10673 - Play with Floor and Ceil
Time limit: 3.000 seconds
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 |
分类讨论~答案其实很简单（见代码）
完整代码：
/*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\n", -x, x);
else
printf("0 %d", k);
}
return 0;
}
PS：若题目要求p，q非负，则p为-x%(x/k+1)+x/k+1
