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 { (x 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.
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
#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 ;
}