POJ 3641

Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6044		Accepted: 2421

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 2
10 3
341 2
341 3
1105 2
1105 3
0 0

Sample Output

no
no
yes
no
yes
yes

Source

Waterloo Local Contest, 2007.9.23

 

 

快速幂

 

 

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>

 using namespace std;

 typedef long long ll;

 int p,a;

 bool judge() {
         for(int i = ; i * i <= p; i++) {
                 if(p % i == ) return false;
         }

         return true;
 }
 bool  mod_pow(ll x,ll n) {
         ll res = ;
         while(n > ) {
                 if(n & ) res = res * x % p;
                 x = x * x % p;
                 n >>= ;
         }

         return res == a;
 }

 int main() {
         //freopen("sw.in","r",stdin);

         while(~scanf("%d%d",&p,&a) && p && a) {
                 if(!judge() && mod_pow(a,p)) printf("yes\n");
                 else printf("no\n");

         }

         return ;
 }