
 return 0;
}

水平有限，如果代码有任何问题或者有不明白的地方，欢迎在留言区评论；也欢迎各位大牛提出宝贵的意见！