UVa 10673 play with Floor and Ceil: number theory

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 all two integers x and K there exists two more integers p and Q such that:

It ' s a fairly easy tasks to prove this theorem, so we ' d ' don't ask for you. 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 t He given equation.

Input

The ' The ' input contains an integer, T (1≤t≤1000) that gives your number of test cases. In each of the following T lines your ' d be given two positive integers x and K. Can safely assume that x and K would always be less than ⁸.

Output

For each of the test cases print two integers: p and Q at one line. These two integers are to is separated by a. If There are multiple pairs of p and Q that satisfy the equation, any one would do. But to help us keep we task simple, please make sure that the values,

and

Fit in a bit signed integer.

3 5 2 2 24444 6

1 1 1 1 0 6

Classification Discussion ~ The answer is actually very simple (see Code)

Complete code:

01./*0.012s*/
.  
#include <cstdio>  
.  
05.int main (void)  
06.{  
An    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;  
18.}

PS: If the problem requires p,q nonnegative, then P is-x% (x/k+1) +x/k+1

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