Dada una array cuadrada de nxn, cuente todas las filas y columnas cuya suma sea igual a la suma de cualquier diagonal principal o diagonal secundaria.
Ejemplos:
Input : n = 3
arr[][] = { {1, 2, 3},
{4, 5, 2},
{7, 9, 10}};
Output : 2
In first example sum of principal diagonal
= (1 + 5 + 10) = 16 and sum of secondary
diagonal = (3 + 5 + 7) = 15.
Input: n = 4
arr[][] = { { 7, 2, 3, 5 },
{ 4, 5, 6, 3 },
{ 7, 9, 10, 12 },
{ 1, 5, 4, 3 } };
Output: 1
Necesitamos contar el número de filas o columnas cuya suma es igual a 16 o 15. Entonces encuentre la suma de todas las filas y columnas y si su suma es igual a 16 o 15, entonces incremente la cuenta.
Por lo tanto, la suma de la columna {2, 5, 9} = 16 y la suma de la columna {3, 2, 10} = 15. Por lo tanto, la cuenta es igual a 2.
C++
// C++ program to Count number of rows and
// columns having sum is equal to sum of
// any diagonal in matrix
#include <iostream>
#define n 4
using namespace std;
// function to count number of rows countnd columns
// whose sum is equal to sum of any diagonal
int count(int arr[][n])
{
int diag1 = 0, diag2 = 0;
int row = 0, col = 0, count = 0;
for (int i = 0, j = n - 1; i < n; i++, j--)
{
// sum of principle diagonal
diag1 += arr[i][i];
// sum of secondary diagonal
diag2 += arr[i][j];
}
// Find the sum of individual
// row and column
for (int i = 0; i < n; i++) {
row = 0, col = 0;
for (int j = 0; j < n; j++) {
row = row + arr[i][j];
}
for (int j = 0; j < n; j++) {
col = col + arr[j][i];
}
if ((row == diag1) || (row == diag2)) {
count++;
}
if ((col == diag1) || (col == diag2))
count++;
}
return count;
}
// Driver code
int main()
{
int arr[n][n] = { { 7, 2, 3, 5 },
{ 4, 5, 6, 3 },
{ 7, 9, 10, 12 },
{ 1, 5, 4, 3 } };
cout << count(arr) << endl;
}
Java
// Java program to Count number of rows and
// columns having sum is equal to sum of
// any diagonal in matrix
import java.io.*;
class GFG {
static int n = 4;
// function to count number of rows countnd
// columns whose sum is equal to sum of any
// diagonal
static int count(int arr[][])
{
int diag1 = 0, diag2 = 0;
int row = 0, col = 0, count = 0;
for (int i = 0, j = n - 1; i < n; i++, j--)
{
// sum of principle diagonal
diag1 += arr[i][i];
// sum of secondary diagonal
diag2 += arr[i][j];
}
// Find the sum of individual
// row and column
for (int i = 0; i < n; i++) {
row = 0;
col = 0;
for (int j = 0; j < n; j++) {
row = row + arr[i][j];
}
for (int j = 0; j < n; j++) {
col = col + arr[j][i];
}
if ((row == diag1) || (row == diag2))
count++;
if ((col == diag1) || (col == diag2))
count++;
}
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[][] = {{7, 2, 3, 5},
{4, 5, 6, 3},
{7, 9, 10, 12},
{1, 5, 4, 3}};
System.out.println(count(arr));
}
}
// This code is contributed by vt_m.
Python3
# Python3 program to Count number of rows # and columns having sum is equal to sum # of any diagonal in matrix n = 4 # Function to count number of rows # countnd columns whose sum is equal # to sum of any diagonal def count(arr): diag1 = 0; diag2 = 0; row = 0 col = 0; count = 0; j = n - 1 for i in range(n): # sum of principle diagonal diag1 += arr[i][i] # sum of secondary diagonal diag2 += arr[i][j] j -= 1 # Find the sum of individual # row and column for i in range(n): row = 0; col = 0 for j in range(n): row += arr[i][j] for j in range(n): col += arr[j][i] if ((row == diag1) or (row == diag2)): count += 1 if ((col == diag1) or (col == diag2)): count += 1 return count # Driver code arr = [[ 7, 2, 3, 5 ], [ 4, 5, 6, 3 ], [ 7, 9, 10, 12 ], [ 1, 5, 4, 3 ] ] print(count(arr)) # This code is contributed by Anant Agarwal.
C#
// C# program to Count number of rows and
// columns having sum is equal to sum of
// any diagonal in matrix
using System;
namespace Matrix
{
public class GFG
{
static int n = 4;
// Function to count number of rows
// countnd columns whose sum is equal
// to sum of any diagonal
static int count(int [,]arr)
{
int diag1 = 0, diag2 = 0;
int row = 0, col = 0, count = 0;
for (int i = 0, j = n - 1; i < n; i++, j--)
{
// sum of principle diagonal
diag1 += arr[i,i];
// sum of secondary diagonal
diag2 += arr[i,j];
}
// Find the sum of individual
// row and column
for (int i = 0; i < n; i++) {
row = 0;
col = 0;
for (int j = 0; j < n; j++) {
row = row + arr[i,j];
}
for (int j = 0; j < n; j++) {
col = col + arr[j,i];
}
if ((row == diag1) || (row == diag2))
count++;
if ((col == diag1) || (col == diag2))
count++;
}
return count;
}
// Driver code
public static void Main()
{
int [,]arr = { { 7, 2, 3, 5 },
{ 4, 5, 6, 3 },
{ 7, 9, 10, 12 },
{ 1, 5, 4, 3 } };
Console.Write(count(arr));
}
}
}
// This code is contributed by Sam007
PHP
<?php
// PHP program to Count number of rows and
// columns having sum is equal to sum of
// any diagonal in matrix
// function to count number of
// rows countnd columns whose
// sum is equal to sum of any diagonal
function countt($arr)
{
$diag1 = 0; $diag2 = 0;
$row = 0; $col = 0;
$count = 0;$n = 4;
for ($i = 0, $j = $n - 1; $i < $n; $i++, $j--)
{
// sum of principle diagonal
$diag1 += $arr[$i][$i];
// sum of secondary diagonal
$diag2 += $arr[$i][$j];
}
// Find the sum of individual
// row and column
for ($i = 0; $i < $n; $i++) {
$row = 0; $col = 0;
for ($j = 0; $j < $n; $j++) {
$row = $row + $arr[$i][$j];
}
for ($j = 0; $j < $n; $j++) {
$col = $col + $arr[$j][$i];
}
if (($row == $diag1) || ($row == $diag2)) {
$count++;
}
if (($col == $diag1) || ($col == $diag2))
$count++;
}
return $count;
}
// Driver code
{
$arr = array(array(7, 2, 3, 5),
array(4, 5, 6, 3),
array(7, 9, 10, 12),
array(1, 5, 4, 3));
echo countt($arr) ;
}
// This code is contributed by nitin mittal.
?>
Producción:
1
Publicación traducida automáticamente
Artículo escrito por shivani.mittal y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA