数论|加法作用下有限循环群的生成器

给定数字n，找到模n下所有循环加性基团的生成器。集合{0，1，…n-1}的生成器是元素x，使得x小于n，并且使用x(和加法运算)，我们可以生成集合中的所有元素。
例子：

Input : 10
Output : 1 3 7 9
The set to be generated is {0, 1, .. 9}
By adding 1, single or more times, we 
can create all elements from 0 to 9.
Similarly using 3, we can generate all
elements.
30 % 10 = 0, 21 % 10 = 1, 12 % 10 = 2, ...
Same is true for 7 and 9.

Input  : 24
Output : 1 5 7 11 13 17 19 23

一个简单的解决方案是运行一个从1到n-1的循环，并检查每个元素是否是生成器。为了检查生成器，我们不断添加元素，并检查是否可以生成所有数字，直到余数开始重复。
一种有效的解决方案基于以下事实：如果x相对于n质数，即gcd(n，x)= 1，则x是生成器。
下面是上述方法的实现：

C++

// A simple C++ program to find all generators
#include 
using namespace std;
 
// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b%a, a);
}
 
// Print generators of n
int printGenerators(unsigned int n)
{
    // 1 is always a generator
    cout << "1 ";
 
    for (int i=2; i < n; i++)
 
        // A number x is generator of GCD is 1
        if (gcd(i, n) == 1)
            cout << i << " ";
}
 
// Driver program to test above function
int main()
{
    int n = 10;
    printGenerators(n);
    return 0;
}

Java

// A simple Java program to find all generators
 
class GFG {
     
 
// Function to return gcd of a and b
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b%a, a);
}
 
// Print generators of n
static void printGenerators(int n)
{
    // 1 is always a generator
    System.out.println("1 ");
 
    for (int i=2; i < n; i++)
 
        // A number x is generator of GCD is 1
        if (gcd(i, n) == 1)
            System.out.println(i +" ");
}
 
// Driver program to test above function
public static void main(String args[])
{
    int n = 10;
    printGenerators(n);
}
}

Python3

# Python3 program to find all generators
 
# Function to return gcd of a and b
def gcd(a, b):
    if (a == 0):
        return b;
    return gcd(b % a, a);
 
# Print generators of n
def printGenerators(n):
     
    # 1 is always a generator
    print("1", end = " ");
 
    for i in range(2, n):
 
        # A number x is generator
        # of GCD is 1
        if (gcd(i, n) == 1):
            print(i, end = " ");
 
# Driver Code
n = 10;
printGenerators(n);
     
# This code is contributed by mits

C#

// A simple C# program to find all generators
using System;
 
class GFG
{
     
// Function to return gcd of a and b
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Print generators of n
static void printGenerators(int n)
{
    // 1 is always a generator
    Console.Write("1 ");
 
    for (int i = 2; i < n; i++)
 
        // A number x is generator of GCD is 1
        if (gcd(i, n) == 1)
            Console.Write(i +" ");
}
 
// Driver code
public static void Main(String []args)
{
    int n = 10;
    printGenerators(n);
}
}
 
// This code contributed by Rajput-Ji

Javascript

 
// A simple Javascript program to
// find all generators
 
// Function to return gcd of a and b
function gcd(a, b)
{
    if (a == 0)
        return b;
         
    return gcd(b % a, a);
}
 
// Print generators of n
function printGenerators(n)
{
     
    // 1 is always a generator
    document.write("1 ");
 
    for(var i = 2; i < n; i++)
 
        // A number x is generator of
        // GCD is 1
        if (gcd(i, n) == 1)
            document.write(i + " ");
}
 
// Driver Code
var n = 10;
 
printGenerators(n);
 
// This code is contributed by Kirti
 

输出 ：

1 3 7 9