[Ybt gold medal navigation 8-6-4] the number of original roots

Number of original roots

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General idea

Give you an odd prime number p, and ask you the number of its primitive roots.

Ideas

First, we need to know what the original root is.

Order

Before talking about the primitive root, we need to know what the order is.

Let $n>1$，$a$ is the same as$n$ is a relatively prime number, then according to the extended Euclid, we can know that there must be many$r$ makes$a^r\equiv 1\left(modn\right)$ . And the smallest one will be between$1\sim$Between$n$ , this value is called$a$ modennThe order of$n$ is denoted as$Or{d}_n(a)$。

It should be noted that after $\gcd (a,n)=When\; it\; is\; 1$ , there will be an order, otherwise its equation will have no solution, that is, there will be no order.

Then let's talk about some of its properties: (We assume that we already know $\gcd (a,n)=1$ )
Known$a$ modennThe order of$n$ is$r$ , if there is a number$N$ makes$a^N\equiv 1\left(modn\right)$ , then$N$ isrrThe multiple of$r$ , which is$r|N$。

After that $a$ modennorder of$n$$r$ it is divisible by$\phi \left(n\right)$ . Then if$n$ is a prime number, then$r$ it divides$n-1$ . (Because theφ \varphi ofprime numbers$The$ value of$\phi$ is minus one)
(This is obtained by Euler's theorem)

Euler's theorem: for mutually prime positive integers a and n, there is a^φ(n) ≡ 1(mod n)

Then there is another one, remember $n=or{d}_m\left(a\right)$ , for greater than$0$ basis$i$，如果 $or{d}_m(a^i)=s$ , then$s=\frac{n}{\gcd (n,i)}$。

Primitive root

Then we look at what the original root is.
The primitive root is to satisfy $Or{d}_m(a)=[Phi]\left(m\right)$ of$a$ , called$a$ ismmThe primitive root of$m$ .
Then you can see that it is mod$m$ remaining class, then we only look at it at$0\sim m$ 's solution.

Then it also has some properties.

1. 首先，$a^0,a^1,a^2,...,a^{Or{d}_m\left(a\right)-1}$ mode$m\; is$ definitely different. That$a$ is modulus$When$ the original root of$m$ , the above must constitute the modulus$m$ simple residual system. Then when the moldmm isformed on it$When$ the simple residual system of$m$ , then this$a$ is the modulusmmThe primitive root of$m$ .
   One special thing is when$When\; m$ is a prime number, the$When\; a$ is modulo its primitive root, it not only constitutes a simple residual system, but it is also like this:$1,2,...,m-1$。
2. All prime numbers have primitive roots.
3. Not all integers have primitive roots.
4. When a few $When\; m$ has primitive roots, it has$\phi \left(\phi \right(m\left)\right)$ primitive roots.
   This can be explained in terms of the last property of the order.

This question

This question is to directly use the fourth nature of the original root, just fine.

Code

#include<cstdio>

using namespace std;

int p;

int phi(int now) {
    
    //求phi值
	int re = now;
	for (int i = 2; i * i <= now; i++)
		if (now % i == 0) {
    
    
			re = re / i * (i - 1);
			while (now % i == 0) now /= i;
		}
	if (now > 1) re = re / now * (now - 1);
	return re;
}

int main() {
    
    
	while (scanf("%d", &p) != EOF) {
    
    
		printf("%d\n", phi(p - 1));//phi(phi(p))，而质数的 phi 值是它减一，所以就省去了一次
	}
	
	return 0;
}