Count of divisors having more set bits than quotient on dividing N
Given a positive integer N. The task is to find the number of divisor (say d) which give quotient (say q) on integer division (i.e q = N/d) such that quotient have less or equal set bit than divisor. In other words, find number of possible value of ‘d’ which will produce ‘q’ on integer dividing ‘n’ (q = N/d) such that ‘q’ have less or equal set bits (atleast 1) than ‘d’.
Examples :
Input : N = 5
Output : 4
for d = 1 (set bit = 1),
q = 5/1 = 5 (set bit = 2), count = 0.
for d = 2 (set bit = 1),
q = 5/2 = 2 (set bit = 1), count = 1.
for d = 3 (set bit = 2),
q = 5/3 = 1 (set bit = 1), count = 2.
for d = 4 (set bit = 1),
q = 5/4 = 1 (set bit = 1), count = 3.
for d = 5 (set bit = 2),
q = 5/5 = 1 (set bit = 1), count = 4.
Input : N = 3
Output : 2
Observe, for all d > n, q have 0 set bit so there is no point to go above n.
Also, for n/2 < d <= n, it will return q = 1, which have 1 set bit which will always be less
than or equal to d. And, minimum possible value of d is 1. So, possible value of d is from 1 to n.
Now, observe on dividing n by d, where 1 <= d <= n, it will give q in sorted order (decreasing).
So, with increasing d we will get decreasing q. So, we get the two sorted sequence, one for increasing d
and other for decreasing q. Now, observe initially the number of set bit of d is greater than q but after
a point number of set bit of d is less than or equal to q.
So, our problem reduce to find that point i.e. ‘x’ for the value of d, from 1 <= d <= n, such that q have less than or equal set bit to d.
So, count of possible d such that q have less than or equal set bit to d will be equal to n – x + 1.
Below is the implementation of this approach :
C++
// C++ Program to find number of Divisors // which on integer division produce quotient // having less set bit than divisor #include <bits/stdc++.h> using namespace std; // Return the count of set bit. int bit(int x) { int ans = 0; while (x) { x /= 2; ans++; } return ans; } // check if q and d have same number of set bit. bool check(int d, int x) { if (bit(x / d) <= bit(d)) return true; return false; } // Binary Search to find the point at which // number of set in q is less than or equal to d. int bs(int n) { int l = 1, r = sqrt(n); // while left index is less than right index while (l < r) { // finding the middle. int m = (l + r) / 2; // check if q and d have same number of // set it or not. if (check(m, n)) r = m; else l = m + 1; } if (!check(l, n)) return l + 1; else return l; } int countDivisor(int n) { return n - bs(n) + 1; } // Driven Program int main() { int n = 5; cout << countDivisor(n) << endl; return 0; } |
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Java
// Java Program to find number // of Divisors which on integer // division produce quotient // having less set bit than divisor import java .io.*; class GFG { // Return the count of set bit. static int bit(int x) { int ans = 0; while (x > 0) { x /= 2; ans++; } return ans; } // check if q and d have // same number of set bit. static boolean check(int d, int x) { if (bit(x / d) <= bit(d)) return true; return false; } // Binary Search to find // the point at which // number of set in q is // less than or equal to d. static int bs(int n) { int l = 1, r = (int)Math.sqrt(n); // while left index is // less than right index while (l < r) { // finding the middle. int m = (l + r) / 2; // check if q and d have // same number of // set it or not. if (check(m, n)) r = m; else l = m + 1; } if (!check(l, n)) return l + 1; else return l; } static int countDivisor(int n) { return n - bs(n) + 1; } // Driver Code static public void main (String[] args) { int n = 5; System.out.println(countDivisor(n)); } } // This code is contributed by anuj_67. |
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Python3
# Python3 Program to find number # of Divisors which on integer # division produce quotient # having less set bit than divisor import math # Return the count of set bit. def bit(x) : ans = 0 while (x) : x /= 2 ans = ans + 1 return ans # check if q and d have # same number of set bit. def check(d, x) : if (bit(x /d) <= bit(d)) : return True return False; # Binary Search to find # the point at which # number of set in q is # less than or equal to d. def bs(n) : l = 1 r = int(math.sqrt(n)) # while left index is less # than right index while (l < r) : # finding the middle. m = int((l + r) / 2) # check if q and d have # same number of # set it or not. if (check(m, n)) : r = m else : l = m + 1 if (check(l, n) == False) : return math.floor(l + 1) else : return math.floor(l) def countDivisor(n) : return n - bs(n) + 1 # Driver Code n = 5print (countDivisor(n)) # This code is contributed by # Manish Shaw (manishshaw1) |
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C#
// C# Program to find number of Divisors // which on integer division produce quotient // having less set bit than divisor using System; class GFG { // Return the count of set bit. static int bit(int x) { int ans = 0; while (x > 0) { x /= 2; ans++; } return ans; } // check if q and d have // same number of set bit. static bool check(int d, int x) { if (bit(x / d) <= bit(d)) return true; return false; } // Binary Search to find // the point at which // number of set in q is // less than or equal to d. static int bs(int n) { int l = 1, r = (int)Math.Sqrt(n); // while left index is // less than right index while (l < r) { // finding the middle. int m = (l + r) / 2; // check if q and d have // same number of // set it or not. if (check(m, n)) r = m; else l = m + 1; } if (!check(l, n)) return l + 1; else return l; } static int countDivisor(int n) { return n - bs(n) + 1; } // Driver Code static public void Main () { int n = 5; Console.WriteLine(countDivisor(n)); } } // This code is contributed by anuj_67. |
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PHP
<?php // PHP Program to find number of Divisors // which on integer division produce quotient // having less set bit than divisor // Return the count of set bit. function bit( $x) { $ans = 0; while ($x) { $x /= 2; $ans++; } return $ans; } // check if q and d have // same number of set bit. function check( $d, $x) { if (bit($x /$d) <= bit($d)) return true; return false; } // Binary Search to find // the point at which // number of set in q is // less than or equal to d. function bs(int $n) { $l = 1; $r = sqrt($n); // while left index is less // than right index while ($l < $r) { // finding the middle. $m = ($l + $r) / 2; // check if q and d have // same number of // set it or not. if (check($m, $n)) $r = $m; else $l = $m + 1; } if (!check($l, $n)) return floor($l + 1); else return floor($l); } function countDivisor( $n) { return $n - bs($n) + 1; } // Driver Code $n = 5; echo countDivisor($n) ; // This code is contributed by anuj_67. ?> |
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Output:
4
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