PAT (Advanced Level)-1011. World Cup Betting (20)

PAT (Advanced Level)-1011. World Cup Betting (20)

1011. World Cup Betting (20)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue


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).

Input

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

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 3.0 1.6
4.1 1.2 1.1

Sample Output

T T W 37.98

解读

这道题目其实就是考察比较大小,找出最大值并记录的问题。
题目中有3×3的数据,我们在每一行找到最大的,并记录所在位置(标识WTL)和数值(运算)。 故而,在下面的代码中,我使用了signNum数组来记录三行中最大值在其所在行中的位置。

完整代码

#include<iostream>
#include<iomanip>

using namespace std;

int main()
{
	double odd[3][3];           //记录原始输入数据

	for (int i = 0; i < 3; i++)
		for (int j = 0; j < 3; j++)
			cin >> odd[i][j];

	char result[3];         //存储输出WTL
	int signNum[3] = { 0,0,0 };         //标记

	for (int i = 0; i < 3; i++)
	{
		double temp = odd[i][0];            //大小比较中的临时量

		for (int j = 1; j < 3; j++)
		{
			if (temp <= odd[i][j])
			{
				temp = odd[i][j];
				signNum[i] = j;
			}
		}	
		if (signNum[i] == 0)
		{
			result[i] = 'W';
		}
		else if (signNum[i] == 1)
		{
			result[i] = 'T';
		}
		else if (signNum[i] == 2)
		{
			result[i] = 'L';
		}
	}

    //计算
	double res = (odd[0][signNum[0]] * odd[1][signNum[1]] * odd[2][signNum[2]] * 0.65 - 1.00) * 2.00;
	
	for (int i = 0; i < 3; i++)
		cout << result[i] << " ";
	cout << fixed << setprecision(2) << res;//这一块我们回到1002题目中,如何限定小数精度并四舍五入。
		

	return 0;
}

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