Codeforces Round #174 (Div. 2) Problem A

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博客介绍了质数p的原始根的概念，即一个整数x（1≤x<p），使得x的幂次从x^2-1到x^p-1-1都不被p整除，但x^p-1-1可以。问题要求帮助计算给定质数p的原始根的数量。输入为一个质数p（2≤p<2000），输出是原始根的个数。示例中，当p=2时，唯一的原始根是2；当p=3时，原始根为2和3。

The cows have just learned what a primitive root is! Given a prime p, a primitive root $\text{[math]}$ is an integer x (1 ≤ x < p) such that none of integers x - 1, x² - 1, ..., x^p - 2 - 1 are divisible by p, but x^p - 1 - 1 is.

Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots $\text{[math]}$ .

Input

The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.

Output

Output on a single line the number of primitive roots $\text{[math]}$ .

Sample test(s)

input
3

output
1

input
5

output
2

Note

The only primitive root $\text{[math]}$ is 2.

The primitive roots $\text{[math]}$ are 2 and 3.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <iomanip>

using namespace std;
int main(){
	
	int p;
	cin>>p;
	int ans=0;
	for(int x = 1; x < p; x++) //遍历x 
	{
		int tmp=1;
		for(int j=1; j<p; j++) //x的j次方  
		{
			tmp*=x;
			tmp%=p;   //防止数据溢出，求余操作分歩 
			if(tmp==1)//即满足（x^tmp - 1）% p == 0 
			{
				if(j==p-1) // 即符合题意 （x^(p-1) - 1）% p == 0 
					ans++;
				break;
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}