(hdu 简单题 128道)hdu 1194 Beat the Spread!(已知两个数的和u两个数的差求这两个数)

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最新推荐文章于 2019-06-16 11:06:05 发布

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分类专栏： acm ACM——夺金之路

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本文介绍了一个名为 BeattheSpread 的赌池问题，参与者需要预测比赛最终得分之和或得分差。通过给出的样例输入输出，文章详细解释了如何使用简单的数学方法来逆推两队的实际得分，并提供了完整的 C++ 实现代码。

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题目：

Beat the Spread!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5169 Accepted Submission(s): 2692

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

Sample Output

  
   30 10
impossible

Source

University of Waterloo Local Contest 2005.02.05

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题目分析：

已知a+b和a-b求a、b。

代码如下：

/*
 * b.cpp
 *
 *  Created on: 2015年3月12日
 *      Author: Administrator
 */

#include <iostream>
#include <cstdio>

using namespace std;


int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int sum;
		int diff;
		scanf("%d%d",&sum,&diff);

		if(sum < diff){//如果两个数的和<两个数的差
			printf("impossible\n");//不可能有结果(a,b均>=0)
		}else if((sum+diff)%2 != 0){//如果(a+b) + (a-b) 不是偶数
			printf("impossible\n");//不可能
		}else{//否则符合正常逻辑
			printf("%d %d\n",(sum+diff)/2,(sum-diff)/2);//计算这两个数都是什么
		}
	}

	return 0;
}