计算总和等于对角线总和的行/列
给定一个 nxn 方阵，计算总和等于任何主对角线或辅助对角线之和的所有行和列。
例子：
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
我们需要计算总和等于 16 或 15 的行数或列数。因此，找到所有行和列的总和，如果它们的总和等于 16 或 15，则增加计数。
因此，{2, 5, 9} 列的总和 = 16，{3, 2, 10} 列的总和 = 15。因此，计数等于2。
C++
// C++ program to Count number of rows and
// columns having sum is equal to sum of
// any diagonal in matrix
#include
#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 Sam007PHP
输出：
1