1011 World Cup Betting （20 分）

最新推荐文章于 2024-06-13 01:45:00 发布

galesaur_wcy

最新推荐文章于 2024-06-13 01:45:00 发布

阅读量196

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT甲级题

本文链接：https://blog.youkuaiyun.com/galesaur_wcy/article/details/83657103

PAT甲级题专栏收录该内容

145 篇文章

订阅专栏

随着2010年FIFA世界杯的进行，全球足球迷的热情高涨，投注爱好者们积极参与，通过投注比赛结果来获取收益。中国足彩提供了一种TripleWinning游戏，参与者需选择三场比赛并预测每场比赛的胜负平结果，赢取赔率的乘积再乘以65%的奖金。示例展示了如何通过选择特定的比赛结果来获得最大利润。

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).

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

水

代码

import java.util.Scanner;
import java.util.TreeMap;
import java.util.Comparator;
import java.util.Map;
public class Main {
	static Scanner sc = new Scanner(System.in);
	public static void main(String[] args) {
		double a[][] = new double[3][3];
		double res = 1;
		char gg[] = {'W','T','L'};
		char o[] = new char[3]; 
		for(int i = 0 ; i < 3; i ++)
		{
			for(int j = 0 ; j < 3; j++)
				a[i][j] = sc.nextDouble();
			double kk = Double.max(Double.max(a[i][0],a[i][1]),a[i][2]);
			res = res * kk;
			for(int j = 0; j < 3; j++)
			{
				if(a[i][j] == kk)
				{
					o[i] = gg[j];
				}
			}
		}
		res = (res * 0.65 - 1) * 2; 
		System.out.printf("%c %c %c %.2f\n",o[0],o[1],o[2],res);
	}
}