Pseudoprime numbers

Pseudoprime numbers

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3408    Accepted Submission(s): 1586

Problem 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 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, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.) 
Given 2 < p ≤ 1,000,000,000 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

 

先判断是否为素数，如果不是再进行快速幂取模．

 

#include <bits/stdc++.h>
#define ll long long
using namespace std;

bool prim(ll x){
    bool flag = true;
    for(int i=2;i<=sqrt(x);i++){
        if(x%i==0){
            flag = false;
            break;
        }
    }
    return flag;
}

ll pow(ll a,ll b,ll mod){
    ll ret = 1;
    while(a){
        if(a&1){
            ret = ret*b%mod;
        }
        b = b*b%mod;
        a>>=1;
    }
    return ret%mod;
}

ll n,m;
int main(){
    while(cin>>n>>m&&!(n==0&&m==0)){
        bool prime = prim(n);
        if(!prime){
            ll ans = pow(n,m,n);
            if(ans==m){
                cout<<"yes"<<endl;
            }else{
                cout<<"no"<<endl;
            }
        }else{
            cout<<"no"<<endl;
        }
    }
    return 0;
}