Given a number n, count total perfect divisors of n. Perfect divisors are those divisors which are square of some integer. For example a perfect divisor of 8 is 4.

Input : n = 16 Output : 3 Explanation : There are only 5 divisor of 16: 1, 2, 4, 8, 16. Only three of them are perfect squares: 1, 4, 16. Therefore the answer is 3 Input : n = 7 Output : 1

**Naive approach**

A brute force is find all the divisors of a number. Count all divisors that are perfect squares.

## C++

// Below is C++ code to count total perfect Divisors #include<bits/stdc++.h> using namespace std; // Utility function to check perfect square number bool isPerfectSquare(int n) { int sq = (int) sqrt(n); return (n == sq * sq); } // Returns count all perfect divisors of n int countPerfectDivisors(int n) { // Initialize result int count = 0; // Consider every number that can be a divisor // of n for (int i=1; i*i <= n; ++i) { // If i is a divisor if (n%i == 0) { if (isPerfectSquare(i)) ++count; if (n/i != i && isPerfectSquare(n/i)) ++count; } } return count; } // Driver code int main() { int n = 16; cout << "Total perfect divisors of " << n << " = " << countPerfectDivisors(n) << "\n"; n = 12; cout << "Total perfect divisors of " << n << " = " << countPerfectDivisors(n); return 0; }

## Python3

# Python3 implementation of Naive method # to count all perfect divisors import math def isPerfectSquare(x) : sq = (int)(math.sqrt(x)) return (x == sq * sq) # function to count all prefect divisors def countPerfectDivisors(n) : # Initialize result cnt = 0 # Consider every number that # can be a divisor of n for i in range(1, (int)(math.sqrt(n)) + 1) : # If i is a divisor if ( n % i == 0 ) : if isPerfectSquare(i): cnt = cnt + 1 if n/i != i and isPerfectSquare(n/i): cnt = cnt + 1 return cnt # Driver program to test above function print("Total perfect divisor of 16 = ", countPerfectDivisors(16)) print("Total perfect divisor of 12 = ", countPerfectDivisors(12)) # This code is contributed by Saloni Gupta

Output:Total Perfect divisors of 16 = 3 Total Perfect divisors of 12 = 2

**Time complexity: **O(sqrt(n))

**Auxiliary space: **O(1)

**Efficient approach (For large number of queries)**

The idea is based on Sieve of Eratosthenes. This approach is beneficial if there are large number of queries. Following is the algorithm to find all perfect divisors up to n numbers.

- Create a list of n consecutive integers from 1 to n:(1, 2, 3, …, n)
- Initially, let d be 2, the first divisor
- Starting from 4(square of 2) increment all the multiples of 4 by 1 in perfectDiv[] array, as all these multiples contain 4 as perfect divisor. These numbers will be 4d, 8d, 12d, … etc
- Repeat the 3
^{rd}step for all other numbers. The final array of perfectDiv[] will contain all the count of perfect divisors from 1 to n

Below is C++ implementation of above steps.

// Below is C++ code to count total perfect // divisors #include<bits/stdc++.h> using namespace std; #define MAX 100001 int perfectDiv[MAX]; // Pre-compute counts of all perfect divisors // of all numbers upto MAX. void precomputeCounts() { for (int i=1; i*i < MAX; ++i) { // Iterate through all the multiples of i*i for (int j=i*i; j < MAX; j += i*i) // Increment all such multiples by 1 ++perfectDiv[j]; } } // Returns count of perfect divisors of n. int countPerfectDivisors(int n) { return perfectDiv[n]; } // Driver code int main() { precomputeCounts(); int n = 16; cout << "Total perfect divisors of " << n << " = " << countPerfectDivisors(n) << "\n"; n = 12; cout << "Total perfect divisors of " << n << " = " << countPerfectDivisors(n); return 0; }

Output:Total Perfect divisors of 16 = 3 Total Perfect divisors of 12 = 2

**Time complexity: **O(MAX * log(log (MAX)))

**Auxiliary space: **O(MAX)

This article is contributed by Shubham Bansal. 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.

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