N x N 矩阵的交替元素的总和

给定一个 NxN 矩阵。任务是找到给定矩阵的替代元素的总和。
例如，在 2 x 2 矩阵中，替代元素是 { A[0][0], A[1, 1] } 和 { A[1][0], A[0][1] }。
例子：

Input: mat[][] = { { 1, 2},
                   { 3, 4} }
Output : Sum of alternate elements : 5, 5
Explanation: The alternate elements are {1, 4} 
and {2, 3} so their sum are 5, 5.

Input : mat[][] = { { 1, 2, 3 },
                    { 4, 5, 6 },
                    { 7, 8, 9 } }
Output : Sum of alternate elements :25, 20

这个想法是遍历矩阵并保留一个计数器。我们将在一个变量中添加计数器值为偶数而在另一个变量中计数器值为奇数的元素，最后在矩阵的完整遍历中打印两个和。
下面是上述方法的实现：

C++

// C++ program to print sum of alternate
// elements of a N x N matrix
#include 
using namespace std;
 
// Function to find the sum of alternate
// elements of a matrix
void sumAlternate(int* arr, int n)
{
    int sum1 = 0, sum2 = 0;
 
    // check the alternate elements
    for (int i = 0; i < n * n; i++) {
 
        // count the elements at even places
        if (i % 2 == 0)
            sum1 += *(arr + i);
 
        else // count the elements at odd places
            sum2 += *(arr + i);
    }
 
    cout << "Sum of alternate elements : " << sum1
         << ", " << sum2 << endl;
}
 
// Driver code
int main()
{
    int mat[3][3] = { { 1, 2, 3 },
                      { 4, 5, 6 },
                      { 7, 8, 9 } };
    int n = 3;
 
    // find the sum of alternate elements
    sumAlternate(&mat[0][0], n);
 
    return 0;
}

Java

// Java program to print sum of alternate
// elements of a N x N matrix
 
class GFG
{
        // Function to find the sum of alternate
        // elements of a matrix
        static void sumAlternate(int [][] mat, int n)
        {
            int sum1 = 0, sum2 = 0;
            int cnt=0;
            // check the alternate elements
            for (int i = 0; i < n; i++) {
         
                for(int j=0;j

---

Python 3

# Python 3 program to print sum of
# alternate elements of a N x N matrix
 
# Function to find the sum of
# alternate elements of a matrix
def sumAlternate(arr, n):
 
    sum1 = 0
    sum2 = 0
 
    # check the alternate elements
    i = 0
    while i < n * n :
 
        # count the elements at
        # even places
        if (i % 2 == 0):
            sum1 += (arr + i)
 
        else: # count the elements
              # at odd places
            sum2 += (arr + i)
             
        i += 1
 
    print("Sum of alternate elements : " +
             str(sum1) + ", " + str(sum2))
 
# Driver code
if __name__ == "__main__":
    mat = [[ 1, 2, 3 ],
        [4, 5, 6 ],
        [7, 8, 9 ]]
    n = 3
 
    # find the sum of alternate elements
    sumAlternate(mat[0][0], n)
 
# This code is contributed
# by ChitraNayal

C#

// C# program to print sum of alternate
// elements of a N x N matrix
using System;
class GFG
{
// Function to find the sum of
// alternate elements of a matrix
static void sumAlternate(int[,] mat,
                         int n)
{
    int sum1 = 0, sum2 = 0;
    int cnt = 0;
     
    // check the alternate elements
    for (int i = 0; i < n; i++)
    {
 
        for(int j = 0; j < n; j++)
        {
            if (cnt % 2 == 0)
                sum1 += mat[i, j];
     
            else // count the elements
                 // at odd places
                sum2 += mat[i, j];
         
        cnt++;
        }
    }
 
    Console.WriteLine("Sum of alternate elements : " +
                                  sum1 + ", " + sum2);
}
 
// Driver code
public static void Main()
{
    int[,] mat = { { 1, 2, 3 },
                    { 4, 5, 6 },
                    { 7, 8, 9 } };
    int n = 3;
 
    // find the sum of alternate elements
    sumAlternate(mat, n);
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Javascript

// Javascript program to print sum of alternate
// elements of a N x N matrix
 
// Function to find the sum of alternate
// elements of a matrix
function sumAlternate(arr, n)
{
    var sum1 = 0, sum2 = 0;
    var cnt = 0;
 
    // check the alternate elements
    for (var i = 0; i < n ; i++) {
 
        for (var j = 0; j < n ; j++) {
        // count the elements at even places
        if (cnt % 2 == 0)
            sum1 += arr[i][j];
 
        else // count the elements at odd places
            sum2 += arr[i][j];
        cnt++;
        }
    }
 
    document.write( "Sum of alternate elements : " + sum1
        + ", " + sum2 );
}
 
// Driver code
var mat = [ [ 1, 2, 3 ],
                [ 4, 5, 6 ],
                [ 7, 8, 9 ] ];
var n = 3;
// find the sum of alternate elements
sumAlternate(mat, n);
 

输出：

Sum of alternate elements : 25, 20

时间复杂度： O(N 2 )


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