UVaOJ10673 - Play with Floor and Ceil

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

Problem setter: Monirul Hasan, Member of Elite Problemsetters' Panel

Special Thanks: Shahriar Manzoor, Member of Elite Problemsetters' Panel

题目大意：

给出x和k，求p*floor(x/k)+q*ceil(x/k)==x的一组解(p,q)

解题方法：

我们知道向上取整和向下取整之差只能为0或1，如果为1，我们显然可以找到p=x，q=-x为符合要求的一组解，如果为0，那么只要(p+q)[x/k]=x，同时注意到在这种情况下x%k==0，所以令p+q=k即可，如果想让p、q有整数解，最简单的办法就是p=0，q=k。

源代码：

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
int T, a, b;
long x, k, p, q;
scanf("%d", &T);
while (T --) {
scanf("%ld%ld", &x, &k);
a = x / k;
b = x / k + (x % k ? 1 : 0);
if (a != b) {
p = -x;
q = x;
} else {
p = 0;
q = k;
}
printf("%ld %ld\n", p, q);
}
return 0;
}