Find numbers with K odd divisors in a given range in C++

In this problem, we are given three integer values, L, R, and k. Our task is to find numbers with K odd divisors in a given range. We will be finding the count of numbers in the range [L, R] that have exactly k divisors.

We will be counting the 1 and the number itself as a divisor.

Let’s take an example to understand the problem,

Input

a = 3, b = 10, k = 3

Output

2

Explanation

Numbers with exactly 3 divisors within the range 3 to 10 are 4 : divisors = 1, 2, 4 9 : divisors = 1, 3, 9

Solution Approach

A simple solution to the problem is by counting the k divisors. So,for k to be an odd number (as depicted in problem), the number has to be a perfect square. So, we will count the number of divisors for only the perfect square number (this will save compilation time). And if the count of divisors in k, we will add 1 to the number count.

Program to illustrate the working of our solution,

Example

 Live Demo

#include<bits/stdc++.h> using namespace std; bool isPerfectSquare(int n) {    int s = sqrt(n);    return (s*s == n); } int countDivisors(int n) {    int divisors = 0;    for (int i=1; i<=sqrt(n)+1; i++) {       if (n%i==0) {          divisors++;          if (n/i != i)             divisors ++;       }    }    return divisors; } int countNumberKDivisors(int a,int b,int k) {    int numberCount = 0;    for (int i=a; i<=b; i++) {       if (isPerfectSquare(i))          if (countDivisors(i) == k)             numberCount++;    }    return numberCount; } int main() {    int a = 3, b = 10, k = 3;    cout<<"The count of numbers with K odd divisors is "<<countNumberKDivisors(a, b, k);    return 0; }

Output

The count of numbers with K odd divisors is 2