POJ 3641 Pseudoprime numbers 测试费马小定理伪素数

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Pseudoprime numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6018 Accepted: 2407

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

Source

Waterloo Local Contest, 2007.9.23

费马小定理：若p是素数且a是正整数，那么a^p=a(mod p)

若a是正整数，p是合数且满足a^p=a(mod p)，那么称p为以a为基的伪素数。

给你p和a，让你判断p是不是伪素数。

//384K16MS#include<stdio.h>#include<math.h>bool isprime(long long x)//判断x是不是素数，如果是素数，肯定不是伪素数{    if(x==1||x==2)return true;    long long tmp=sqrt(x);    for(long long i=2;i<=tmp;i++)        if(x%i==0)return false;    return true;}long long quick_mod(long long a,long long b,long long m)//快速幂求a^b%m{    long long ans=1;    while(b)    {        if(b&1){ans=(ans*a)%m;b--;}        b/=2;        a=a*a%m;    }    return ans;}int main(){    long long p,a;    while(scanf("%lld%lld",&p,&a),p|a)    {        if(isprime(p)){printf("no\n");continue;}        long long mol=quick_mod(a,p,p);        if(mol==a%p)printf("yes\n");        else printf("no\n");    }    return 0;}