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2013
11-13

POJ 3641 Pseudoprime numbers [解题报告] Java

Pseudoprime numbers

问题描述 :

Fermat’s theorem states that for any prime number p and for any integer a > 1, ap = 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 ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.

输入:

Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.

输出:

For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".

样例输入:

3 2
10 3
341 2
341 3
1105 2
1105 3
0 0

样例输出:

no
no
yes
no
yes
yes

解题代码:

//* @author: [email protected]
import java.util.Scanner;
public class Main
{
 public static void main(String[] args)
 {
  Scanner in=new Scanner(System.in);
  while(true)
  {
	long p=in.nextLong();
	long a=in.nextLong();
	if(a==0&&p==0) break;
	if(isPrime(p)){
		System.out.println("no");
		continue;
	}
	System.out.println(sum(a,p,p)==a?"yes":"no");
   }
  }

 public static boolean isPrime(long a)
 {
	if(a==2) return true;
	for(int i=3;i<=Math.sqrt(a);i+=2)
		if(a%i==0) return false;
	return true;
  }

 public static long sum(long a,long n,long p)
 {
	if(n==0) return 1;
	long w=sum(a*a%p,n/2,p);
	if(n%2==1) w=(w*a)%p;
	return w;
 }
}