# Find the number of divisors of all numbers in the range [1, n]

Given an integer N. The task is to find the number of divisors of all the numbers in the range [1, N].

Examples:

Input: N = 5
Output: 1 2 2 3 2
divisors(1) = 1
divisors(2) = 1 and 2
divisors(3) = 1 and 3
divisors(4) = 1, 2 and 4
divisors(5) = 1 and 5

Input: N = 10
Output: 1 2 2 3 2 4 2 4 3 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Create an array arr[] of the size (N + 1) where arr[i] stores the number of divisors of i. Now for every j from the range [1, N], increment all the elements which are divisible by j.
For example, if j = 3 then update arr[3], arr[6], arr[9], …

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the number of divisors ` `// of all numbers in the range [1, n] ` `void` `findDivisors(``int` `n) ` `{ ` ` `  `    ``// Array to store the count ` `    ``// of divisors ` `    ``int` `div``[n + 1]; ` `    ``memset``(``div``, 0, ``sizeof` `div``); ` ` `  `    ``// For every number from 1 to n ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` ` `  `        ``// Increase divisors count for ` `        ``// every number divisible by i ` `        ``for` `(``int` `j = 1; j * i <= n; j++) ` `            ``div``[i * j]++; ` `    ``} ` ` `  `    ``// Print the divisors ` `    ``for` `(``int` `i = 1; i <= n; i++) ` `        ``cout << ``div``[i] << ``" "``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 10; ` `    ``findDivisors(n); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach  ` ` `  `class` `GFG  ` `{  ` `     `  `    ``// Function to find the number of divisors  ` `    ``// of all numbers in the range [1, n]  ` `    ``static` `void` `findDivisors(``int` `n)  ` `    ``{  ` `     `  `        ``// Array to store the count  ` `        ``// of divisors  ` `        ``int``[] div = ``new` `int``[n + ``1``];  ` `     `  `        ``// For every number from 1 to n  ` `        ``for` `(``int` `i = ``1``; i <= n; i++)  ` `        ``{  ` `     `  `            ``// Increase divisors count for  ` `            ``// every number divisible by i  ` `            ``for` `(``int` `j = ``1``; j * i <= n; j++)  ` `                ``div[i * j]++;  ` `        ``}  ` `     `  `        ``// Print the divisors  ` `        ``for` `(``int` `i = ``1``; i <= n; i++)  ` `            ``System.out.print(div[i]+``" "``);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String args[])  ` `    ``{  ` `        ``int` `n = ``10``;  ` `        ``findDivisors(n);  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Ryuga `

 `# Python3 implementation of the approach ` `# Function to find the number of divisors ` `# of all numbers in the range [1,n] ` `def` `findDivisors(n): ` `     `  `    ``# List to store the count ` `    ``# of divisors ` `    ``div ``=` `[``0` `for` `i ``in` `range``(n ``+` `1``)] ` `     `  `    ``# For every number from 1 to n ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``): ` `         `  `        ``# Increase divisors count for ` `        ``# every number divisible by i ` `        ``for` `j ``in` `range``(``1``, n ``+` `1``): ` `            ``if` `j ``*` `i <``=` `n: ` `                ``div[i ``*` `j] ``+``=` `1` ` `  `    ``# Print the divisors ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``): ` `        ``print``(div[i], end ``=` `" "``) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``n ``=` `10` `    ``findDivisors(n) ` ` `  `# This code is contributed by ` `# Vivek Kumar Singh `

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to find the number of divisors ` `// of all numbers in the range [1, n] ` `static` `void` `findDivisors(``int` `n) ` `{ ` ` `  `    ``// Array to store the count ` `    ``// of divisors ` `    ``int``[] div = ``new` `int``[n + 1]; ` ` `  `    ``// For every number from 1 to n ` `    ``for` `(``int` `i = 1; i <= n; i++)  ` `    ``{ ` ` `  `        ``// Increase divisors count for ` `        ``// every number divisible by i ` `        ``for` `(``int` `j = 1; j * i <= n; j++) ` `            ``div[i * j]++; ` `    ``} ` ` `  `    ``// Print the divisors ` `    ``for` `(``int` `i = 1; i <= n; i++) ` `        ``Console.Write(div[i]+``" "``); ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``int` `n = 10; ` `    ``findDivisors(n); ` `} ` `} ` ` `  `// This code is contributed by mits `

 ` `

Output:
```1 2 2 3 2 4 2 4 3 4
```

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :