hdu 4320 Arcane Numbers 1 多校联合赛（三）第一题

 

“对于所有的 x|a, 有 x|A”。
能够在进制 A 下用有限数位表示的任何一个有理数.
其最简分母 (a) 和进制数 (A) 肯定能满足上面的“充要条件”。
也就是说，这些有理数的最简分母，都可以表示成 A 的若干个因子相乘的形式
（相同因子可乘多次）。
那么如果要保证这些有理数在进制 B 下也能顺利用有限数位表示的话，
B 这个数的因子必须至少要包含 A 中的因子。

看到这题目谁想的到啊！

#include <stdio.h>
#include <string.h>
int main()
{
    __int64 a,b,ans,i,t,cas;
    scanf("%I64d",&t);
    for(cas=1;cas<=t;cas++)
    {
        printf("Case #%d: ",cas);
        ans=1;
        scanf("%I64d%I64d",&a,&b);
        for(i=2;i*i<=a;i++)
        {
            if(a%i==0)
            {
                ans*=i;
                while(a%i==0)
                    a/=i;
            }
        }
        ans*=a;
        if(b%ans==0)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

 

 

Arcane Numbers 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1719 Accepted Submission(s): 539

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