Dada una array cuadrada mat[][] de dimensiones impares, la tarea es cambiar el valor del elemento más central de la array a la suma de los elementos que lo rodean.
Ejemplos:
Input:
mat[][] = { {2, 1, 7},
{3, 7, 2},
{5, 4, 9} }
Output:
2 1 7
3 10 2
5 4 9
Input:
mat[][] = {{1, 3, 5, 6, 7},
{3, 5, 3, 2, 1},
{1, 2, 3, 4, 5},
{7, 9, 2, 1, 6},
{9, 1, 5, 3, 2} }
Output:
1 3 5 6 7
3 5 3 2 1
1 2 11 4 5
7 9 2 1 6
9 1 5 3 2
Aproximación: El elemento medio de la array de orden impar estará ubicado en la posición mat[n/2][n/2] .
Por lo tanto, actualice directamente el elemento como:
mat[n / 2][n / 2] = mat[n / 2 - 1][n / 2]
+ mat[n / 2][n / 2 - 1]
+ mat[n / 2 + 1][n / 2]
+ mat[n / 2][n / 2 + 1]
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to replace the value of the middle element
// of the matrix with the sum of surrounding elements
#include <iostream>
using namespace std;
const int MAX = 100;
// Function to print the matrix
void print(int mat[][MAX], int n)
{
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
cout << mat[i][j] << " ";
}
cout << endl;
}
}
// Function to change the value of the middle element
// of the matrix to the sum of surrounding elements
void changemiddle(int mat[][MAX], int n)
{
// Change the middle element
mat[n / 2][n / 2]
= mat[n / 2 - 1][n / 2]
+ mat[n / 2][n / 2 - 1]
+ mat[n / 2 + 1][n / 2]
+ mat[n / 2][n / 2 + 1];
// Function call to print the matrix
print(mat, n);
}
// Driver code
int main()
{
int mat[][MAX] = { { 2, 1, 7 },
{ 3, 7, 2 },
{ 5, 4, 9 } };
changemiddle(mat, 3);
return 0;
}
C
// C program to replace the value of the middle element
// of the matrix with the sum of surrounding elements
#include <stdio.h>
#define MAX 100
// Function to print the matrix
void print(int mat[][MAX], int n)
{
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
printf("%d ",mat[i][j]);
}
printf("\n");
}
}
// Function to change the value of the middle element
// of the matrix to the sum of surrounding elements
void changemiddle(int mat[][MAX], int n)
{
// Change the middle element
mat[n / 2][n / 2]
= mat[n / 2 - 1][n / 2]
+ mat[n / 2][n / 2 - 1]
+ mat[n / 2 + 1][n / 2]
+ mat[n / 2][n / 2 + 1];
// Function call to print the matrix
print(mat, n);
}
// Driver code
int main()
{
int mat[][MAX] = { { 2, 1, 7 },
{ 3, 7, 2 },
{ 5, 4, 9 } };
changemiddle(mat, 3);
return 0;
}
// This code is contributed by kothavvsaakash.
Java
// Java program to replace the value of the middle element
// of the matrix with the sum of surrounding elements\
public class GFG{
// Function to print the matrix
static void print(int mat[][], int n)
{
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
System.out.print(mat[i][j] + " ");
}
System.out.println();
}
}
// Function to change the value of the middle element
// of the matrix to the sum of surrounding elements
static void changemiddle(int mat[][], int n)
{
// Change the middle element
mat[n / 2][n / 2]
= mat[n / 2 - 1][n / 2]
+ mat[n / 2][n / 2 - 1]
+ mat[n / 2 + 1][n / 2]
+ mat[n / 2][n / 2 + 1];
// Function call to print the matrix
print(mat, n);
}
// Driver code
public static void main(String []args)
{
int mat[][] = { { 2, 1, 7 },
{ 3, 7, 2 },
{ 5, 4, 9 } };
changemiddle(mat, 3);
}
// This code is contributed by Ryuga
}
Python3
# Python3 program to replace the value # of the middle element of the matrix # with the sum of surrounding elements MAX = 100 # Function to print the matrix def printMatrix(mat, n): for i in range(n): for j in range(n): print(mat[i][j], end = " ") print() # Function to change the value of # the middle element of the matrix # to the sum of surrounding elements def changemiddle(mat, n): # Change the middle element mat[n // 2][n // 2] = (mat[n // 2 - 1][n // 2] + mat[n // 2][n // 2 - 1] + mat[n // 2 + 1][n // 2] + mat[n // 2][n // 2 + 1]) # Function call to print the matrix printMatrix(mat, n) # Driver Code if __name__ == "__main__": mat = [ [ 2, 1, 7 ], [ 3, 7, 2 ], [ 5, 4, 9 ]] changemiddle(mat, 3) # This code is contributed # by rituraj_jain
C#
// C# program to replace the value of the middle element
// of the matrix with the sum of surrounding elements
using System;
public class GFG{
// Function to print the matrix
static void print(int[,] mat, int n)
{
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
Console.Write(mat[i,j] + " ");
}
Console.Write("\n");
}
}
// Function to change the value of the middle element
// of the matrix to the sum of surrounding elements
static void changemiddle(int[,] mat, int n)
{
// Change the middle element
mat[n / 2,n / 2]
= mat[n / 2 - 1,n / 2]
+ mat[n / 2,n / 2 - 1]
+ mat[n / 2 + 1,n / 2]
+ mat[n / 2,n / 2 + 1];
// Function call to print the matrix
print(mat, n);
}
// Driver code
public static void Main()
{
int[,] mat = { { 2, 1, 7 },
{ 3, 7, 2 },
{ 5, 4, 9 } };
changemiddle(mat, 3);
}
}
PHP
<?php
// PHP program to replace the value of
// the middle element of the matrix
// with the sum of surrounding elements
// Function to print the matrix
function printmat(&$mat, $n)
{
for ($i = 0; $i < $n; $i++)
{
for ($j = 0; $j < $n; $j++)
{
echo ($mat[$i][$j]);
echo (" ");
}
echo ("\n");
}
}
// Function to change the value of the
// middle element of the matrix to the
// sum of surrounding elements
function changemiddle(&$mat, $n)
{
// Change the middle element
$mat[$n / 2][$n / 2] = $mat[$n / 2 - 1][$n / 2] +
$mat[$n / 2][$n / 2 - 1] +
$mat[$n / 2 + 1][$n / 2] +
$mat[$n / 2][$n / 2 + 1];
// Function call to print the matrix
printmat($mat, $n);
}
// Driver code
$mat = array(array(2, 1, 7),
array(3, 7, 2),
array(5, 4, 9));
changemiddle($mat, 3);
// This code is contributed
// by Shivi_Aggarwal
?>
Javascript
<script>
// JavaScript program to replace the
// value of the middle element
// of the matrix with the sum of
// surrounding elements
// Function to print the matrix
function print(mat,n)
{
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
document.write(mat[i][j] + " ");
}
document.write("<br>");
}
}
// Function to change the value of the middle element
// of the matrix to the sum of surrounding elements
function changemiddle(mat,n)
{
// Change the middle element
mat[Math.floor(n / 2)][Math.floor(n / 2)]
= mat[Math.floor(n / 2) - 1][Math.floor(n / 2)]
+ mat[Math.floor(n / 2)][Math.floor(n / 2) - 1]
+ mat[Math.floor(n / 2) + 1][Math.floor(n / 2)]
+ mat[Math.floor(n / 2)][Math.floor(n / 2) + 1];
// Function call to print the matrix
print(mat, n);
}
// Driver code
let mat = [ [ 2, 1, 7 ],
[ 3, 7, 2 ],
[ 5, 4, 9 ]] ;
changemiddle(mat, 3)
// This code is contributed by rag2127
</script>
Producción:
2 1 7 3 10 2 5 4 9
Complejidad de tiempo: O (n * m), aquí n es el número de filas y m es el número de columna.
Publicación traducida automáticamente
Artículo escrito por gfg_sal_gfg y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA