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YTUOJ-Pseudoprime numbers

时间：2015-04-03 09:33:09 阅读：93 评论：0 收藏：0 [点我收藏+]

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 contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a. For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

Input

Output

Sample Input

Sample Output

no
no
yes
no
yes
yes

HINT

题意：

给定p，a两个数字2<=p<=1000000000,1<a<p;若p为素数，则输出no，否则（若a^p%p=a 输出yes，否则输出no）

代码如下：

#include <iostream>
#include <cmath>
using namespace std;
bool isPrime(long long n)
{
    long long i;
    for (i=2;i<=sqrt(n);i++)
    {
        if (n%i==0)
            return false;
    }
    return true;
}

long long mod(long long a,long long n,long long m)
{
    long long d=1,t=a;
    while (n>0)                        //通过n来求a^p%p;
    {
        if (n%2==1)                    //当n为奇数时先乘一个a并%m;
            d=(d*t)%m;
        n/=2;
        t=(t*t)%m;                     //缩短计算过程
    }
    return d;
}

int main()
{
    long long a,p;
    while (cin>>p>>a)
    {
        if (a==0&&p==0)
            break;
        if (isPrime(p))                   
            cout<<"no"<<endl;
        else
        {
            if (a==mod(a,p,p))
                cout<<"yes"<<endl;
            else
                cout<<"no"<<endl;
        }
    }
    return 0;
}

运行结果：

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