POJ 3641

Pseudoprime numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7043 Accepted: 2881

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

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <algorithm>#include <cmath>#define N 1000009#define ll __int64using namespace std;ll vis[N];ll fun(ll a){    for(ll i=2;i*i<=a;i++)    {        if(a%i==0) return 1;    }    return 0;}ll pow_mod(ll m,ll n){    ll b=1;    ll mod=n;    while(n)    {        if(n&1)         b=(b*m)%mod;         n=n>>1;         m=m*m%mod;    }    return b;}int main(){    ll p,a;    while(~scanf("%I64d %I64d",&p,&a))    {       if(a+p==0) break;       if(fun(p)==0)       {           cout<<"no"<<endl;           continue;       }        ll ans=pow_mod(a,p);        //cout<<"ans="<<ans<<endl;        if(ans==a)        cout<<"yes"<<endl;        else        cout<<"no"<<endl;    }    return 0;}