POJ Pseudoprime numbers

Pseudoprime numbers

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3641

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

#include<stdio.h>
int f[35000],prime[5000],count = 0 ;
typedef __int64 LL ;
void getPrime(){
    for(int i=2;i<=34000;i++)
        if(!f[i]){
            prime[count++] = i;
            for(int j=i*i;j<=34000;j+=i)
                f[j] = 1;
        }
}
bool Isprime(int n){
    for(int i=0;i<count && prime[i]*prime[i]<=n;i++)
        if(n%prime[i] == 0)    return 0;
    return 1;
}
bool func(LL p,LL a){
    LL temp = a,mod = p;
    LL ans = 1;
    while(p){
        if(p % 2)    ans = (ans * temp) % mod ;
        p /= 2;    
        temp = (temp * temp) % mod ;
    }
    if(ans == a)    return 1;
    return 0;
}
int main(){
    getPrime();
    int p,a;
    while(scanf("%d%d",&p,&a)){
        if(p == 0 && a== 0)    break;
        if(Isprime(p))
            printf("no\n");
        else    {
            if(func(p,a))
                printf("yes\n");
            else
                printf("no\n") ;
        }
    }
    return 0;
}