Given a positive integer N, the task is to find the count of all the numbers M such that when the number N is divided by M, the quotient is equal to its remainder i.e (⌊N/M⌋ = N mod M) where ⌊ ⌋ denotes the floor value of a given number.
Examples:
Input: N = 8
Output: 2
Explanation: When 8 is divided by 3 and 7, it returns the same Quotient and Remainder.
8 / 3 = 8 % 3, 8 / 7 = 8 % 7. Therefore, the required answer is 2.Input: N = 1000000000000
Output: 84
Naive Approach: The simplest approach is based on the fact that M can not be greater than N as for any number greater than N, the quotient would be zero. Whereas, its modulo with N will always be N. Therefore, iterate through all the numbers from 1 to N and count all such numbers satisfying the given condition.
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach: The optimal idea is based on the observation stated below:
If (⌊N/M⌋ = N mod M), then M + 1 must be a divisor of N.
Proof for the observation:
If ⌊N/M⌋ = N mod M, then let N / M be equal to K.
Therefore, N must be equal to K * M + K as K is the quotient as well as the remainder.
N = K * M + K
N = K * (M + 1)
Therefore, for M to be in our answer set, it is necessary that M + 1 is a divisor of N.
Note that M + 1 must be a divisor of N is a necessary condition but not a sufficient condition for ⌊N/M⌋ = N mod M.
Follow the below steps to solve the problem:
- Find all divisors of N and store them in an array. This can be computed in O(√ N) complexity.
- Check for all divisors by iterating through the array, and if divisor minus 1 satisfy the given condition ⌊N / M⌋ = N mod M, increase the count.
Below is the implementation of the above approach:
C++
// C++ program of the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the// count of numbers such that it// gives same quotient and remaindervoid countDivisors(long long int n){ int count = 0; // Stores divisor of number N. vector<long long int> divisor; // Iterate through numbers from 2 // to sqrt(N) and store divisors of N for (int i = 2; i <= sqrt(n); i++) { if (n % i == 0) { if (n / i == i) divisor.push_back(i); else { divisor.push_back(i); divisor.push_back(n / i); } } } // As N is also divisor of itself divisor.push_back(n); // Iterate through divisors for (auto x : divisor) { x -= 1; // Checking whether x satisfies the // required condition if ((n / x) == (n % x)) count++; } // Print the count cout << count;}// Driver Codeint main(){ // Given N long long int N = 1000000000000; // Function Call countDivisors(N); return 0;} |
Java
// Java program of the above approachclass GFG { // Function to calculate the // count of numbers such that it // gives same quotient and remainder static void countDivisors(int n) { int count = 0; int j = 0; // Stores divisor of number N. int divisor[] = new int[n]; // Iterate through numbers from 2 // to sqrt(N) and store divisors of N for (int i = 2; i <= Math.sqrt(n); i++) { if (n % i == 0) { if (n / i == i){ divisor[j] = i; j += 1; } else { divisor[j] = i; divisor[j + 1] = n / i; j += 2; } } } // As N is also divisor of itself divisor[j] = n; // Iterate through divisors for (int i = 0; i <= j; i++) { int x = divisor[i]; x -= 1; // Checking whether x satisfies the // required condition if ((n / x) == (n % x)) count++; } // Print the count System.out.print(count); } // Driver Code public static void main (String[] args) { // Given N int N = 10000000; // Function Call countDivisors(N); }}// This code is contributed by AnkThon |
Python3
# Python3 program of the above approachfrom math import floor, sqrt# Function to calculate the# count of numbers such that it# gives same quotient and remainderdef countDivisors(n): count = 0 # Stores divisor of number N. divisor = [] # Iterate through numbers from 2 # to sqrt(N) and store divisors of N for i in range(2, floor(sqrt(n))): if (n % i == 0): if (n // i == i): divisor.append(i) else: divisor.append(i) divisor.append(n // i) # As N is also divisor of itself divisor.append(n) # Iterate through divisors for x in divisor: x -= 1 # Checking whether x satisfies the # required condition if ((n // x) == (n % x)): count += 1 # Print the count print(count) # Driver Codeif __name__ == '__main__': # Given N N = 1000000000000 # Function Call countDivisors(N)# This code is contributed by mohit kumar 29 |
C#
// C# program of the above approachusing System;class GFG { // Function to calculate the // count of numbers such that it // gives same quotient and remainder static void countDivisors(int n) { int count = 0; int j = 0; // Stores divisor of number N. int []divisor = new int[n]; // Iterate through numbers from 2 // to sqrt(N) and store divisors of N for (int i = 2; i <= Math.Sqrt(n); i++) { if (n % i == 0) { if (n / i == i){ divisor[j] = i; j += 1; } else { divisor[j] = i; divisor[j + 1] = n / i; j += 2; } } } // As N is also divisor of itself divisor[j] = n; // Iterate through divisors for (int i = 0; i <= j; i++) { int x = divisor[i]; x -= 1; // Checking whether x satisfies the // required condition if ((n / x) == (n % x)) count++; } // Print the count Console.Write(count); } // Driver Code public static void Main(String[] args) { // Given N int N = 10000000; // Function Call countDivisors(N); }}// This code contributed by shikhasingrajput |
84
Time Complexity: O(sqrt(N))
Auxiliary Space: O(sqrt(N))
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