2012 #3 Arcane Numbers

Arcane Numbers 1

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4320

Description

Vance and Shackler like playing games. One day, they are playing a game called "arcane numbers". The game is pretty simple, Vance writes down a finite decimal under base A, and then Shackler translates it under base B. If Shackler can translate it into a finite decimal, he wins, else it will be Vance’s win. Now given A and B, please help Vance to determine whether he will win or not. Note that they are playing this game using a mystery language so that A and B may be up to 10^12.

 

Input

The first line contains a single integer T, the number of test cases. 
For each case, there’s a single line contains A and B. 

 

Output

For each case, output “NO” if Vance will win the game. Otherwise, print “YES”. See Sample Output for more details.

 

Sample Input

3
5 5
2 3
1000 2000

 

Sample Output

Case #1: YES
Case #2: NO
Case #3: YES

 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 using namespace std;

 long long gcd(long long a,long long b)
 {
     if(a<b)
         return gcd(b,a);
     else if(b==)
         return a;
     else
         return gcd(b,a%b);
 }

 int main()
 {
     int T,ca=;
     long long A,B;
     int i,j,k,flg;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%I64d %I64d",&A,&B);
         long long C=gcd(A,B),D;
         A=A/C,B=B/C;
         flg=;
         if(B>)
         {
             while()
             {
                 if(B==)
                     break;
                 D=gcd(B,C);
                 if(D==)
                 {
                     flg=;
                     break;
                 }
                 B=B/D;
             }
         }
         if(A>)
         {
             while()
             {
                 if(A==)
                     break;
                 D=gcd(A,C);
                 if(D==)
                 {
                     flg=;
                     break;
                 }
                 A=A/D;
             }
         }
         if(flg==)
             printf("Case #%d: YES\n",ca);
         else
             printf("Case #%d: NO\n",ca);
         ca++;
     }
     return ;
 }