hdu 4320 Arcane Numbers 1（小数进制转化后是否有限位）

Arcane Numbers 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2380 Accepted Submission(s): 758



Problem 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

Author

BJTU

Source

2012 Multi-University Training Contest 3

Recommend

zhoujiaqi2010

题意：将a进制的有限小数转化为b进制的小数是否是有限位的

题解：若b有a的所有质因子，则可以，反之不可以

#include<stdio.h>
long long gcd(long long a,long long b)
{
if(b==0) return a;
else return gcd(b,a%b);
}
int main()
{
int t,cas=1;
long long x,y,temp;

scanf("%d",&t);
while(t--)
{
scanf("%I64d%I64d",&x,&y);
temp=x;
while(temp>1)
{
temp=gcd(x,y);
x/=temp;
}
if(x>1) printf("Case #%d: NO\n",cas++);
else printf("Case #%d: YES\n",cas++);
}

return 0;
}