UVa 10673 Play with Floor and Ceil：数论

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.

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\n", -x, x);
14.        else
15.            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

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