矩阵的中心元素等于半对角线之和
给定一个奇数阶矩阵，即（5*5）。任务是检查矩阵的中心元素是否等于所有半对角线的单个总和。
例子：
Input : mat[][] = { 2 9 1 4 -2
6 7 2 11 4
4 2 9 2 4
1 9 2 4 4
0 2 4 2 5 }
Output :Yes
Explanation :
Sum of Half Diagonal 1 = 2 + 7 = 9
Sum of Half Diagonal 2 = 9 + 0 = 9
Sum of Half Diagonal 3 = 11 + -2 = 9
Sum of Half Diagonal 4 = 5 + 4 = 9
Here, All the sums equal to the center element
that is mat[2][2] = 9简单的方法：
迭代两个循环，找到所有半对角线和，然后检查所有和是否等于矩阵的中心元素。如果其中任何一个不等于中心元素，则打印“否”，否则打印“是”。
时间复杂度：O(N*N)
高效方法：基于在 O(N) 中找到对角线和的高效方法。
下面是这种方法的实现
C++
// C++ Program to check if the center
// element is equal to the individual
// sum of all the half diagonals
#include
#include
using namespace std;
const int MAX = 100;
// Function to Check center element
// is equal to the individual
// sum of all the half diagonals
bool HalfDiagonalSums(int mat[][MAX], int n)
{
// Find sums of half diagonals
int diag1_left = 0, diag1_right = 0;
int diag2_left = 0, diag2_right = 0;
for (int i = 0, j = n - 1; i < n; i++, j--) {
if (i < n/2) {
diag1_left += mat[i][i];
diag2_left += mat[j][i];
}
else if (i > n/2) {
diag1_right += mat[i][i];
diag2_right += mat[j][i];
}
}
return (diag1_left == diag2_right &&
diag2_right == diag2_left &&
diag1_right == diag2_left &&
diag2_right == mat[n/2][n/2]);
}
// Driver code
int main()
{
int a[][MAX] = { { 2, 9, 1, 4, -2},
{ 6, 7, 2, 11, 4},
{ 4, 2, 9, 2, 4},
{ 1, 9, 2, 4, 4},
{ 0, 2, 4, 2, 5} };
cout << ( HalfDiagonalSums(a, 5) ? "Yes" : "No" );
return 0;
} Java
// Java program to find maximum elements
// that can be made equal with k updates
import java.util.Arrays;
public class GFG {
static int MAX = 100;
// Function to Check center element
// is equal to the individual
// sum of all the half diagonals
static boolean HalfDiagonalSums(int mat[][],
int n)
{
// Find sums of half diagonals
int diag1_left = 0, diag1_right = 0;
int diag2_left = 0, diag2_right = 0;
for (int i = 0, j = n - 1; i < n;
i++, j--)
{
if (i < n/2) {
diag1_left += mat[i][i];
diag2_left += mat[j][i];
}
else if (i > n/2) {
diag1_right += mat[i][i];
diag2_right += mat[j][i];
}
}
return (diag1_left == diag2_right &&
diag2_right == diag2_left &&
diag1_right == diag2_left &&
diag2_right == mat[n/2][n/2]);
}
// Driver code
public static void main(String args[])
{
int a[][] = { { 2, 9, 1, 4, -2},
{ 6, 7, 2, 11, 4},
{ 4, 2, 9, 2, 4},
{ 1, 9, 2, 4, 4},
{ 0, 2, 4, 2, 5} };
System.out.print ( HalfDiagonalSums(a, 5)
? "Yes" : "No" );
}
}
// This code is contributed by Sam007Python 3
# Python 3 Program to check if the center
# element is equal to the individual
# sum of all the half diagonals
MAX = 100
# Function to Check center element
# is equal to the individual
# sum of all the half diagonals
def HalfDiagonalSums( mat, n):
# Find sums of half diagonals
diag1_left = 0
diag1_right = 0
diag2_left = 0
diag2_right = 0
i = 0
j = n - 1
while i < n:
if (i < n//2) :
diag1_left += mat[i][i]
diag2_left += mat[j][i]
elif (i > n//2) :
diag1_right += mat[i][i]
diag2_right += mat[j][i]
i += 1
j -= 1
return (diag1_left == diag2_right and
diag2_right == diag2_left and
diag1_right == diag2_left and
diag2_right == mat[n//2][n//2])
# Driver code
if __name__ == "__main__":
a = [[2, 9, 1, 4, -2],
[6, 7, 2, 11, 4],
[ 4, 2, 9, 2, 4],
[1, 9, 2, 4, 4 ],
[ 0, 2, 4, 2, 5]]
print("Yes") if (HalfDiagonalSums(a, 5)) else print("No" )C#
// C# program to find maximum
// elements that can be made
// equal with k updates
using System;
class GFG
{
// Function to Check
// center element is
// equal to the individual
// sum of all the half
// diagonals
static bool HalfDiagonalSums(int [,]mat,
int n)
{
// Find sums of
// half diagonals
int diag1_left = 0,
diag1_right = 0;
int diag2_left = 0,
diag2_right = 0;
for (int i = 0, j = n - 1;
i < n; i++, j--)
{
if (i < n / 2)
{
diag1_left += mat[i, i];
diag2_left += mat[j, i];
}
else if (i > n / 2)
{
diag1_right += mat[i, i];
diag2_right += mat[j, i];
}
}
return (diag1_left == diag2_right &&
diag2_right == diag2_left &&
diag1_right == diag2_left &&
diag2_right == mat[n / 2, n / 2]);
}
// Driver code
static public void Main ()
{
int [,]a = {{ 2, 9, 1, 4, -2},
{ 6, 7, 2, 11, 4},
{ 4, 2, 9, 2, 4},
{ 1, 9, 2, 4, 4},
{ 0, 2, 4, 2, 5}};
Console.WriteLine(HalfDiagonalSums(a, 5)?
"Yes" : "No" );
}
}
// This code is contributed by ajitPHP
$n / 2)
{
$diag1_right += $mat[$i][$i];
$diag2_right += $mat[$j][$i];
}
}
return ($diag1_left == $diag2_right &&
$diag2_right == $diag2_left &&
$diag1_right == $diag2_left &&
$diag2_right == $mat[$n / 2][$n / 2]);
}
// Driver code
$a = array(array(2, 9, 1, 4, -2),
array(6, 7, 2, 11, 4),
array(4, 2, 9, 2, 4),
array(1, 9, 2, 4, 4),
array(0, 2, 4, 2, 5));
if(HalfDiagonalSums($a, 5) == 0)
echo "Yes" ;
else
echo "No" ;
// This code is contributed
// by akt_mit
?>Javascript
输出：
Yes时间复杂度： O(N)