Beat the Spread!

垮掉的传播

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Superbowl Sunday is nearly here. In order to pass the time waiting for the half-time commercials and wardrobe malfunctions,

美国橄榄球超级杯大赛的星期天突然来临了,为了消磨时间等待下半场广告和衣柜故障,

the local hackers have organized a betting pool on the game. Members place their bets on the sum of the two final scores,

当地的黑客组织打赌游戏。人们不是押了两个决赛分数的和（求s+d）,

or on the absolute difference between the two scores.

就是押了两数之差的绝对值（求| s-d |）。

Given the winning numbers for each type of bet, can you deduce the final scores?

对于每一个赌注给一个赢得号码,你能推断出决赛分数吗?（用s和d求出决赛分数）

Input

The first line of input contains n, the number of test cases. n lines follow, each representing a test case.

输入的第一行是n,表示测试的个数。接下来是n行，每一行代表一个测试事件。

Each test case gives s and d, non-negative integers representing the sum and (absolute) difference between the two final scores.

对于每一个测试用例，给s和d。和是非负整数，差的绝对值。

Output

For each test case, output a line giving the two final scores, largest first. If there are no such scores,

对于每一个测试用例，在一行输出两个决赛分数，大的数在第一个。如果没有这样的分数,

output a line containing "impossible". Recall that football scores are always non-negative integers.

在单独的一行输出"impossible。（回想一下）记住,足球成绩总是非负整数  {(s+d)/2是整数}。

Sample Input

2
40 20   s   d
20 40

Sample Output

30 10     非负整数（ s>d ），**不能是小数，也不能是负数，只能是0和正整数

目的：提取公因式，让公因式做判断语句，使代码看起来有很高的水平

因为答案不能为负数和小数，所以我们要排除这些答案

正真输出的只有0和正整数***

(s+d)/2     -----    (s-d)/2 

（ s+d  + (s-d) -  (s-d) ）/2 -------    (s-d)/2

d + (s-d)/2  ------    (s-d)/2   

目的完成：  公因式 (s-d)/2   ------ 

{ 改进：(s-d)&0x01  让 （s-d）与16进制1进行 异或 } 例：0011 0001 & 0000 0001 = 0000 0001

如果异或等于1 代表(s-d)是奇数,那么结果就是小数

impossible   s-d<0

Source

University of Waterloo Local Contest 2005.02.05

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		while (n-- > 0) {
			int s = sc.nextInt();
			int d = sc.nextInt();
			int temp = s - d;
			if (temp < 0 || (temp & 0x01) == 1) {
				System.out.println("impossible");
			} else {
				System.out.println((d + temp / 2) + " " + temp / 2);
			}
		}
	}

}