poj 1284 Primitive Roots（未完）

Primitive Roots

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

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

 #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 ;
 }