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

5 2

40 2

24444 6

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：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。。。。

View Code

 1 #include<stdio.h>
 2 #include<math.h>
 3 typedef struct
 4 {
 5     long d;
 6     long x;
 7     long y;
 8 }NODE;
 9 NODE gcd(long a,long b)
10 {
11     NODE s,p;
12     if(!b)
13     {
14         s.x=1;
15         s.y=0;
16         s.d=a;
17         return s;
18     }
19     s=gcd(b,a%b);
20     p.x=s.x;
21     s.x=s.y;
22     s.y=p.x-(a/b)*s.y;
23     return s;
24 }
25 int main(void)
26 {
27     long x,k,a,b;
28     int t;
29     NODE s;
30     scanf("%d",&t);
31     while(t--)
32     {
33         scanf("%ld%ld",&x,&k);
34         a=floor((double)x/k);
35         b=ceil((double)x/k);
36         s=gcd(a,b);
37         printf("%ld %ld\n",s.x*(x/s.d),s.y*(x/s.d));
38 
39     }
40     return 0;
41 }