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=2;
    int ans=a;
    if(a%2==0){
        //cout<<2<<endl;
        ans=ans/i*(i-1);
        while(a%2==0){
              a/=2;
        }
    }
    for(i=3;i<=a;i+=2){//优化 
    if(a%i==0){
        //cout<<i<<endl;
        ans=ans/i*(i-1);
        while(a%i==0){
              a/=i;
            }
        }
    }
    return ans;
}
int main()//23, 28, and 33
{
    int p;
    while(cin>>p){
        cout<<euler(p-1)<<endl;
    }
    return 0;
}