1 #include <iostream>
2 #include<cstdio>
3 using namespace std;
4 #define LL long long
5 LL a,b,m,n,d;
6 void ex_gcd(LL a,LL b,LL &x,LL &y,LL &d)
7 {
8 if(b==0){
9 d=a,x=1,y=0;
10 }
11 else{
12 ex_gcd(b,a%b,x,y,d);
13 LL t=x;
14 x=y,y=t-a/b*y;
15 }
16 }
17 int main()
18 {
19 LL T;
20 cin>>T;
21 for(int i=0;i<T;i++)
22 {
23 LL x,y;
24 cin>>x>>y;
25 if(x%y==0){
26 cout<<1<<‘ ‘<<y-1<<endl;
27 }
28 else{
29 a=x/y,b=a+1;
30 ex_gcd(a,b,m,n,d);
31 cout<<m*x<<‘ ‘<<n*x<<endl;
32 }
33 }
34 return 0;
35 }
Theorem
For any two integers x and k there exists two more integers p and q such that:
<!--[if !vml]--><!--[endif]-->
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.
<!--[if !supportEmptyParas]--> <!--[endif]-->
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, <!--[if !vml]--><!--[endif]--> and <!--[if !vml]--><!--[endif]-->fit in a 64 bit signed integer.
对于这道题目来说，要注意上下界的问题，当x%k==0时，它的上界和下界是一样的，因为答案有多种，输出一个即可，所以此时将答案定位1和k-1即可。
在x%k！=0时，它的上界和下界相差1，那么很自然的想到它们的最大公约数为1，所以可以直接用扩展欧几里德算法。
当然因为x是最大公约数的x倍，所以最后答案要乘上x
代码如下：