2012 #3 Arcane Numbers

Arcane Numbers 1
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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 
 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 long long gcd(long long a,long long b) 
 7 { 
 8     if(a<b) 
 9         return gcd(b,a); 
10     else if(b==0) 
11         return a; 
12     else
13         return gcd(b,a%b); 
14 }
15 
16 int main()
17 {
18     int T,ca=1;
19     long long A,B;
20     int i,j,k,flg;
21     scanf("%d",&T);
22     while(T--)
23     {
24         scanf("%I64d %I64d",&A,&B);
25         long long C=gcd(A,B),D;
26         A=A/C,B=B/C;
27         flg=1;
28         if(B>1)
29         {
30             while(1)
31             {
32                 if(B==1)
33                     break;
34                 D=gcd(B,C);
35                 if(D==1)
36                 {
37                     flg=0;
38                     break;
39                 }
40                 B=B/D;
41             }
42         }
43         if(A>1)
44         {
45             while(1)
46             {
47                 if(A==1)
48                     break;
49                 D=gcd(A,C);
50                 if(D==1)
51                 {
52                     flg=0;
53                     break;
54                 }
55                 A=A/D;
56             }
57         }
58         if(flg==1)
59             printf("Case #%d: YES\n",ca);
60         else
61             printf("Case #%d: NO\n",ca);
62         ca++;
63     }
64     return 0;
65 }
View Code

 

posted @ 2015-08-30 19:37  cyd2014  阅读(130)  评论(0)  编辑  收藏  举报