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

35 52 31000 2000

 

Sample Output

Case #1: YESCase #2: NOCase #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;}