POJ3641 (快速幂）判断a^p = a (mod p)是否成立

最新推荐文章于 2025-04-30 22:23:57 发布

转载最新推荐文章于 2025-04-30 22:23:57 发布 · 194 阅读

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原文链接：http://www.cnblogs.com/yexiaozi/p/5698795.html

本文介绍了一种利用费马小定理来判断特定条件下整数是否为伪素数的方法，并提供了一个C++实现示例，用于检测给定数值是否满足特定条件下的伪素数性质。

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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

如果p是素数，输出no；如果p不是素数，判断a^p对p取余是否等于a。

 1 #include<cstdio>
 2 #include<math.h>
 3 __int64 f(__int64 a,__int64 b)
 4 {
 5     __int64 c=b,t=1;
 6     while(b)
 7     {
 8         if(b % 2 != 0)
 9         {
10             t=t*a%c;
11         }
12         a=a*a%c;
13         b/=2;
14     }
15     return t%c;
16 }
17 __int64 f2(__int64 a)
18 {
19     __int64 i;
20     if(a <= 1 || a % 2 == 0) return 0;
21     for(i=3;i<=sqrt(a);i++)
22     {
23         if(a % i == 0) return 0;
24     }
25     return 1;
26 }
27 int main()
28 {
29     
30     __int64 p,a;
31     while(scanf("%I64d %I64d",&p,&a) && p && a)
32     {
33         if(f2(p) == 1) printf("no\n");
34         else
35         {
36             if(f(a,p) == a) printf("yes\n");
37             else 
38             printf("no\n");
39         }
40         
41     }
42 }

转载于:https://www.cnblogs.com/yexiaozi/p/5698795.html