求模根的本底根数

给定素数$p$ 。任务是计算所有的原始根$p$ 。
基本根是整数x(1 <= x ，因此整数x – 1，x ² – 1，…。，x ^{p – 2} – 1都不可被整数整除。 $p$但是x ^{p – 1} – 1可被整除$p$ 。
例子：

Input: P = 3 
Output: 1 
The only primitive root modulo 3 is 2. 
Input: P = 5 
Output: 2 
Primitive roots modulo 5 are 2 and 3. 
 

方法：所有素数始终至少有一个原始根。因此，使用Eulers上位函数，我们可以说f(p-1)是必需的答案，而f(n)是上位上函数。
下面是上述方法的实现：

C++

// CPP program to find the number of
// primitive roots modulo prime
#include 
using namespace std;
 
// Function to return the count of
// primitive roots modulo p
int countPrimitiveRoots(int p)
{
    int result = 1;
    for (int i = 2; i < p; i++)
        if (__gcd(i, p) == 1)
            result++;
 
    return result;
}
 
// Driver code
int main()
{
    int p = 5;
 
    cout << countPrimitiveRoots(p - 1);
 
    return 0;
}

Java

// Java program to find the number of
// primitive roots modulo prime
 
import java.io.*;
 
class GFG {
 // Recursive function to return gcd of a and b
    static int __gcd(int a, int b)
    {
        // Everything divides 0 
        if (a == 0)
          return b;
        if (b == 0)
          return a;
        
        // base case
        if (a == b)
            return a;
        
        // a is greater
        if (a > b)
            return __gcd(a-b, b);
        return __gcd(a, b-a);
    }
 
// Function to return the count of
// primitive roots modulo p
static int countPrimitiveRoots(int p)
{
    int result = 1;
    for (int i = 2; i < p; i++)
        if (__gcd(i, p) == 1)
            result++;
 
    return result;
}
 
// Driver code
    public static void main (String[] args) {
            int p = 5;
 
    System.out.println( countPrimitiveRoots(p - 1));
    }
}
// This code is contributed by anuj_67..

Python3

# Python 3 program to find the number
# of primitive roots modulo prime
from math import gcd
 
# Function to return the count of
# primitive roots modulo p
def countPrimitiveRoots(p):
    result = 1
    for i in range(2, p, 1):
        if (gcd(i, p) == 1):
            result += 1
 
    return result
 
# Driver code
if __name__ == '__main__':
    p = 5
 
    print(countPrimitiveRoots(p - 1))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find the number of
// primitive roots modulo prime
   
using System;
   
class GFG {
 // Recursive function to return gcd of a and b 
    static int __gcd(int a, int b) 
    { 
        // Everything divides 0  
        if (a == 0) 
          return b; 
        if (b == 0) 
          return a; 
          
        // base case 
        if (a == b) 
            return a; 
          
        // a is greater 
        if (a > b) 
            return __gcd(a-b, b); 
        return __gcd(a, b-a); 
    } 
   
// Function to return the count of
// primitive roots modulo p
static int countPrimitiveRoots(int p)
{
    int result = 1;
    for (int i = 2; i < p; i++)
        if (__gcd(i, p) == 1)
            result++;
   
    return result;
}
   
// Driver code
     static public void Main (String []args) {
            int p = 5;
   
    Console.WriteLine( countPrimitiveRoots(p - 1));
    }
}
// This code is contributed by Arnab Kundu

PHP

 $b)
        return __gcd($a - $b, $b);
    return __gcd($a, $b - $a);
}
 
// Function to return the count of
// primitive roots modulo p
function countPrimitiveRoots($p)
{
    $result = 1;
    for ($i = 2; $i < $p; $i++)
        if (__gcd($i, $p) == 1)
            $result++;
 
    return $result;
}
 
// Driver code
$p = 5;
 
echo countPrimitiveRoots($p - 1);
 
// This code is contributed by anuj_67
?>

Javascript

 
// Javascript program to find the number of
// primitive roots modulo prime
 
 // Recursive function to return gcd of a and b 
 function __gcd( a,  b) 
 { 
     // Everything divides 0  
     if (a == 0) 
       return b; 
     if (b == 0) 
       return a; 
       
     // base case 
     if (a == b) 
         return a; 
       
     // a is greater 
     if (a > b) 
         return __gcd(a-b, b); 
     return __gcd(a, b-a); 
 } 
   
 
// Function to return the count of
// primitive roots modulo p
function countPrimitiveRoots(p)
{
    var result = 1;
    for (var i = 2; i < p; i++)
        if (__gcd(i, p) == 1)
            result++;
 
    return result;
}
 
// Driver code
var p = 5;
document.write( countPrimitiveRoots(p - 1));
 

输出：

2

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