Given an integer N. The task is to find the number of integers 1 < x < N, for which x and x + 1 have the same number of positive divisors.
Examples:
Input: N = 3
Output: 1
Divisors(1) = 1
Divisors(2) = 1 and 2
Divisors(3) = 1 and 3
Only valid x is 2.Input: N = 15
Output: 2
Approach: Find the number of divisors of all numbers below N and store them in an array. And count the number of integers x such that x and x + 1 have the same number of positive divisors by running a loop.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define N 100005 // To store number of divisors and // Prefix sum of such numbers int d[N], pre[N]; // Function to find the number of integers // 1 < x < N for which x and x + 1 have // the same number of positive divisors void Positive_Divisors() { // Count the number of divisors for (int i = 1; i < N; i++) { // Run a loop upto sqrt(i) for (int j = 1; j * j <= i; j++) { // If j is divisor of i if (i % j == 0) { // If it is perfect square if (j * j == i) d[i]++; else d[i] += 2; } } } int ans = 0; // x and x+1 have same number of // positive divisors for (int i = 2; i < N; i++) { if (d[i] == d[i - 1]) ans++; pre[i] = ans; } } // Driver code int main() { // Function call Positive_Divisors(); int n = 15; // Required answer cout << pre[n] << endl; return 0; } |
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Java
// Java implementation of the approach class GFG { static int N =100005; // To store number of divisors and // Prefix sum of such numbers static int d[] = new int[N], pre[] = new int[N]; // Function to find the number of integers // 1 < x < N for which x and x + 1 have // the same number of positive divisors static void Positive_Divisors() { // Count the number of divisors for (int i = 1; i < N; i++) { // Run a loop upto sqrt(i) for (int j = 1; j * j <= i; j++) { // If j is divisor of i if (i % j == 0) { // If it is perfect square if (j * j == i) d[i]++; else d[i] += 2; } } } int ans = 0; // x and x+1 have same number of // positive divisors for (int i = 2; i < N; i++) { if (d[i] == d[i - 1]) ans++; pre[i] = ans; } } // Driver code public static void main(String[] args) { // Function call Positive_Divisors(); int n = 15; // Required answer System.out.println(pre[n]); } } /* This code contributed by PrinciRaj1992 */ |
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Python3
# Python3 implementation of the above approach from math import sqrt; N = 100005 # To store number of divisors and # Prefix sum of such numbers d = [0] * N pre = [0] * N # Function to find the number of integers # 1 < x < N for which x and x + 1 have # the same number of positive divisors def Positive_Divisors() : # Count the number of divisors for i in range(N) : # Run a loop upto sqrt(i) for j in range(1, int(sqrt(i)) + 1) : # If j is divisor of i if (i % j == 0) : # If it is perfect square if (j * j == i) : d[i] += 1 else : d[i] += 2 ans = 0 # x and x+1 have same number of # positive divisors for i in range(2, N) : if (d[i] == d[i - 1]) : ans += 1 pre[i] = ans # Driver code if __name__ == "__main__" : # Function call Positive_Divisors() n = 15 # Required answer print(pre[n]) # This code is contributed by Ryuga |
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C#
// C# implementation of the approach using System; class GFG { static int N =100005; // To store number of divisors and // Prefix sum of such numbers static int []d = new int[N]; static int []pre = new int[N]; // Function to find the number of integers // 1 < x < N for which x and x + 1 have // the same number of positive divisors static void Positive_Divisors() { // Count the number of divisors for (int i = 1; i < N; i++) { // Run a loop upto sqrt(i) for (int j = 1; j * j <= i; j++) { // If j is divisor of i if (i % j == 0) { // If it is perfect square if (j * j == i) d[i]++; else d[i] += 2; } } } int ans = 0; // x and x+1 have same number of // positive divisors for (int i = 2; i < N; i++) { if (d[i] == d[i - 1]) ans++; pre[i] = ans; } } // Driver code public static void Main(String[] args) { // Function call Positive_Divisors(); int n = 15; // Required answer Console.WriteLine(pre[n]); } } // This code has been contributed by 29AjayKumar |
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PHP
<?php // PHP implementation of the approach $N = 100005; // To store number of divisors and // Prefix sum of such numbers $d = array_fill(0,$N,NULL); $pre = array_fill(0,$N,NULL); // Function to find the number of integers // 1 < x < N for which x and x + 1 have // the same number of positive divisors function Positive_Divisors() { global $N,$d,$pre; // Count the number of divisors for ($i = 1; $i < $N; $i++) { // Run a loop upto sqrt(i) for ($j = 1; $j * $j <= $i; $j++) { // If j is divisor of i if ($i % $j == 0) { // If it is perfect square if ($j * $j == $i) $d[$i]++; else $d[$i] += 2; } } } $ans = 0; // x and x+1 have same number of // positive divisors for ($i = 2; $i < $N; $i++) { if ($d[$i] == $d[$i - 1]) $ans++; $pre[$i] = $ans; } } // Driver code // Function call Positive_Divisors(); $n = 15; // Required answer echo $pre[$n] ; return 0; // This code is contributed by ChitraNayal ?> |
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Output:
2
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