生成二次对角线之和等于完美平方的矩阵
原文:https://www . geeksforgeeks . org/generate-a-matrix-having-sum-secondary-对角线等于完美平方/
给定一个整数 N ，任务是使用范围【1，N】中的正整数生成维度矩阵 N x N ，使得次对角线之和为完美平方。
示例:
输入: N = 3 输出: 1 2 3 2 3 1 3 2 1 说明: 二次对角线之和= 3 + 3 + 3 = 9(= 3 2 )。
输入: N = 7 输出: 1 2 3 4 5 6 7 2 3 4 5 6 7 1 3 4 5 6 7 1 2 4 5 6 7 1 2 3 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 7 1 2 3 4 5 6 说明: 二次对角线之和= 7 + 7 + 7 + 7 + 7 + 7
方法:由于生成的矩阵需要为维度 N x N ，因此，为了使次对角线中的元素之和成为完美的正方形，想法是在次对角线的每个索引处分配 N 。所以这条对角线上所有 N 元素之和为 N 2 ，是一个完美的正方形。按照以下步骤解决问题:
- 初始化一个矩阵 mat[][] 的维度 N x N 。
- 将矩阵的第一行初始化为{1 2 3 … N}。
- 现在对于矩阵的剩余行，用 1 将矩阵前一行的排列按循环左移来填充每一行。
- 完成上述步骤后，打印矩阵。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the matrix whose sum
// of element in secondary diagonal is a
// perfect square
void diagonalSumPerfectSquare(int arr[], int N)
{
// Iterate for next N - 1 rows
for(int i = 0; i < N; i++)
{
// Print the current row after
// the left shift
for(int j = 0; j < N; j++)
{
cout << (arr[(j + i) % 7]) << " ";
}
cout << endl;
}
}
// Driver Code
int main()
{
// Given N
int N = 7;
int arr[N];
// Fill the array with elements
// ranging from 1 to N
for(int i = 0; i < N; i++)
{
arr[i] = i + 1;
}
// Function Call
diagonalSumPerfectSquare(arr, N);
}
// This code is contributed by gauravrajput1
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 print the matrix whose sum
// of element in secondary diagonal is a
// perfect square
static void diagonalSumPerfectSquare(int[] arr,
int N)
{
// Iterate for next N - 1 rows
for (int i = 0; i < N; i++)
{
// Print the current row after
// the left shift
for (int j = 0; j < N; j++)
{
System.out.print(arr[(j + i) % 7] + " ");
}
System.out.println();
}
}
// Driver Code
public static void main(String[] srgs)
{
// Given N
int N = 7;
int[] arr = new int[N];
// Fill the array with elements
// ranging from 1 to N
for (int i = 0; i < N; i++)
{
arr[i] = i + 1;
}
// Function Call
diagonalSumPerfectSquare(arr, N);
}
}
// This code is contributed by Amit Katiyar
Python 3
# Python3 program for the above approach
# Function to print the matrix whose sum
# of element in secondary diagonal is a
# perfect square
def diagonalSumPerfectSquare(arr, N):
# Print the current row
print(*arr, sep =" ")
# Iterate for next N - 1 rows
for i in range(N-1):
# Perform left shift by 1
arr = arr[i::] + arr[:i:]
# Print the current row after
# the left shift
print(*arr, sep =" ")
# Driver Code
# Given N
N = 7
arr = []
# Fill the array with elements
# ranging from 1 to N
for i in range(1, N + 1):
arr.append(i)
# Function Call
diagonalSumPerfectSquare(arr, N)
C
// C# program for the
// above approach
using System;
class GFG {
// Function to print the matrix whose sum
// of element in secondary diagonal is a
// perfect square
static void diagonalSumPerfectSquare(int[] arr,
int N)
{
// Iterate for next N - 1 rows
for (int i = 0; i < N; i++)
{
// Print the current row after
// the left shift
for (int j = 0; j < N; j++)
{
Console.Write(arr[(j + i) % 7] + " ");
}
Console.WriteLine();
}
}
// Driver Code
public static void Main(String[] srgs)
{
// Given N
int N = 7;
int[] arr = new int[N];
// Fill the array with elements
// ranging from 1 to N
for (int i = 0; i < N; i++) {
arr[i] = i + 1;
}
// Function Call
diagonalSumPerfectSquare(arr, N);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript program to implement
// the above approach
// Function to print the matrix whose sum
// of element in secondary diagonal is a
// perfect square
function diagonalSumPerfectSquare( arr,N)
{
// Iterate for next N - 1 rows
for (let i = 0; i < N; i++)
{
// Print the current row after
// the left shift
for (let j = 0; j < N; j++)
{
document.write(arr[(j + i) % 7] + " ");
}
document.write("<br/>");
}
}
// Driver Code
// Given N
let N = 7;
let arr = new Array(N).fill(0);
// Fill the array with elements
// ranging from 1 to N
for (let i = 0; i < N; i++)
{
arr[i] = i + 1;
}
// Function Call
diagonalSumPerfectSquare(arr, N);
// This code is contributed by avijitmondal1998.
</script>
Output
1 2 3 4 5 6 7
2 3 4 5 6 7 1
3 4 5 6 7 1 2
4 5 6 7 1 2 3
5 6 7 1 2 3 4
6 7 1 2 3 4 5
7 1 2 3 4 5 6
时间复杂度:O(N2) 辅助空间: O(N)
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