Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3381		Accepted: 1980

Description

We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (x

ⁱ mod p) | 1 <= i <= p-1 } is equal to { 1, ..., p-1 }. For example, the consecutive powers of 3 modulo 7 are 3, 2, 6, 4, 5, 1, and thus 3 is a primitive root modulo 7. 

Write a program which given any odd prime 3 <= p < 65536 outputs the number of primitive roots modulo p. 

Input

Each line of the input contains an odd prime numbers p. Input is terminated by the end-of-file seperator.

Output

For each p, print a single number that gives the number of primitive roots in a single line.

Sample Input

233179

Sample Output

10824

一个数m的1次方，2次方，3次方到n-1次方 mod n 得到的数值各不相同，就说m是n的原根。

一个数是数n的原根就必然与n-1互质，所以求n的原根的数量即是求欧拉函数n-1。

代码：

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #pragma warning(disable:4996)using namespace std;long long euler(long long n){ long long res = n, a = n; for (long long i = 2; i*i <= a; i++) { if (a%i == 0) { res = res / i*(i - 1); while (a%i == 0)a /= i; } } if (a > 1)res = res / a*(a - 1); return res;}int main(){ long long n; while (cin >> n) { cout << euler(n-1) << endl; } return 0;}
         
        
       
      
     
    

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