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

In this problem, we are given a number N. Our task is to find the number of divisors of all numbers in the range [1, n].

Let’s take an example to understand the problem,

Input : N = 7
Output : 1 2 2 3 2 4 2

Solution Approach

A simple solution to the problem is by starting from 1 to N and for every number count the number of divisors and print them.

Example 1

Program to illustrate the working of our solution

#include <iostream>
using namespace std;
int countDivisor(int N){
   int count = 1;
   for(int i = 2; i <= N; i++){
      if(N%i == 0)
         count++;
   }
   return count;
}
int main(){
   int N = 8;
   cout<<"The number of divisors of all numbers in the range are \t";
   cout<<"1 ";
   for(int i = 2; i <= N; i++){
      cout<<countDivisor(i)<<" ";
   }
   return 0;
}

Output

The number of divisors of all numbers in the range are 1 2 2 3 2 4 2 4

Another approach to solve the problem is using increments of values. For this, we will create an array of size (N+1). Then, starting from 1 to N, we will check for each value i, we will increment the array value for all multiples of i less than n.

Example 2

Program to illustrate the working of our solution,

#include <iostream>
using namespace std;
void countDivisors(int N){
   int arr[N+1];
   for(int i = 0; i <= N; i++)
      arr[i] = 1;
      for (int i = 2; i <= N; i++) {
         for (int j = 1; j * i <= N; j++)
            arr[i * j]++;
      }
      for (int i = 1; i <= N; i++)
         cout<<arr[i]<<" ";
}
int main(){
   int N = 8;
   cout<<"The number of divisors of all numbers in the range are \t"; countDivisors(N);
   return 0;
}

Output

The number of divisors of all numbers in the range are 1 2 2 3 2 4 2 4

Sudhir sharma

Published on 24-Jan-2022 12:45:08

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