Poj - 3641 - Pseudoprime numbers-（快速幂）

最新推荐文章于 2018-07-19 17:06:23 发布

菜圾

最新推荐文章于 2018-07-19 17:06:23 发布

阅读量265

点赞数

CC 4.0 BY-SA版权

分类专栏：数论文章标签： ACM poj 数论算法

本文链接：https://blog.youkuaiyun.com/sdau20163942/article/details/78572190

数论专栏收录该内容

47 篇文章

订阅专栏

本文介绍了一种利用费马小定理来判断特定整数是否为base-a伪素数的方法，并提供了相应的C++实现代码。通过快速幂取模运算，可以有效地计算出(a^p)%p是否等于a，以此来判断p是否为base-a伪素数。

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-apseudoprimes for all a.)

Given 2 < p ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-apseudoprime.

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

Sample Output

no
no
yes
no
yes
yes

给出a，p，如果(a^p)%p==a则称p为base-a pseudoprimes（注意这里p需为素数），判断p是不是一个base-a pseudoprimes

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<queue>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;

ll p,a;

ll pow_mod(ll n,ll k,ll m) //快速幂求(n^p)%m的值
{
    ll res=1;
    while(k>0)
    {
        if(k&1)
            res=res*n%m;
        n=n*n%m;
        k>>=1;
    }
    return res;
}

bool is_prime(ll x)
{
    double tx=sqrt(1.0*x);
    ll it=2;
    while(it<=tx&&x%it!=0)
        it++;
    if(it>tx)
        return true;
    else
        return false;
}

int main()
{
    while(scanf("%lld%lld",&p,&a)!=EOF)
    {
        if(p==0&&a==0)
            break;

        if(is_prime(p)==true)
        {
            printf("no\n");
            continue;
        }
        ll temp=pow_mod(a,p,p);
        if(temp==a)
        {
            printf("yes\n");
        }else{
            printf("no\n");
        }
    }
    return 0;
}