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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
Updated on: 24-Jan-2022
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