UVa 10673 Play with Floor and Ceil：数论

UVa 10673 Play with Floor and Ceil：数论2014-07-10 synapse7 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

分类讨论~答案其实很简单（见代码）

完整代码：

01./*0.012s*/02.03.#include<cstdio>04.05.int main（void）06.{07.int t, x, k;08.scanf（"%d", &t）;09.while （t--）10.{11.scanf（"%d%d", &x, &k）;12.if （x % k）13.printf（"%d %d
", -x, x）;14.else15.printf（"0 %d", k）;16.}17.return 0;18.}

PS：若题目要求p，q非负，则p为-x%（x/k+1）+x/k+1