POJ3641Pseudoprime numbers-快速幂的使用

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于 2018-08-10 19:18:05 发布

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本文介绍了一个程序，该程序基于费马小定理来判断一个数是否为特定基数的伪素数。通过计算给定的a^p mod p 是否等于 a 来确定p是否为a的伪素数，并排除真正的素数。

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Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12457		Accepted: 5363

Description

Fermat's theorem states that for any prime number p and for any integer a > 1, ap = 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 ≤ 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

Sample Output

no
no
yes
no
yes
yes

Source

Waterloo Local Contest, 2007.9.23

题目大意：输入p，n若p为素数，输出no，否则若n^p%p=n 输出yes 不然输出no

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define ll long long
using namespace std;
int a,p;
int su(int p)
{
    if(p < 2)
        return 0;
    for(int i=2; i<=sqrt(p+0.0); i++)
    {
        if(p%i==0)
        {
            return 0;
        }
    }
    return 1;
}

int power(int p, int a)
{
    ll r = p, t = 1, mod = p;
    while(r > 0)
    {
        if(r&1)
            t = ((t%mod)*(a%mod))%mod;
        a = ((a%mod)*(a%mod))%mod;
        r >>= 1;
    }
    return t%mod;
}


int main()
{
    int p, a;
    while(scanf("%d%d", &p, &a), p||a)
    {
        if(su(p))
        {
            printf("no\n");
            continue;
        }
        else if(power(p, a) == a)
        {
            printf("yes\n");
        }
        else
            printf("no\n");
    }
    return 0;
}