创建一个 nxn 方阵，其中所有子矩阵的对角元素之和为偶数
给定一个整数 N。任务是生成(nxn)的方阵，其元素范围为 1 到 n^2，条件如下：
- 矩阵的元素应该是不同的，即只使用一次
- 范围从 1 到 n^2 的数字
- 您选择的每个子矩阵都应该具有对角元素的总和，即左上角和右下角的总和应该是偶数，右上角和左下角元素的总和应该是偶数
此属性应适用于矩阵的所有子矩阵。您需要生成一个偶数子矩阵
例子：
Input: 2
Output: 1 2
4 3
Explanation: Here sum of 1+3=4 is even and 2+4=6 is even
Input: 4
Output: 1 2 3
4 5 6
7 8 9
Explanation: The sub matrix [1 2 4 5], [2 3 5 6], [4 5 7 8], [5 6 8 9], [1 2 3 4 5 6 7 8 9] satisfies the condition of opposite corner
elements having even sum
方法：
正如我们所知，任何两个元素的总和是偶数，它可以是奇数和奇数的总和或偶数和偶数的总和。在这两种情况下，角元素总和为偶数，我们需要确保对角线图案排列的元素应该是奇数或偶数。因此，我们使对角线为所有奇数或所有偶数的二维数组，以找到任何具有角元素总和的子矩阵。可以遵循以下方法。
- 当 n 为奇数时，对角线已经在所有奇数或偶数元素中，因此我们无需修改并生成简单的二维数组
- 当 n 为偶数时，生成的矩阵不满足子矩阵的对角元素之和为偶数的性质，因此我们反转交替的行元素，使得每个子矩阵的对角线要么全为奇数，要么全为偶数。
下面是实现。
C++
// C++ program for
// the above approach
#include
using namespace std;
void sub_mat_even(int N)
{
// Counter to initialize
// the values in 2-D array
int K = 1;
// To create a 2-D array
// from to 1 to N*2
int A[N][N];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
A[i][j] = K;
K++;
}
}
// If found even we reverse
// the alternate row elements
// to get all diagonal elements
// as all even or all odd
if(N % 2 == 0)
{
for(int i = 0; i < N; i++)
{
if(i % 2 == 1)
{
int s = 0;
int l = N - 1;
// Reverse the row
while(s < l)
{
swap(A[i][s],
A[i][l]);
s++;
l--;
}
}
}
}
// Print the formed array
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
cout << A[i][j] << " ";
}
cout << endl;
}
}
// Driver code
int main()
{
int N = 4;
// Function call
sub_mat_even(N);
}
// This code is contributed by mishrapriyanshu557 Java
// Java program for
// the above approach
import java.io.*;
class GFG{
static void sub_mat_even(int N)
{
// Counter to initialize
// the values in 2-D array
int K = 1;
// To create a 2-D array
// from to 1 to N*2
int[][] A = new int[N][N];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
A[i][j] = K;
K++;
}
}
// If found even we reverse
// the alternate row elements
// to get all diagonal elements
// as all even or all odd
if (N % 2 == 0)
{
for(int i = 0; i < N; i++)
{
if (i % 2 == 1)
{
int s = 0;
int l = N - 1;
// Reverse the row
while (s < l)
{
swap(A[i], s, l);
s++;
l--;
}
}
}
}
// Print the formed array
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
System.out.print(A[i][j] + " ");
}
System.out.println();
}
}
private static void swap(int[] A, int s, int l)
{
int temp = A[s];
A[s] = A[l];
A[l] = temp;
}
// Driver code
public static void main(String[] args)
{
int N = 4;
// Function call
sub_mat_even(N);
}
}
// This code is contributed by jithinPython3
# Python3 program for
# the above approach
import itertools
def sub_mat_even(n):
temp = itertools.count(1)
# create a 2d array ranging
# from 1 to n^2
l = [[next(temp)for i in range(n)]for i in range(n)]
# If found even we reverse the alternate
# row elements to get all diagonal elements
# as all even or all odd
if n%2 == 0:
for i in range(0,len(l)):
if i%2 == 1:
l[i][:] = l[i][::-1]
# Printing the array formed
for i in range(n):
for j in range(n):
print(l[i][j],end=" ")
print()
n = 4
sub_mat_even(n)C#
// C# program for
// the above approach
using System;
class GFG {
static void sub_mat_even(int N)
{
// Counter to initialize
// the values in 2-D array
int K = 1;
// To create a 2-D array
// from to 1 to N*2
int[,] A = new int[N, N];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
A[i, j] = K;
K++;
}
}
// If found even we reverse
// the alternate row elements
// to get all diagonal elements
// as all even or all odd
if (N % 2 == 0)
{
for(int i = 0; i < N; i++)
{
if (i % 2 == 1)
{
int s = 0;
int l = N - 1;
// Reverse the row
while (s < l)
{
int temp = A[i, s];
A[i, s] = A[i, l];
A[i, l] = temp;
s++;
l--;
}
}
}
}
// Print the formed array
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
Console.Write(A[i, j] + " ");
}
Console.WriteLine();
}
}
static void Main() {
int N = 4;
// Function call
sub_mat_even(N);
}
}
// This code is contributed by divyeshrabadiya07输出：
1 2 3 4
8 7 6 5
9 10 11 12
16 15 14 13这种方法需要 O(n*2) 时间复杂度。