UVA11388 GCD LCM

Description of the title

The GCD of two positive integers is the largest integer that divides both the integers without any remainder.

The LCM of two positive integers is the smallest positive integer that is divisible by both the integers.

A positive integer can be the GCD of many pairs of numbers.

Similarly, it can be the LCM of many pairs of numbers.

In this problem, you will be given two positive integers.

You have to output a pair of numbers whose GCD is the first number and LCM is the second number.

Input

The first line of input will consist of a positive integer T.

T denotes the number of cases.

Each of the next T lines will contain two positive integer, G and L.

Output

For each case of input, there will be one line of output.

It will contain two positive integers a and b, a ≤ b, which has a GCD of G and LCM of L.

In case there is more than one pair satisfying the condition, output the pair for which a is minimized.

In case there is no such pair, output ‘-1’.

Constraints

• T ≤ 100

• Both G and L will be less than 231 .

Sample Input

2 1 2 3 4

Sample Output

1 2 -1

Solution

This is a Chinese puzzle!

请记住，看到gcd/lcm，先想数论，而不是暴力！

想不出来再打表找规律！

普及- 的题，不是模拟就是数论。 ——佚名（我）

Algorithm（Naive）

暴力代码

1 #include
      
        2 using namespace std; 3 inline unsigned long long gcd(unsigned long long a,unsigned long long b) 4 { 5     if(b==0) return a; 6     if(b==1) return 1; 7     if(a%2==0&&b%2==0) return 2*gcd(a/2,b/2); 8     if(a%2==1&&b%2==0) return gcd(a,b/2); 9     if(a%2==0&&b%2==1) return gcd(a/2,b);10     if(a%2==1&&b%2==1) return gcd(b,a%b);11 //    return (b==0)?a:gcd(b,a%b);12 }13 unsigned long long t,a,b,j,g,l,flag;14 int main()15 {16     cin>>t;17     while(t--)18     {19         cin>>g>>l;20         j=g*l;21         flag=1;22         for(int a=1;a*a<=j;a++)23         {24             if(j%a==0)25             if(gcd(a,j/a)==g){26                 cout<
       <<" "<

朴素算法Code

时间复杂度

O（t sigma log（sqrt（1 to g*j）））

understand

没脑子的我都懂怎么暴力了……

i*j==g*l

Algorithm（Arithmetical）

1 #include
     
       2 using namespace std; 3 unsigned long long t,g,l; 4 int main() 5 { 6     cin>>t; 7     while(t--) 8     { 9         cin>>g>>l;10         if(l%g==0)11             cout<
      
       <<" "<
       
        <

时间复杂度

O（t）