一、題目

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=39.31yuans (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

二、題目大意

給定三個比賽的賠率，求可以獲取的最大收益。

三、考點

模擬

四、解題思路

1、題目比較簡單，直接比較大小就可以了。

2、考慮到題目的擴充套件性，使用陣列儲存；也可以聯機演算法，變讀取，變處理，減小一個迴圈，但笨方法，思路更加清晰，降低出錯的概率。

五、程式碼

#include<iostream>
using namespace std;

int main() {
	float game[3][3];
	char int2char[3] = { 'W','T','L' };
	//讀取資料
	for (int i = 0; i < 3; i++) {
		for (int j = 0; j < 3; j++) {
			cin >> game[i][j];
		}
	}

	//獲得每次比賽的最大值
	float ans_win = 1.0;
	for (int i = 0; i < 3; i++) {
		float ans_max = 0.0;
		int ans_id ;
		for (int j = 0; j < 3; j++) {
			if (game[i][j] > ans_max) {
				ans_max = game[i][j];
				ans_id = j;
			}
		}
		//輸出結果
		cout << int2char[ans_id] << " ";
		ans_win *= ans_max;
	}
	printf("%.2f", (ans_win*0.65 - 1) * 2);
	system("pause");
	return 0;
}