在棋s中计算具有奇数边长的正方形

给定N * N的棋盘，任务是计算边长为奇数的方格数。
例子：

Input: N = 3 
Output: 10 
9 squares are possible whose sides are 1 
and a single square with side = 3 
9 + 1 = 10
Input: N = 8 
Output: 120 
 

方法：对于从1到N的所有奇数，然后计算可以形成具有该奇数边的平方数。对于^第i^个面，正方形的计数等于(N – I + ^1)2。此外，添加所有此类平方数。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count
// of odd length squares possible
int count_square(int n)
{
 
    // To store the required count
    int count = 0;
 
    // For all odd values of i
    for (int i = 1; i <= n; i = i + 2) {
 
        // Add the count of possible
        // squares of length i
        int k = n - i + 1;
        count += (k * k);
    }
 
    // Return the required count
    return count;
}
 
// Driver code
int main()
{
    int N = 8;
 
    cout << count_square(N);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function to return the count
    // of odd length squares possible
    static int count_square(int n)
    {
 
        // To store the required count
        int count = 0;
 
        // For all odd values of i
        for (int i = 1; i <= n; i = i + 2) {
 
            // Add the count of possible
            // squares of length i
            int k = n - i + 1;
            count += (k * k);
        }
 
        // Return the required count
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 8;
 
        System.out.println(count_square(N));
    }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python implementation of the approach
 
# Function to return the count
# of odd length squares possible
def count_square(n):
 
    # To store the required count
    count = 0;
 
    # For all odd values of i
    for i in range(1, n + 1, 2):
 
        # Add the count of possible
        # squares of length i
        k = n - i + 1;
        count += (k * k);
 
    # Return the required count
    return count;
 
# Driver code
N = 8;
print(count_square(N));
 
# This code has been contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to return the count
    // of odd length squares possible
    static int count_square(int n)
    {
 
        // To store the required count
        int count = 0;
 
        // For all odd values of i
        for (int i = 1; i <= n; i = i + 2) {
 
            // Add the count of possible
            // squares of length i
            int k = n - i + 1;
            count += (k * k);
        }
 
        // Return the required count
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int N = 8;
 
        Console.WriteLine(count_square(N));
    }
}
 
// This code is contributed by Code_Mech.

Javascript

 
// Javascript implementation of the approach
 
// Function to return the count
// of odd length squares possible
function count_square(n)
{
 
    // To store the required count
    let count = 0;
 
    // For all odd values of i
    for (let i = 1; i <= n; i = i + 2) {
 
        // Add the count of possible
        // squares of length i
        let k = n - i + 1;
        count += (k * k);
    }
 
    // Return the required count
    return count;
}
 
// Driver code
    let N = 8;
 
    document.write(count_square(N));
 

输出：

120

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