给定一个整数N ，任务是使用质数分解来找到所有除数。
例子：
Input: N = 6
Output: 1 2 3 6
Input: N = 10
Output: 1 2 5 10
方法：由于每个大于1的数字都可以在质数分解中表示为p 1 a 1 * p 2 a 2 *……* p k a k ，这里p i是质数，k≥1，a i是a正整数。
现在，如果知道每个素数n的出现次数，则可以递归生成所有可能的除数。对于每个素数因子p i ，可以将其包括x次，其中0≤x≤a i 。首先，使用这种方法找到n的素因式分解，并针对每个素因数，将其与它的出现次数一起存储。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include "iostream"
#include "vector"
using namespace std;
struct primeFactorization {
// to store the prime factor
// and its highest power
int countOfPf, primeFactor;
};
// Recursive function to generate all the
// divisors from the prime factors
void generateDivisors(int curIndex, int curDivisor,
vector& arr)
{
// Base case i.e. we do not have more
// primeFactors to include
if (curIndex == arr.size()) {
cout << curDivisor << ' ';
return;
}
for (int i = 0; i <= arr[curIndex].countOfPf; ++i) {
generateDivisors(curIndex + 1, curDivisor, arr);
curDivisor *= arr[curIndex].primeFactor;
}
}
// Function to find the divisors of n
void findDivisors(int n)
{
// To store the prime factors along
// with their highest power
vector arr;
// Finding prime factorization of n
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
int count = 0;
while (n % i == 0) {
n /= i;
count += 1;
}
// For every prime factor we are storing
// count of it's occurenceand itself.
arr.push_back({ count, i });
}
}
// If n is prime
if (n > 1) {
arr.push_back({ 1, n });
}
int curIndex = 0, curDivisor = 1;
// Generate all the divisors
generateDivisors(curIndex, curDivisor, arr);
}
// Driver code
int main()
{
int n = 6;
findDivisors(n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static class primeFactorization
{
// to store the prime factor
// and its highest power
int countOfPf, primeFactor;
public primeFactorization(int countOfPf,
int primeFactor)
{
this.countOfPf = countOfPf;
this.primeFactor = primeFactor;
}
}
// Recursive function to generate all the
// divisors from the prime factors
static void generateDivisors(int curIndex, int curDivisor,
Vector arr)
{
// Base case i.e. we do not have more
// primeFactors to include
if (curIndex == arr.size())
{
System.out.print(curDivisor + " ");
return;
}
for (int i = 0; i <= arr.get(curIndex).countOfPf; ++i)
{
generateDivisors(curIndex + 1, curDivisor, arr);
curDivisor *= arr.get(curIndex).primeFactor;
}
}
// Function to find the divisors of n
static void findDivisors(int n)
{
// To store the prime factors along
// with their highest power
Vector arr = new Vector<>();
// Finding prime factorization of n
for (int i = 2; i * i <= n; ++i)
{
if (n % i == 0)
{
int count = 0;
while (n % i == 0)
{
n /= i;
count += 1;
}
// For every prime factor we are storing
// count of it's occurenceand itself.
arr.add(new primeFactorization(count, i ));
}
}
// If n is prime
if (n > 1)
{
arr.add(new primeFactorization( 1, n ));
}
int curIndex = 0, curDivisor = 1;
// Generate all the divisors
generateDivisors(curIndex, curDivisor, arr);
}
// Driver code
public static void main(String []args)
{
int n = 6;
findDivisors(n);
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 implementation of the approach
# Recursive function to generate all the
# divisors from the prime factors
def generateDivisors(curIndex, curDivisor, arr):
# Base case i.e. we do not have more
# primeFactors to include
if (curIndex == len(arr)):
print(curDivisor, end = ' ')
return
for i in range(arr[curIndex][0] + 1):
generateDivisors(curIndex + 1, curDivisor, arr)
curDivisor *= arr[curIndex][1]
# Function to find the divisors of n
def findDivisors(n):
# To store the prime factors along
# with their highest power
arr = []
# Finding prime factorization of n
i = 2
while(i * i <= n):
if (n % i == 0):
count = 0
while (n % i == 0):
n //= i
count += 1
# For every prime factor we are storing
# count of it's occurenceand itself.
arr.append([count, i])
# If n is prime
if (n > 1):
arr.append([1, n])
curIndex = 0
curDivisor = 1
# Generate all the divisors
generateDivisors(curIndex, curDivisor, arr)
# Driver code
n = 6
findDivisors(n)
# This code is contributed by SHUBHAMSINGH10C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
public class primeFactorization
{
// to store the prime factor
// and its highest power
public int countOfPf, primeFactor;
public primeFactorization(int countOfPf,
int primeFactor)
{
this.countOfPf = countOfPf;
this.primeFactor = primeFactor;
}
}
// Recursive function to generate all the
// divisors from the prime factors
static void generateDivisors(int curIndex, int curDivisor,
List arr)
{
// Base case i.e. we do not have more
// primeFactors to include
if (curIndex == arr.Count)
{
Console.Write(curDivisor + " ");
return;
}
for (int i = 0; i <= arr[curIndex].countOfPf; ++i)
{
generateDivisors(curIndex + 1, curDivisor, arr);
curDivisor *= arr[curIndex].primeFactor;
}
}
// Function to find the divisors of n
static void findDivisors(int n)
{
// To store the prime factors along
// with their highest power
List arr = new List();
// Finding prime factorization of n
for (int i = 2; i * i <= n; ++i)
{
if (n % i == 0)
{
int count = 0;
while (n % i == 0)
{
n /= i;
count += 1;
}
// For every prime factor we are storing
// count of it's occurenceand itself.
arr.Add(new primeFactorization(count, i ));
}
}
// If n is prime
if (n > 1)
{
arr.Add(new primeFactorization( 1, n ));
}
int curIndex = 0, curDivisor = 1;
// Generate all the divisors
generateDivisors(curIndex, curDivisor, arr);
}
// Driver code
public static void Main(String []args)
{
int n = 6;
findDivisors(n);
}
}
// This code is contributed by PrinciRaj1992 输出：
1 3 2 6