POJ Pseudoprime numbers

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原文链接：http://www.cnblogs.com/jun930123/archive/2012/08/16/2642219.html

本文介绍了一种用于判断特定范围内整数是否为基于a的伪素数的算法，并提供了完整的C语言实现代码。该算法利用了费马小定理的概念，通过快速幂运算来检查给定的整数p是否满足对于所有a>1且a<p，a^p ≡ a (mod p) 的条件。

Pseudoprime numbers

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3641

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

View Code

 1 #include<stdio.h>
 2 int f[35000],prime[5000],count = 0 ;
 3 typedef __int64 LL ;
 4 void getPrime(){
 5     for(int i=2;i<=34000;i++)
 6         if(!f[i]){
 7             prime[count++] = i;
 8             for(int j=i*i;j<=34000;j+=i)
 9                 f[j] = 1;
10         }
11 }
12 bool Isprime(int n){
13     for(int i=0;i<count && prime[i]*prime[i]<=n;i++)
14         if(n%prime[i] == 0)    return 0;
15     return 1;
16 }
17 bool func(LL p,LL a){
18     LL temp = a,mod = p;
19     LL ans = 1;
20     while(p){
21         if(p % 2)    ans = (ans * temp) % mod ;
22         p /= 2;    
23         temp = (temp * temp) % mod ;
24     }
25     if(ans == a)    return 1;
26     return 0;
27 }
28 int main(){
29     getPrime();
30     int p,a;
31     while(scanf("%d%d",&p,&a)){
32         if(p == 0 && a== 0)    break;
33         if(Isprime(p))
34             printf("no\n");
35         else    {
36             if(func(p,a))
37                 printf("yes\n");
38             else
39                 printf("no\n") ;
40         }
41     }
42     return 0;
43 }

转载于:https://www.cnblogs.com/jun930123/archive/2012/08/16/2642219.html