Hdoj 5391 Zball in Tina town

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5391

Test instructions: give you a N, ask you (N-1)! The value of% n.

This problem is a Wilson theorem, the content of Wilson's theorem is that if n is a prime number, then there must be (n-1)! % n = =-1, in fact, if the comparison wit know to play table students should also be able to see this law, then small series here to prove this Wilson theorem, first in this mod n multiplication, there are n-1 elements, if n is a prime number, then each element must exist an inverse element, that thinking, If each element has an inverse then the 22 pairing result should not be 1 ... Of course, there is no way to consider the 22 pairing, that is, the inverse is its own, x^2 MOD n = 1, the solution x = 1 or x = n-1, then we in the hierarchy in addition to the two special just good can be 22 paired, and because (n-1)%n = = N-1, so the result is n-1.

Then the prime number is solved, we look at the non-prime, think carefully, if there is a factor x, then X * (n/x) = = N, so if it is a non-prime, then there must be two factors, so that the product is n, then the whole class must be a multiple of n, so the answer is 0. But is it true that every non-prime is like this? Of course, we have to consider the special case, that is, there is only one factor X, and x^2 = = N, this obviously can not constitute a multiple of n, the value of the solution is 4.

So this problem we only need to determine whether it is a prime number, if the number of non-prime and then a special sentence of 4 is good.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define LL __int64
int prime (int n)
{for
	(int i=2; i*i<=n; i++)
		if (n% i = = 0)
			return 0;
	return 1;
}
int main ()
{
	int T;
	scanf ("%d", &t);
	while (t--)
	{
		int n;
		scanf ("%d", &n);
		if (n = = 4) printf ("2\n");
		else if (!prime (n)) printf ("0\n");
		else printf ("%d\n", n-1);
	}
}