UVa 10812 - Beat the Spread!

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?
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. 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

Output for Sample Input

30 10
impossible

Analysis

This problem is about getting a and b with (a+b) and (a-b). It looks simple but there are tricky parts: Given that a and b are non-negative integers, it is necessary to check whether the given input
is possible. Since a can be calculated by [(a+b)+(a-b)]/2, (a+b)+(a-b) must be divisible by 2(to give an integer a). Since a and b are non-negative, the sum must be greater than their absolute difference.

#include <iostream>
using namespace std;

int main()
{
int n; // number of test cases
int s, d; // sum and difference
int a, b;

cin >> n;
for( int i=0; i<n; i++ )
{
cin >> s >> d;
if( (s+d)%2 == 0 && s>=d ) // trap: check condition!
cout << (s+d)/2 << ' ' << (s-d)/2 << endl;
else
cout << "impossible" << endl;
}
return 0;
}