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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Its a fairly easy task to prove this theorem, so wed not ask you to do that. Wed ask for something even easier! Given the values of x and k, youd 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 youd 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：