hdoj 1905 Pseudoprime numbers

最新推荐文章于 2019-04-10 19:15:00 发布

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#Pseudoprime numbers #hdoj 1905 #快速幂取模 #数论

本文提供了一个关于判断给定范围内的数是否为伪素数的算法实现，涉及快速幂取模运算，适用于数学和计算机科学领域。

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Pseudoprime numbers

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2446 Accepted Submission(s): 1000

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

Sample Output

no
no
yes
no
yes
yes

快速幂取模：
题意：两个数，q,a.,如果q是素数，输出no；如果(a^q)%q==a,输出yes否则输出no!
代码：
#include<stdio.h>
bool judge(__int64 n)//判断素数 
{
	__int64 i;
	for(i=2;i*i<=n;i++)
	if(n%i==0)
	return false;
	return true;
}
int  Pseudoprime(__int64 a,__int64 b,__int64 k)//判断是否符合条件 
{
	__int64 ans=1;
    a=a%k;
	while(b>0)
	{
		if(b%2)
		ans=(ans*a)%k;
		b=b/2;
		a=(a*a)%k;
	}
	return ans;
}
int main()
{
   __int64 n,m,i;
   while(scanf("%I64d%I64d",&n,&m),n,m)
   {
   	  if(judge(n))
   	  {
		 printf("no\n");
	     continue;
	  }
	 __int64 s=Pseudoprime(m,n,n);
	 if(s==m)
	 printf("yes\n");
	 else
	 printf("no\n");
   }
   return 0;
}