zoj2388----------------Beat the Spread!_uperbowl sunday is nearly here. in order to pass t-优快云博客

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在等待超级碗中场休息的广告和意外事故时，当地黑客组织了一个赌博池。参与者对两个最终分数的和或绝对差进行下注。输入给出每种赌注的获胜数字，需要求解可能的最终比分。如果不存在这样的比分，则输出'不可能'。题目给出了样例输入和输出，并指出了出现'不可能'的情况。

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Superbowl Sunday is nearly here. In order to pass the time waiting for the half-time commercials and wardrobe malfunctions, the local hackers have organized a betting pool on the game. Members place their bets on the sum of the two final scores, or on the absolute difference between the two scores.

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

Input

The first line of input contains n, the number of test cases. n lines follow, each representing a test case. Each test case gives s and d, non-negative integers representing the sum and (absolute) difference between the two final scores.

Output

For each test case, output a line giving the two final scores, largest first. If there are no such scores, output a line containing "impossible". Recall that football scores are always non-negative integers.

Sample Input

2
40 20
20 40

Sample Output

30 10
impossible

考虑三种情况是IMPOSSIBLE

1、和比差还小

2、和与差的和不能被2 整除

3、和与差的差不能被2整除

代码：

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		int s,d;
		cin>>s>>d;
		int x=(s+d)/2;
		double y=abs(double((s-d)/2));
		if(s-d<0||(s+d)%2!=0||(s-d)%2!=0)
			cout<<"impossible"<<endl;
		else
			cout<<x<<" "<<y<<endl;		
	}
	return 0;	
}