[PAT]1011. World Cup Betting (20)@Java解题报告

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    L1.1  2.5  1.71.2  3.0  1.64.1  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*3.0*2.5*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 1.71.2 3.0 1.64.1 1.2 1.1

Sample Output

T T W 37.98

易错点：不能申请一个double sum=1；然后在循环中用sum去乘以每行的最大值，这样会损失精度。

package go.jacob.day829;import java.text.DecimalFormat;import java.util.Scanner;public class Demo1 {public static void main(String[] args) {Scanner in = new Scanner(System.in);float[] max = new float[3];String[] num = new String[3];String[] odd = { "W", "T", "L" };int tmp = 0;for (int i = 0; i < 3; ++i) {for (int j = 0; j < 3; ++j) {float now = in.nextFloat();if (max[i] < now) {max[i] = now;tmp = j;}}num[i] = odd[tmp];}in.close();//只进行一次double运算，如果是进行多次，结果精度会有问题double out = (max[0] * max[1] * max[2] * (0.65) - 1) * 2;// 表示成只有两位小数DecimalFormat f = new DecimalFormat("0.00");System.out.print(num[0] + " " + num[1] + " " + num[2] + " " + f.format(out));}}