HDU -&gt; ACM -&gt; Pseudoprime numbers-优快云博客

本文介绍了一种基于费马小定理的伪素数判断算法，并通过C++代码实现了一个快速幂运算函数来验证给定的数值是否为特定基数下的伪素数。该算法适用于在大范围内快速筛选出可能的伪素数。

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Pseudoprime numbers

Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 21 Accepted Submission(s) : 13

Problem 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

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 2

   10 3 
 

   341 2 
 

   341 3 
 

   1105 2 
 

   1105 3 
 

0 0

Sample Output

no

no

yes

no

yes

yes

#include<cstdio>  
int quickpow(__int64 n,__int64 m,__int64 mod)//n是底数，m是幂数，mod是取余数  
{  
    __int64 ans=1,base=n;  
    while(m)  
    {  
        if(m&1)//如果幂数是奇数,则将底数保存在ans中,这样幂数-1变成偶数就可以继续降幂了。 base是幂数。  
  
        {  
            ans=(base*ans)%mod;//保存并更新多幂底数  
        }  
        base=(base*base)%mod;//更新底数  
        m/=2;//更新幂数   
    }  
    return ans;  
}  
int main()  
{  
    __int64 n,m;  
    __int64 mod;  
    while(scanf("%I64d%I64d",&n,&m)&&n||m)  
    {  
        int sum=0;  
        for(int i=2;i*i<n;i++)  
        {  
            if(n%i==0)  
            sum++;  
        }  
        if(sum==0)  
        printf("no\n");  
        else  
        {  
        mod=n;  
        if(quickpow(m,n,mod)==m)  
        printf("yes\n");  
        else  
        printf("no\n");       
        }     
    }  
    return 0;  
}