HDU 1194 Beat the Spread!（数学）

Beat the Spread!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5882    Accepted Submission(s): 3097

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

240 2020 40

 

Sample Output

30 10impossible

 

题解：

题意:输入两个数的和与差,求这两个数,按从大到小的输出.
假设两个数的和与差分别为a,b,如果这两个数存在的话,这两个数分别为x=(a+b)/2; y=(a-b)/2;

AC代码：

#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<cstdlib>#include<algorithm>typedef long long LL;using namespace std;int main(){int T,x,y;int n,m;cin>>T;while(T--){scanf("%d%d",&n,&m);if(n<m)printf("impossible\n");else if((n+m)%2!=0){printf("impossible\n");//这两个数的和与差同奇偶}   else {    x=(n+m)/2;y=(n-m)/2;if(x<y)swap(x,y);printf("%d %d\n",x,y);}}   return 0;}