poj 3641 Pseudoprime numbers

Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7000		Accepted: 2855

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

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
using namespace std;
#define max 1000000000
typedef long long ll;
bool isprime(int n){//判断素数，该题max过大，不适合用打表法
    if(n==0||n==1){
        return false;
    }
    if(n==2){
        return true;
    }
    if(n%2==0)
        return false;
    int i=3,half=sqrt(n*1.0);
    for(;i<=half;i+=2){
        if(n%i==0){
            return false;
        }
    }
    return true;
}
ll work(ll a,ll b,ll k){//a^b mod k  快速幂模板
    if(a==0){
        return 0;
    }
    if(b==0){
        return 1;
    }
    ll x=a%k,ans=1;
    for(;b>0;b/=2){
        if(b%2){
            ans=(ans*x)%k;//这歩一定是会最后到达 
        }
        x=x*x%k;
    }
    return ans;
}
int main()
{
    int p,a;
    while(cin>>p>>a&&p&&a){
         //cout<<isprime(p)<<endl;
         if(!isprime(p)&&(work(a,p,p)==a%p)){//注意素数判断
             cout<<"yes"<<endl;
         }        
         else{
             cout<<"no"<<endl;
         }
    }
    return 0;
}