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:

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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.

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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：