HDU 2504 see GCD again

Question: http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 2504

It means that gcd (a, c) = B, and a, B are known, and the minimum c is obtained, where c! = B. I thought that since c! = B, the minimum is 2b, of course, but WA is submitted first. Looking back, we found that it was not that simple because a may be a multiple of c. For example, if a = 6b, c = 2b, 3b, and 4b are not the answer, because gcd (a, c) = 2b, 3b, 2b at this time. In this case, I had to try it one by one, starting from 2b, increasing B at a time until gcd (a, c) = B. The Code is as follows:

Code

  1 #include <iostream>
 2  using namespace std;
 3 
 4 int gcd(int a,int b)
 5 {
 6     while(a%b)
 7     {
 8         int r=a%b;
 9         a=b;
10         b=r;
11     }
12     return b;
13 }
14 
15 int main()
16 {
17     int n,a,b,c,i,j;
18     cin>>n;
19     for (i=0;i<n;i++)
20     {
21         cin>>a>>b;
22         c=2*b;
23         while(gcd(a,c)!=b)
24         {
25             c+=b;
26         }
27         cout<<c<<endl;
28     }
29     return 0;
30 }