POJ3641 (Fast power) to determine whether a^p = A (mod p) is established

Description

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

If P is a prime number, output no; If P is not a prime number, determine if a^p is equal to P.

1#include <cstdio>2#include <math.h>3 __int64 F (__int64 A,__int64 B)4 {5__int64 c=b,t=1;6      while(b)7     {8         if(b%2!=0)9         {Tent=t*a%C; One         } Aa=a*a%C; -B/=2; -     } the     returnt%C; - } - __int64 F2 (__int64 a) - { + __int64 i; -     if(A <=1|| A2==0)return 0; +      for(i=3; I<=sqrt (a); i++) A     { at         if(a% i = =0)return 0; -     } -     return 1; - } - intMain () - { in      - __int64 p,a; to      while(SCANF ("%i64d%i64d", &p,&a) && p &&a) +     { -         if(F2 (p) = =1) printf ("no\n"); the         Else *         { $             if(f (a,p) = = a) printf ("yes\n");Panax Notoginseng             Else  -printf"no\n"); the         } +          A     } the}

POJ3641 (Fast power) to determine whether a^p = A (mod p) is established