UVa 10812 - Beat the Spread!

最新推荐文章于 2019-06-30 01:30:45 发布

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本文介绍了一个简单的算法，用于根据已知的最终比分之和及绝对差值来推断超级碗比赛的具体比分。通过检查输入的有效性和计算两个比分值，算法能够判断出可能的比赛结果或者输出‘impossible’表示无法得出合理结论。

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;
}