杭电oj1194（Beat the Spread!）

最新推荐文章于 2020-02-29 15:50:51 发布

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最新推荐文章于 2020-02-29 15:50:51 发布

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文章标签： oj

本文链接：https://blog.youkuaiyun.com/qq_35225958/article/details/86768156

本文提供了一道名为“BeattheSpread!”的编程题目解决方案，该题目来源于杭电在线评测系统（OJ）。文章详细解析了如何通过输入两队最终得分的总和与绝对差值，来推算出每队的具体得分。代码采用C++实现，通过简单的数学运算解决了这一问题，并确保了得分的非负整数特性。

杭电oj1194（Beat the Spread!）

Problem Description

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

答案

water题~
输入两个数的和与差，输出这两个数。只要注意答案不能是负数或者小数就行了。

#include <iostream>
#include <math.h>
#include <string.h>
#include <ctype.h>

//#define NDEBUG

#include <assert.h>

using namespace std;

int main() {
	int n, sum, diff, larger, smller;
	cin >> n;
	while (n--) {
		cin >> sum >> diff;

		((sum + diff) / 2 > (sum - diff) / 2) ?
			(larger = (sum + diff) / 2, smller = (sum - diff) / 2) :
			(smller = (sum + diff) / 2, larger = (sum - diff) / 2);

		if (larger < 0 || smller < 0 || (sum + diff) % 2 != 0)
			cout << "impossible" << endl;
		else
			cout << larger << " " << smller << endl;
	}
}