UVA10673 上下界问题

 #include <iostream>
 #include<cstdio>
 using namespace std;
 #define LL long long
 LL a,b,m,n,d;
 void ex_gcd(LL a,LL b,LL &x,LL &y,LL &d)
 {
     if(b==){
         d=a,x=,y=;
     }
     else{
         ex_gcd(b,a%b,x,y,d);
         LL t=x;
         x=y,y=t-a/b*y;
     }
 }
 int main()
 {
     LL T;
     cin>>T;
     for(int i=;i<T;i++)
     {
         LL x,y;
         cin>>x>>y;
         if(x%y==){
             cout<<<<' '<<y-<<endl;
         }
         else{
             a=x/y,b=a+;
             ex_gcd(a,b,m,n,d);
             cout<<m*x<<' '<<n*x<<endl;
         }
     }
     return ;
 }

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：