District 9

Problem D: 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

解題思考:
根據題易比賽只有勝負，可以不考慮平手情況，因此勝方的分數必定是大於敗方。
假設勝方分數為x，敗方分數為y。 ( x, y >= 0; x >= y 且各為整數)
令分數總和為: sum; 分數差為diff。
sum = x + y;
diff = x - y;

=> x = ( sum + diff ) / 2
=> y = ( sum - diff ) / 2
根據題意可知x, y >= 0 故sum >= diff
且x, y皆為正整數，故 兩者皆為偶數(可被2整除)

解題:

/*
* 2009.4.11
* UVa Online Judge
* Root :: Contest Volumes :: Volume CVIII 
* 10812
* Problem D: Beat the Spread!
*/

#include <cstdlib>
#include <iostream>

using namespace std;

int main()
{
    int case_no;
    
    while( ( cin >> case_no ) != 0 )
    {          
           for (int counter = 0; counter != case_no; counter++)
           {
               int sum, diff;
               cin >> sum >> diff;
               
               if( sum >= diff && (sum - diff) % 2 == 0)
                   cout << (sum + diff) / 2 << 
                   " " << (sum - diff) / 2 << endl;
               
               else
                   cout << "impossible" << endl;
           }
    }
               
    system("PAUSE");
    return EXIT_SUCCESS;
}