构造对角线和的奇偶性与矩阵大小相同的方阵
原文:https://www . geesforgeks . org/construct-a-square-matrix-其对角线和的奇偶性与矩阵的大小相同/
给定一个表示矩阵大小的整数 N ，任务是构造一个正方形矩阵 N * N ，它具有从 1 到 N2T7】的元素，使得它的对角线之和的奇偶性等于整数 N 的奇偶性。
示例:
输入: N = 4 输出: 1 2 3 4 8 7 6 5 9 10 11 12 16 15 14 13 说明: 对角线= 32 和 36 之和与整数 N = 4，所有的数都是偶数即相同的奇偶性。
输入: N = 3 输出: 1 2 3 6 5 4 7 8 9 说明: 对角线之和= 15，整数 N = 3，所有数字都是奇数即相同奇偶性。
方法:想法是观察在以替代方式填充矩阵中的元素时， N 的奇偶性和对角线的和是相同的。从 1 开始计数，然后从 0 到 N–1依次填充第一行，然后从索引N–1 到 0 填充第二行，以此类推。以这种交替的方式从数值 1 到数值 2 填充每个元素，以获得所需的矩阵。 以下是上述方法的实施:
C++
// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to construct a N * N
// matrix based on the given condition
void createMatrix(int N)
{
// Matrix with given sizM
int M[N][N];
// Counter to insert elements
// from 1 to N * N
int count = 1;
for (int i = 0; i < N; i++) {
// Check if it is first row
// or odd row of the matrix
if (i % 2 == 0) {
for (int j = 0; j < N; j++) {
M[i][j] = count;
count++;
}
}
else
{
// Insert elements from
// right to left
for (int j = N - 1;j >= 0; j--){
M[i][j] = count;
count += 1;
}
}
}
// Print the matrix
for (int i = 0; i < N; i++) {
// Traverse column
for (int j = 0; j < N; j++) {
cout << M[i][j] << " ";
}
cout << endl;
}
}
// Driver Code
int main()
{
// Given size of matrix N
int N = 3;
// Function Call
createMatrix(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
class GFG{
// Function to construct a N * N
// matrix based on the given condition
static void createMatrix(int N)
{
// Matrix with given sizM
int M[][] = new int[N][N];
// Counter to insert elements
// from 1 to N * N
int count = 1;
for (int i = 0; i < N; i++)
{
// Check if it is first row
// or odd row of the matrix
if (i % 2 == 0)
{
for (int j = 0; j < N; j++)
{
M[i][j] = count;
count++;
}
}
else
{
// Insert elements from
// right to left
for(int j = N - 1; j >= 0; j--){
M[i][j] = count;
count += 1 ;
}
}
}
// Print the matrix
for (int i = 0; i < N; i++)
{
// Traverse column
for (int j = 0; j < N; j++)
{
System.out.print(M[i][j] + " ");
}
System.out.println();
}
}
// Driver Code
public static void main(String[] args)
{
// Given size of matrix N
int N = 3;
// Function Call
createMatrix(N);
}
}
// This code is contributed by Ritik Bansal
Python 3
# Python3 program for the above approach
# Function to construct a N * N
# matrix based on the given condition
def createMatrix(N):
# Matrix with given size
M = [[0] * N for i in range(N)]
# Counter to insert elements
# from 1 to N * N
count = 1
for i in range(N):
# Check if it is first row
# or odd row of the matrix
if (i % 2 == 0):
for j in range(N):
# Insert elements from
# left to right
M[i][j] = count
count += 1
# Condition if it is second
# row or even row
else:
# Insert elements from
# right to left
for j in range(N - 1, -1, -1):
M[i][j] = count
count += 1
# Print the matrix
for i in range(N):
# Traverse column
for j in range(N):
print(M[i][j], end = " ")
print()
# Driver Code
if __name__ == '__main__':
# Given size of matrix N
N = 3
# Function call
createMatrix(N)
# This code is contributed by mohit kumar 29
C
// C# program for
// the above approach
using System;
class GFG{
// Function to construct a N * N
// matrix based on the given condition
static void createMatrix(int N)
{
// Matrix with given sizM
int[,] M = new int[N, N];
// Counter to insert elements
// from 1 to N * N
int count = 1;
for (int i = 0; i < N; i++)
{
// Check if it is first row
// or odd row of the matrix
if (i % 2 == 0)
{
for (int j = 0; j < N; j++)
{
M[i, j] = count;
count++;
}
}
else
{
// Insert elements from
// right to left
for(int j = N - 1; j >= 0; j--)
{
M[i, j] = count;
count += 1;
}
}
}
// Print the matrix
for (int i = 0; i < N; i++)
{
// Traverse column
for (int j = 0; j < N; j++)
{
Console.Write(M[i, j] + " ");
}
Console.WriteLine();
}
}
// Driver Code
public static void Main()
{
// Given size of matrix N
int N = 3;
// Function Call
createMatrix(N);
}
}
// This code is contributed by Chitranayal
java 描述语言
<script>
// javascript program for the above approach
// Function to construct a N * N
// matrix based on the given condition
function createMatrix(N)
{
// Matrix with given sizM
var M = Array(N).fill(0).map(x => Array(N).fill(0));
// Counter to insert elements
// from 1 to N * N
var count = 1;
for (i = 0; i < N; i++)
{
// Check if it is first row
// or odd row of the matrix
if (i % 2 == 0)
{
for (j = 0; j < N; j++)
{
M[i][j] = count;
count++;
}
}
else
{
// Insert elements from
// right to left
for(j = N - 1; j >= 0; j--){
M[i][j] = count;
count += 1 ;
}
}
}
// Prvar the matrix
for (i = 0; i < N; i++)
{
// Traverse column
for (j = 0; j < N; j++)
{
document.write(M[i][j] + " ");
}
document.write("<br>");
}
}
// Driver Code
var N = 3;
// Function Call
createMatrix(N);
// This code is contributed by 29AjayKumar
</script>
Output:
1 2 3
6 5 4
7 8 9
时间复杂度: O(NN)* 辅助空间: O(1)
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