Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4864		Accepted: 1872

Description

Fermat's theorem states that for any prime number p and for any integer a > 1, a^p = a (mod p). That is, if we raise a to the p^th power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)

Given 2 < p ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

Input

Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.

Output

For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

Sample Input

3 210 3341 2341 31105 21105 30 0

Sample Output

nonoyesnoyesyes

Source

, 2007.9.23

 

//给你一个数 p和一个数 a， 若p是素数输出no，a的p次方模p等于a输出yes，否则输出no

#include 
     
      #include 
      
       #include 
       
        using namespace std;bool is_prime(int n){
          int i,m=sqrt(double(n));	 for(i=2;i<=m;i++)		 if(n%i==0)			 return 0;	 return 1;}int main(){
     	int q,a,tq;	__int64 ta,i;    while(scanf("%d%d",&q,&a),q||a)	{
     		if(is_prime(q))		{
     			printf("no\n");continue;		}         tq=q; i=a;	    for(ta=1;tq>0;tq=tq>>1,i=(i*i)%q)		{
      		   if(tq&1) ta=(ta*i)%q;		}		if(ta==a)		   printf("yes\n");		else			printf("no\n");	}    return 0;}