Diagonals (Editorial)

Let $m$ be a given two-dimensional array with zero-based indexing. The element $m[i][j]$ belongs to:

• the main diagonal if i = j;

• the secondary diagonal if i + j = n - 1;

Example

Let’s consider the given example.

The sum of the integers on the main diagonal is 1 + 6 + 5 + 3 = 15.

The sum of the integers on the secondary diagonal is 8 + 2 + 7 + 4 = 21.

Algorithm implementation

Read the input data. The matrix elements are processed on the fly, without creating a two-dimensional array beforehand.

scanf("%d",&n);
a = b = 0;
for(i = 0; i < n; i++)
for(j = 0; j < n; j++)
{
  scanf("%d",&val);

In the variable $a$, compute the sum of the elements on the main diagonal.

In the variable $b$, compute the sum of the elements on the secondary diagonal.

  if (i == j) a += val;
  if (i + j == n - 1) b += val;
}

Print the answer.

printf("%d %d\n",a,b);

Algorithm implementation — two-dimensional array

Declare a two-dimensional array.

int m[510][510];

Read the input data.

scanf("%d", &n);
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
  scanf("%d", &m[i][j]);

In the variable $a$, compute the sum of the elements on the main diagonal.

In the variable $b$, compute the sum of the elements on the secondary diagonal.

a = b = 0;

Iterate through all the elements of the two-dimensional array.

for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
{

The element $m[i][j]$ is on the main diagonal if i = j;

  if (i == j) a += m[i][j];

The element $m[i][j]$ is on the secondary diagonal if i + j = n - 1;

  if (i + j == n - 1) b += m[i][j];
}

Print the answer.

printf("%d %d\n", a, b);

Java implementation

import java.util.*;
 
public class Main
{
  public static void main(String[] args)
  {
    Scanner con = new Scanner(System.in);
    int n = con.nextInt();
    int m[][] = new int[n][n];
    
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
      m[i][j] = con.nextInt();
 
    int a = 0, b = 0;
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
    {
      if (i == j) a += m[i][j];
      if (i + j == n - 1) b += m[i][j];
    }
        
    System.out.println(a + " " + b);
    con.close();
  }
}

Java implementation — on the fly

import java.util.*;
 
public class Main
{
  public static void main(String[] args)
  {
    Scanner con = new Scanner(System.in);
    int n = con.nextInt();

    int a = 0, b = 0;
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
    {
      int val = con.nextInt();        
      if (i == j) a += val;
      if (i + j == n - 1) b += val;
    }

    System.out.println(a + " " + b);
    con.close();
  }
}

Python implementation

Read the input data.

n = int(input())
m = [list(map(int, input().split())) for _ in range(n)]

In the variable $a$, compute the sum of the elements on the main diagonal.

In the variable $b$, compute the sum of the elements on the secondary diagonal.

a = b = 0

Iterate through all the elements of the two-dimensional array.

for i in range(n):
  for j in range(n):

The element $m[i][j]$ is on the main diagonal if i = j;

    if i == j: a += m[i][j]

The element $m[i][j]$ is on the secondary diagonal if i + j = n - 1;

    if i + j == n - 1: b += m[i][j]

Print the answer.

print(a, b)

Python implementation — list comprehension

Read the input data.

n = int(input())
m = [list(map(int, input().split())) for _ in range(n)]

In the variable $a$, compute the sum of the elements on the main diagonal.

In the variable $b$, compute the sum of the elements on the secondary diagonal.

a = sum(m[i][i] for i in range(n))
b = sum(m[i][n - 1 - i] for i in range(n))

Print the answer.

print(a, b)