POJ-3641 Pseudoprime numbers

最新推荐文章于 2018-08-10 16:34:36 发布

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本文介绍了一种用于判断特定条件下伪素数的算法，并通过快速幂运算实现对给定数值的有效验证。文中提供了完整的C++代码示例，涵盖质数检测、快速幂计算等关键步骤。

Pseudoprime numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8638		Accepted: 3627

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

Sample Output

no
no
yes
no
yes
yes

Source

Waterloo Local Contest, 2007.9.23

#include<stdio.h>  
#include<string.h>   
bool sushu(long long n)  
{  
    long long  i;  
    for(i=2;i*i<n;++i)  
    {  
        if(n%i==0)  
            return false;  
    }  
    return true;  
}   
long long quictpow(long long n,long long m)  
{  
    long long ans=1,cnt=m;  
    while(cnt>0)  
    {  
        if(cnt&1)  
            ans=(ans*n)%m;    
        n=(n*n)%m; 
		cnt=cnt/2; 
    }  
    return ans;  
}   
int main()  
{  
    long long p,a;  
    while(scanf("%lld%lld",&p,&a)&&p||a)  
    {  
        if(sushu(p))  
            printf("no\n");  
        else if(a==quictpow(a,p))  
            printf("yes\n");  
        else if(a!=quictpow(a,p))  
            printf("no\n");  
    }  
    return 0;  
}