UVA10673 Play with Floor and Ceil

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 uskeep our task simple, please make sure that the values,andfit 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

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

题目大意：对于任意两个整数x,k，存在两个整数p和q满足：

                                            

输入x,k,找出符合要求的整数p和q,如果有多组，输出任意一组即可。

题解：赤果果的扩展欧几里得。。。。不过我提交几次都是WA，改成long就AC了，题目严重误导人啊！！！！坑死了。。。。

#include<stdio.h>
#include<math.h>
typedef struct
{
    long d;
    long x;
    long y;
}NODE;
NODE gcd(long a,long b)
{
    NODE s,p;
    if(!b)
    {
        s.x=1;
        s.y=0;
        s.d=a;
        return s;
    }
    s=gcd(b,a%b);
    p.x=s.x;
    s.x=s.y;
    s.y=p.x-(a/b)*s.y;
    return s;
}
int main(void)
{
    long x,k,a,b;
    int t;
    NODE s;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%ld%ld",&x,&k);
        a=floor((double)x/k);
        b=ceil((double)x/k);
        s=gcd(a,b);
        printf("%ld %ld\n",s.x*(x/s.d),s.y*(x/s.d));

    }
    return 0;
}