CSES Solutions – Counting Divisors

Given N integers as an array arr[], your task is to report for each integer the number of its divisors.

For example, if x=18, the correct answer is 6 because its divisors are 1,2,3,6,9,18.

Examples:

Input: N = 3, arr[] = {16, 17, 18}
Output:
5
2
6
Explanation:

• There are 5 divisors of 16: 1, 2, 4, 8 and 16.
• There are 2 divisors of 17: 1 and 17.
• There are 6 divisors of 18: 1, 2, 3, 6, 9 and 18.

Input: N = 3, arr[] = {5, 6, 7}
Output:
2
4
2
Explanation:

• There are 2 divisors of 5: 1 and 5.
• There are 4 divisors of 6: 1, 2, 3 and 6.
• There are 2 divisors of 7: 1 and 7.

Approach: To solve the problem, follow the below idea:

The problem can be solved by Precomputing the number of divisors for each number up to 1,000,004. Then, uses a nested loop to iterate over each number and its multiples, incrementing a counter for each multiple. This results in an array where the value at each index is the number of divisors for that index.

Step-by-step algorithm:

• Create an array numDivisors to store the number of divisors for each number up to 1000004.
• Iterate through each number i from 1 to 1000004.
• For each i, iterate through multiples of i from j = i to 1000004.
• Increment numDivisors[j] by 1 for each multiple.

Below is the implementation of the algorithm:

C++

#include <iostream>
using namespace std;

// Function to calculate the number of divisors for each
// number up to 1000004
void calculateDivisors(long long numDivisors[])
{
    for (int i = 1; i < 1000005; i++) {
        for (int j = i; j < 1000005; j += i) {
            numDivisors[j]++;
        }
    }
}

int main()
{
    // Array to store the number of divisors for each number
    // up to 1000004
    long long numDivisors[1000005] = {};

    // Call the function to calculate the number of divisors
    calculateDivisors(numDivisors);

    // Read the number of integers
    long long n = 3;
      long long arr[] = {16, 17, 18};
    for(int i=0; i < n; i++) {
        cout << numDivisors[arr[i]] << "\n";
    }
}

Java

public class Main {
    // Function to calculate the number of divisors for each
    // number up to 1000004
    static void calculateDivisors(long[] numDivisors) {
        for (int i = 1; i < 1000005; i++) {
            for (int j = i; j < 1000005; j += i) {
                numDivisors[j]++;
            }
        }
    }

    public static void main(String[] args) {
        // Array to store the number of divisors for each number
        // up to 1000004
        long[] numDivisors = new long[1000005];

        // Call the function to calculate the number of divisors
        calculateDivisors(numDivisors);

        // Read the number of integers
        int n = 3;
        long[] arr = {16, 17, 18};
        for (int i = 0; i < n; i++) {
            System.out.println(numDivisors[(int)arr[i]]);
        }
    }
}

Python3

def calculate_divisors(num_divisors):
    for i in range(1, 1000005):
        for j in range(i, 1000005, i):
            num_divisors[j] += 1

def main():
    # Array to store the number of divisors for each number
    # up to 1000004
    num_divisors = [0] * 1000005

    # Call the function to calculate the number of divisors
    calculate_divisors(num_divisors)

    # Read the number of integers
    n = 3
    arr = [16, 17, 18]
    for num in arr:
        print(num_divisors[num])

if __name__ == "__main__":
    main()

JavaScript

// Function to calculate the number of divisors for each
// number up to 1000004
function calculateDivisors(numDivisors) {
    for (let i = 1; i < 1000005; i++) {
        for (let j = i; j < 1000005; j += i) {
            numDivisors[j]++;
        }
    }
}

// Array to store the number of divisors for each number
// up to 1000004
let numDivisors = new Array(1000005).fill(0);

// Call the function to calculate the number of divisors
calculateDivisors(numDivisors);

// Read the number of integers
let n = 3;
let arr = [16, 17, 18];
for (let i = 0; i < n; i++) {
    console.log(numDivisors[arr[i]]);
}

Output

5
2
6

Time complexity: O(N * log(max(arr[i]))), where N is the count of numbers given as input array arr[].
Auxiliary Space: O(N)