Find Integers x in Range 1 to n with Same Divisors as x+1 in C++

Suppose, we have an integer N, we have to find the number of integers 1 < x < N, for which x and x + 1 has same number of positive divisors. So if N = 3, then output will be 1, as divisor of 1 is 1, divisor of 2 is 1 and 2, and divisor of 3 is 1 and 3.

To solve this, we will find the number of divisors of all numbers below N, and store them in an array. Then count number of integers x such that x, such that x + 1 have the same number of positive divisors by running the loop.

Example

 Live Demo

#include<iostream>
#include<cmath>
#define N 100005
using namespace std;
int table[N], pre[N];
void findPositiveDivisor() {
   for (int i = 1; i < N; i++) {
      for (int j = 1; j * j <= i; j++) {
         if (i % j == 0) {
            if (j * j == i)
               table[i]++;
            else
               table[i] += 2;
         }
      }
   }
   int ans = 0;
   for (int i = 2; i < N; i++) {
      if (table[i] == table[i - 1])
      ans++;
      pre[i] = ans;
   }
}
int main() {
   findPositiveDivisor();
   int n = 15;
   cout << "Number of integers: " << pre[n] << endl;
}

Output

Number of integers: 2