POJ1284(SummerTrainingDay04-K 原根)

Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4505		Accepted: 2652

Description

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

23
31
79

Sample Output

10
8
24

Source

贾怡@pku

 

 //2017-08-04
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 
 using namespace std;
 
 int phi(int n){
     int ans = n;
     for(int i = ; i*i <= n; i++){
         if(n%i==){
             ans -= ans/i;
             while(n%i==)
                 n /= i;
         }
     }
     if(n > )ans = ans - ans/n;
     return ans;
 }
 
 int main()
 {
     int num;
     while(scanf("%d", &num)!=EOF){
         printf("%d\n", phi(num-));
     }
 
     return ;
 }