Given a perfect square natural number N. The task is to find all the factors of N.
Examples
Input: N = 100
Output: 1 2 4 5 10 20 25 50 100Input: N = 900
Output: 1 2 4 3 6 12 9 18 36 5 10 20 15 30 60 45 90 180 25 50 100 75 150 300 225 450 900
Approach:
- Find the square root of N in temp.
- Find all the prime factors of temp in O(sqrt(temp)) using the approach discussed in this article.
- Initialise an array factor[] with element 1 in it.
- Store all the prime factors of temp obtained in above step twice in an array factor[].
- Initialise a matrix M such that for every element in factor[] starting from index 1:
- If factor[i] is equals to factor[i-1], then store factor[i]*factor[i-1] in matrix M in row i – 1 .
- Else factor[i] is not equals to factor[i-1], then store factor[i]*factor[i-1] in matrix M in row i.
- Initialise two arrays arr1[] and arr2[] with the element 1 in both the array.
- Iterate over every row of matrix M such that the product of every element in arr1[] with every element of current row must be stored in arr2[].
- After above step, copy every element of arr2[] in arr1[].
- Repeat above two steps, till all the element of matrixM is traverse.
- The array arr2[] contains all the factors of number N.
Below is the implementation of the above approach:
CPP
// C++ program to find the factors // of large perfect square number // in O(sqrt(sqrt(N))) time #include "bits/stdc++.h" using namespace std;
int MAX = 100000;
// Function that find all the prime // factors of N void findFactors(int N)
{ // Store the sqrt(N) in temp
int temp = sqrt(N);
// Initialise factor array with
// 1 as a factor in it
int factor[MAX] = { 1 };
int i, j, k;
int len1 = 1;
// Check divisibility by 2
while (temp % 2 == 0) {
// Store the factors twice
factor[len1++] = 2;
factor[len1++] = 2;
temp /= 2;
}
// Check for other prime
// factors other than 2
for (j = 3; j < sqrt(temp); j += 2) {
// If j is a prime factor
while (temp % j == 0) {
// Store the prime
// factor twice
factor[len1++] = j;
factor[len1++] = j;
temp /= j;
}
}
// If j is prime number left
// other than 2
if (temp > 2) {
// Store j twice
factor[len1++] = temp;
factor[len1++] = temp;
}
// Intialise Matrix M to
// to store all the factors
int M[len1][MAX] = { 0 };
// tpc for rows
// tpr for column
int tpc = 0, tpr = 0;
// Initialise M[0][0] = 1 as
// it also factor of N
M[0][0] = 1;
j = 1;
// Traversing factor array
while (j < len1) {
// If current and previous
// factors are not same then
// move to next row and
// insert the current factor
if (factor[j] != factor[j - 1]) {
tpr++;
M[tpr][0] = factor[j];
j++;
tpc = 1;
}
// If current and previous
// factors are same then,
// Insert the factor with
// previous factor inserted
// in matrix M
else {
M[tpr][tpc]
= M[tpr][tpc - 1] * factor[j];
j++;
tpc++;
}
}
// The arr1[] and arr2[] used to
// store all the factors of N
int arr1[MAX], arr2[MAX];
int l1, l2;
l1 = l2 = 1;
// Initialise arrays as 1
arr1[0] = arr2[0] = 1;
// Traversing the matrix M
for (i = 1; i < tpr + 1; i++) {
// Traversing till column
// element doesn't become 0
for (j = 0; M[i][j] != 0; j++) {
// Store the product of
// every element of current
// row with every element
// in arr1[]
for (k = 0; k < l1; k++) {
arr2[l2++]
= arr1[k] * M[i][j];
}
}
// Copying every element of
// arr2[] in arr1[]
for (j = l1; j < l2; j++) {
arr1[j] = arr2[j];
}
// length of arr2[] and arr1[]
// are equal after copying
l1 = l2;
}
// Print all the factors
for (i = 0; i < l2; i++) {
cout << arr2[i] << ' ';
}
} // Drivers Code int main()
{ int N = 900;
findFactors(N);
return 0;
} |
Java
// Java program to find the factors // of large perfect square number // in O(Math.sqrt(Math.sqrt(N))) time import java.util.*;
class GFG{
static int MAX = 100000;
// Function that find all the prime // factors of N static void findFactors(int N)
{ // Store the Math.sqrt(N) in temp
int temp = (int) Math.sqrt(N);
// Initialise factor array with
// 1 as a factor in it
int []factor = new int[MAX];
Arrays.fill(factor, 1);
int i, j, k;
int len1 = 1;
// Check divisibility by 2
while (temp % 2 == 0) {
// Store the factors twice
factor[len1++] = 2;
factor[len1++] = 2;
temp /= 2;
}
// Check for other prime
// factors other than 2
for (j = 3; j < Math.sqrt(temp); j += 2) {
// If j is a prime factor
while (temp % j == 0) {
// Store the prime
// factor twice
factor[len1++] = j;
factor[len1++] = j;
temp /= j;
}
}
// If j is prime number left
// other than 2
if (temp > 2) {
// Store j twice
factor[len1++] = temp;
factor[len1++] = temp;
}
// Intialise Matrix M to
// to store all the factors
int [][]M = new int[len1][MAX];
// tpc for rows
// tpr for column
int tpc = 0, tpr = 0;
// Initialise M[0][0] = 1 as
// it also factor of N
M[0][0] = 1;
j = 1;
// Traversing factor array
while (j < len1) {
// If current and previous
// factors are not same then
// move to next row and
// insert the current factor
if (factor[j] != factor[j - 1]) {
tpr++;
M[tpr][0] = factor[j];
j++;
tpc = 1;
}
// If current and previous
// factors are same then,
// Insert the factor with
// previous factor inserted
// in matrix M
else {
M[tpr][tpc]
= M[tpr][tpc - 1] * factor[j];
j++;
tpc++;
}
}
// The arr1[] and arr2[] used to
// store all the factors of N
int []arr1 = new int[MAX];
int []arr2 = new int[MAX];
int l1, l2;
l1 = l2 = 1;
// Initialise arrays as 1
arr1[0] = arr2[0] = 1;
// Traversing the matrix M
for (i = 1; i < tpr + 1; i++) {
// Traversing till column
// element doesn't become 0
for (j = 0; M[i][j] != 0; j++) {
// Store the product of
// every element of current
// row with every element
// in arr1[]
for (k = 0; k < l1; k++) {
arr2[l2++]
= arr1[k] * M[i][j];
}
}
// Copying every element of
// arr2[] in arr1[]
for (j = l1; j < l2; j++) {
arr1[j] = arr2[j];
}
// length of arr2[] and arr1[]
// are equal after copying
l1 = l2;
}
// Print all the factors
for (i = 0; i < l2; i++) {
System.out.print(arr2[i] + " ");
}
} // Drivers Code public static void main(String[] args)
{ int N = 900;
findFactors(N);
} } // This code is contributed by sapnasingh4991 |
Python3
# Python 3 program to find the factors # of large perfect square number # in O(sqrt(sqrt(N))) time import math
MAX = 100000
# Function that find all the prime # factors of N def findFactors( N):
# Store the sqrt(N) in temp
temp = int(math.sqrt(N))
# Initialise factor array with
# 1 as a factor in it
factor = [1]*MAX
len1 = 1
# Check divisibility by 2
while (temp % 2 == 0) :
# Store the factors twice
factor[len1] = 2
len1 += 1
factor[len1] = 2
len1 += 1
temp //= 2
# Check for other prime
# factors other than 2
sqt = math.sqrt(temp)
for j in range(3, math.ceil(sqt), 2):
# If j is a prime factor
while (temp % j == 0):
# Store the prime
# factor twice
factor[len1] = j
len1 += 1
factor[len1] = j
len1 += 1
temp //= j
# If j is prime number left
# other than 2
if (temp > 2) :
# Store j twice
factor[len1] = temp
len1 += 1
factor[len1] = temp
len1 += 1
# Intialise Matrix M to
# to store all the factors
M = [ [ 0 for x in range(MAX)] for y in range(len1)]
# tpc for rows
# tpr for column
tpc , tpr = 0 , 0
# Initialise M[0][0] = 1 as
# it also factor of N
M[0][0] = 1
j = 1
# Traversing factor array
while (j < len1):
# If current and previous
# factors are not same then
# move to next row and
# insert the current factor
if (factor[j] != factor[j - 1]):
tpr+=1
M[tpr][0] = factor[j]
j += 1
tpc = 1
# If current and previous
# factors are same then,
# Insert the factor with
# previous factor inserted
# in matrix M
else :
M[tpr][tpc]= M[tpr][tpc - 1] * factor[j]
j += 1
tpc += 1
# The arr1[] and arr2[] used to
# store all the factors of N
arr1 = [0]*MAX
arr2 = [0]*MAX
l1 = l2 = 1
# Initialise arrays as 1
arr1[0] = 1
arr2[0] = 1
# Traversing the matrix M
# print("tpr ",tpr)
for i in range(1 , tpr + 1) :
# Traversing till column
# element doesn't become 0
j = 0
while M[i][j] != 0:
# Store the product of
# every element of current
# row with every element
# in arr1[]
for k in range(l1):
arr2[l2]= arr1[k] * M[i][j]
l2 += 1
j += 1
# Copying every element of
# arr2[] in arr1[]
for j in range(l1, l2):
arr1[j] = arr2[j]
# length of arr2[] and arr1[]
# are equal after copying
l1 = l2
# Print all the factors
for i in range(l2):
print(arr2[i] ,end= " ")
# Drivers Code if __name__ == "__main__":
N = 900
findFactors(N)
# This code is contributed by chitranayal |
C#
// C# program to find the factors // of large perfect square number // in O(Math.Sqrt(Math.Sqrt(N))) time using System;
class GFG{
static int MAX = 100000;
// Function that find all the prime // factors of N static void findFactors(int N)
{ // Store the Math.Sqrt(N) in temp
int temp = (int) Math.Sqrt(N);
// Initialise factor array with
// 1 as a factor in it
int []factor = new int[MAX];
for(int l= 0; l < MAX; l++)
factor[l] = 1;
int i, j, k;
int len1 = 1;
// Check divisibility by 2
while (temp % 2 == 0) {
// Store the factors twice
factor[len1++] = 2;
factor[len1++] = 2;
temp /= 2;
}
// Check for other prime
// factors other than 2
for (j = 3; j < Math.Sqrt(temp); j += 2) {
// If j is a prime factor
while (temp % j == 0) {
// Store the prime
// factor twice
factor[len1++] = j;
factor[len1++] = j;
temp /= j;
}
}
// If j is prime number left
// other than 2
if (temp > 2) {
// Store j twice
factor[len1++] = temp;
factor[len1++] = temp;
}
// Intialise Matrix M to
// to store all the factors
int [,]M = new int[len1, MAX];
// tpc for rows
// tpr for column
int tpc = 0, tpr = 0;
// Initialise M[0,0] = 1 as
// it also factor of N
M[0, 0] = 1;
j = 1;
// Traversing factor array
while (j < len1) {
// If current and previous
// factors are not same then
// move to next row and
// insert the current factor
if (factor[j] != factor[j - 1]) {
tpr++;
M[tpr, 0] = factor[j];
j++;
tpc = 1;
}
// If current and previous
// factors are same then,
// Insert the factor with
// previous factor inserted
// in matrix M
else {
M[tpr,tpc]
= M[tpr,tpc - 1] * factor[j];
j++;
tpc++;
}
}
// The arr1[] and arr2[] used to
// store all the factors of N
int []arr1 = new int[MAX];
int []arr2 = new int[MAX];
int l1, l2;
l1 = l2 = 1;
// Initialise arrays as 1
arr1[0] = arr2[0] = 1;
// Traversing the matrix M
for (i = 1; i < tpr + 1; i++) {
// Traversing till column
// element doesn't become 0
for (j = 0; M[i, j] != 0; j++) {
// Store the product of
// every element of current
// row with every element
// in arr1[]
for (k = 0; k < l1; k++) {
arr2[l2++]
= arr1[k] * M[i, j];
}
}
// Copying every element of
// arr2[] in arr1[]
for (j = l1; j < l2; j++) {
arr1[j] = arr2[j];
}
// length of arr2[] and arr1[]
// are equal after copying
l1 = l2;
}
// Print all the factors
for (i = 0; i < l2; i++) {
Console.Write(arr2[i] + " ");
}
} // Drivers Code public static void Main(String[] args)
{ int N = 900;
findFactors(N);
} } // This code is contributed by sapnasingh4991 |
Output:
1 2 4 3 6 12 9 18 36 5 10 20 15 30 60 45 90 180 25 50 100 75 150 300 225 450 900
Time Complexity: O(sqrt(sqrt(N)))
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