UVa

先上题目

Problem A
Play with Floor and Ceil
Input: standard input
Output: standard output
Time Limit: 1 second
 

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

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

　　题意很简单，就是给你一个x一个k，求出x/k的向上取整和向下取整的结果a,b然后求出任意一组p,q满足x==pa+qb;这里直接用扩展欧几里得就可以得出答案，所以这其实就是一个模板题。

上代码：

#include <stdio.h>
#include <math.h>
#define LL long long
#define I64 ll
using namespace std;

LL gcd(LL a,LL b)
{
    return b==0 ?  a : gcd(b,a%b);
}

void ex_gcd(LL a,LL b,LL &x,LL &y)
{
    if(b==0) {x=1;y=0;return ;}
    ex_gcd(b,a%b,x,y);
    LL t=y;
    y=x-(a/b)*y;
    x=t;
}

int main()
{
    int t;
    LL x,k,p,q,a,b,g;
    //freopen("data.txt","r",stdin);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d %I64d",&x,&k);
        a=floor(x*1.0/k);
        b=ceil(x*1.0/k);
        g=gcd(a,b);
        a/=g;
        b/=g;
        x/=g;
        ex_gcd(a,b,p,q);
        printf("%I64d %I64d
",p*x,q*x);
    }
    return 0;
}