📜  求模根的本底根数

📅  最后修改于: 2021-04-23 05:27:08             🧑  作者: Mango

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:

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
?>


输出:
2