螺旋奇数阶方阵的两条对角线之和

我们给出了一个奇数阶的螺旋矩阵，其中我们以数字 1 为中心，顺时针向右移动。
例子：

Input : n = 3 
Output : 25
Explanation : spiral matrix = 
7 8 9
6 1 2
5 4 3
The sum of diagonals is 7+1+3+9+5 = 25

Input : n = 5
Output : 101
Explanation : spiral matrix of order 5
21 22 23 23 25
20  7  8  9 10
19  6  1  2 11
18  5  4  3 12
17 16 15 14 13
The sum of diagonals is 21+7+1+3+13+
25+9+5+17 = 101

如果我们仔细观察 nxn 的螺旋矩阵，我们可以注意到右上角元素的值为 n ² 。左上角的值为 (n^2) – (n-1) [为什么？不是我们在螺旋矩阵中逆时针移动，因此我们从右上角减去 n-1 后得到左上角的值]。类似地，左下角的值为 (n^2) – 2(n-1)，右下角的值为 (n^2) – 3(n-1)。添加所有四个角后，我们得到 4[(n^2)] – 6(n-1)。
令 f(n) 为 anxn 矩阵的对角线元素之和。使用上述观察，我们可以递归地将 f(n) 写为：

f(n) = 4[(n^2)] – 6(n-1) + f(n-2)

由上述关系式，我们可以借助迭代法求出一个螺旋矩阵的所有对角线元素之和。

spiralDiaSum(n)
{
    if (n == 1)
       return 1;

    // as order should be only odd
    // we should pass only odd-integers
    return (4*n*n - 6*n + 6 + spiralDiaSum(n-2));
}

下面是实现。

C++

// C++ program to find sum of
// diagonals of spiral matrix
#include
using namespace std;
 
// function returns sum of diagonals
int spiralDiaSum(int n)
{
    if (n == 1)
        return 1;
 
    // as order should be only odd
    // we should pass only odd-integers
    return (4*n*n - 6*n + 6 + spiralDiaSum(n-2));
}
 
// Driver program
int main()
{
    int n = 7;
    cout <<  spiralDiaSum(n);
    return 0;
}

Java

// Java program to find sum of
// diagonals of spiral matrix
 
class GFG
{
    // function returns sum of diagonals
    static int spiralDiaSum(int n)
    {
        if (n == 1)
            return 1;
     
        // as order should be only odd
        // we should pass only odd-integers
        return (4 * n * n - 6 * n + 6 +
                     spiralDiaSum(n - 2));
    }
     
    // Driver program to test
    public static void main (String[] args)
    {
        int n = 7;
        System.out.print(spiralDiaSum(n));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find sum of
# diagonals of spiral matrix
 
# function returns sum of diagonals
def spiralDiaSum(n):
     
    if n == 1:
        return 1
 
    # as order should be only odd
    # we should pass only odd
    # integers
    return (4 * n*n - 6 * n + 6 +
               spiralDiaSum(n-2))
     
# Driver program
n = 7;
print(spiralDiaSum(n))
 
# This code is contributed by Anant Agarwal.

C#

// C# program to find sum of
// diagonals of spiral matrix
using System;
 
class GFG  {
     
    // function returns sum of diagonals
    static int spiralDiaSum(int n)
    {
        if (n == 1)
            return 1;
     
        // as order should be only odd
        // we should pass only odd-integers
        return (4 * n * n - 6 * n + 6 +
                spiralDiaSum(n - 2));
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int n = 7;
        Console.Write(spiralDiaSum(n));
    }
}
 
// This code is contributed by parashar...

PHP

Javascript

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