Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3155		Accepted: 1817

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. 

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

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

23

31

79

10

8

24

 #include<iostream>

 using namespace std;

 int euler(int a){

     int i=;

     int ans=a;

     if(a%==){

         //cout<<2<<endl;

         ans=ans/i*(i-);

         while(a%==){

               a/=;

         }

     }

     for(i=;i<=a;i+=){//优化

     if(a%i==){

         //cout<<i<<endl;

         ans=ans/i*(i-);

         while(a%i==){

               a/=i;

             }

         }

     }

     return ans;

 }

 int main()//23, 28, and 33

 {

     int p;

     while(cin>>p){

         cout<<euler(p-)<<endl;

     }

     return ;

 }