UVa 10673 Play with Floor and Ceil(数論)

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10673 - Play with Floor and Ceil
Time limit: 3.000 seconds
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=115&page=show_problem&problem=1614
Theorem
For any two integers x and k there exists two more integers p and q such that:
It’s a fairly easy task to prove this theorem, so we’d not ask you to do that. We’d ask for something even easier! Given the values of x and k, you’d only need to find integers p and q that satisfies the given equation.
 
Input
The first line of the input contains an integer, T (1≤T≤1000) that gives you the number of test cases. In each of the following T lines you’d be given two positive integers x and k. You can safely assume that x and k will always be less than 108.
 
Output
For each of the test cases print two integers: p and q in one line. These two integers are to be separated by a single space. If there are multiple pairs of p and q that satisfy the equation, any one would do. But to help us keep our task simple, please make sure that the values,  and fit in a 64 bit signed integer.
 
Sample Input      Output for Sample Input
3 5 2 40 2 24444 6           
1 1 1 1 0 6                           
分類討論~答えは実は簡単です(コードを参照)
完全なコード:
/*0.012s*/

#include<cstdio>

int main(void)
{
	int t, x, k;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%d", &x, &k);
		if (x % k)
			printf("%d %d
", -x, x); else printf("0 %d", k); } return 0; }

PS:タイトル要求p,q非負、pは-x%(x/k+1)+x/k+1