Codeforces Round #174 (Div. 2) Cows and Primitive Roots（数论）

Cows and Primitive Roots

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The cows have just learned what a primitive root is! Given a prime p, a primitive root 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 .

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 .

Sample test(s)

Input

3

Output

1

Input

5

Output

2

Note

The only primitive root is 2.

The primitive roots are 2 and 3.

可使用蛮力法，但需要注意两点：

1、不要每次都用pow计算x的n次方，由于x^n=x*X^(n-1)，设一个变量m储存x^(n-1)，那么x^n=m*x.

2、如果直接算出m，就算用64位也会溢出，利用性质（a-b）%p=(a%p-b%p)%p，则只需保留m%p的值。

AC Code：

#include <iostream>
#include <fstream>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <utility>
using namespace std;
#define ll long long
#define cti const int
#define ctll const long long
#define dg(i) cout << '*' << i << endl;

int main()
{
    ll p, x;
    ll m;
    int ans;
    bool ok;
    while(scanf("%I64d", &p) != EOF)
    {
        ans = 0;
        for(x = 1; x < p; x++)
        {
            m = 1;
            ok = true;
            for(int i = 1; i < p - 1; i++)
            {
                m *= x;
                m %= p;
                if((m - 1) % p == 0)
                {
                    ok = false;
                    break;
                }
            }
            if(ok && ((m * x) - 1) % p == 0) ans++;
        }
        printf("%d\n", ans);
    }
    return 0;
}