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Find the number of cells in the table contains X
  • Last Updated : 01 Feb, 2021

Given two integer N and X. N represents the number of rows and columns of a table. And the element at the ith row and the jth column in the table is i*j. The task is to find the number of cells in the table contains X

Examples: 

Input : N = 6, X = 12 
Output :
Cells {2, 6}, {3, 4}, {4, 3}, {6, 2} contains the number 12 

Input : N = 5, X = 11 
Output :

Approach: 
It’s easy to see that number x can appear only once in a row. If x contains in the ith row then the column number will be x/i. x contains in the ith row if x is divisible by i. let’s check that x divides i and x/i <= n. If these conditions met then update the answer.



Below is the implementation of the above approach : 

C++




// CPP program to find number of
// cells in the table contains X
#include <bits/stdc++.h>
using namespace std;
 
// Function to find number of
// cells in the table contains X
int Cells(int n, int x)
{
    int ans = 0;
    for (int i = 1; i <= n; i++)
        if (x % i == 0 && x / i <= n)
            ans++;
 
    return ans;
}
 
// Driver code
int main()
{
    int n = 6, x = 12;
 
    // Function call
    cout << Cells(n, x);
 
    return 0;
}


Java




// Java program to find number of
// cells in the table contains X
class GFG
{
 
    // Function to find number of
    // cells in the table contains X
    public static int Cells(int n, int x)
    {
        int ans = 0;
        for (int i = 1; i <= n; i++)
            if (x % i == 0 && x / i <= n)
                ans++;
 
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 6, x = 12;
 
        // Function call
        System.out.println(Cells(n, x));
    }
}
 
// This code is contributed by sanjeev2552


Python3




# Python3 program to find number of
# cells in the table contains X
 
# Function to find number of
# cells in the table contains X
def Cells(n, x):
 
    ans = 0;
    for i in range(1, n + 1):
        if (x % i == 0 and x / i <= n):
            ans += 1;
 
    return ans;
 
# Driver code
if __name__ == '__main__':
 
    n = 6; x = 12;
 
    # Function call
    print(Cells(n, x));
 
# This code is contributed by 29AjayKumar


C#




// C# program to find number of
// cells in the table contains X
using System;
 
class GFG
{
    // Function to find number of
    // cells in the table contains X
    static int Cells(int n, int x)
    {
        int ans = 0;
        for (int i = 1; i <= n; i++)
            if (x % i == 0 && x / i <= n)
                ans++;
 
        return ans;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 6, x = 12;
         
        // Function call
        Console.WriteLine(Cells(n,x));
    }
}
 
// This code is contributed by nidhiva


Output

4

Approach:

 Ignore the cases with negative squares. If n is 0, there won’t be any numbers in the square and if x is 0 it won’t appear in a square, so return 0 in both cases. If x is greater than n^2, it won’t be in the square, so return 0 in that case as well. 
Next, loop through all numbers i from 1 to the square root of x, If i is a factor of x and x/i <= n, there are at least two additional places x is on the n-squared chart, due to the associative property: i*(x/i) and (x/i)*i, e.g., if x is 12 and i is 3: 3*4 and 4*3. 
Lastly, find out if the square root of x is whole. If it is, it appears one additional time on the diagonal of the n-square table, which is the list of all squares.
| 1 |   2 |   3 |   4 |   5 |   6 |

| 2 |   4 |   6 |   8 | 10 | 12 |

| 3 |   6 |   9 | 12 | 15 | 18 |

| 4 |   8 | 12 | 16 | 20 | 24 |

| 5 | 10 | 15 | 20 | 25 | 30 |

| 6 | 12 | 18 | 24 | 30 | 36 |

Below is the implementation of the above approach : 

Java




// Java program to find number of
// cells in the table contains X
class GFG {
 
    // Function to find number of
    // cells in the table contains X
    public static int Cells(int n, int x)
    {
        if (n <= 0 || x <= 0 || x > n * n)
            return 0;
        int i = 0, count = 0;
        while (++i * i < x)
            if (x % i == 0 && x <= n * i)
                count += 2;
        return i * i == x ? count + 1 : count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 6, x = 12;
 
        // Function call
        System.out.println(Cells(n, x));
    }
}
 
// This code is contributed by stephenbrasel


Output

4

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