打印给定中心的矩形图案
给定 3 正整数 c1，c2 和 n，其中 n 是二维正方形矩阵的大小。任务是打印填充有矩形图案的矩阵，该矩形图案具有中心坐标 c1、c2 ，使得 0 < = c1、c2 < n.
示例:
输入 : c1 = 2，c2 = 2，n = 5 输出: 2 2 2 2 2 2 2 2 1 1 2 2 1 0 1 2 2 1 1 2 2 2 2 2 2 2 2 2 2
输入: c1 = 3，c2 = 4，n = 7 T3】输出:T5】4 3 3 3 3 3 3 3 3 4 3 2 2 2 2 4 3 2 1 1 1 2 4 3 2 1 0 1 2 4 3 2 1 1 2 4 3 2 2 2 2 2 2 4 3 3 3 3 3 3 3 3 3
方法:这个问题可以通过使用两个嵌套循环来解决。按照以下步骤解决此问题:
- 使用变量 I 在[0，N-1]范围内迭代，并执行以下步骤:
- 使用变量 j 在[0，N-1]范围内迭代，并执行以下步骤:
- 打印 ABS(C1–I)和 ABS(C2–j)的最大值。
- 打印新行。
- 使用变量 j 在[0，N-1]范围内迭代，并执行以下步骤:
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 filled
// with rectangle pattern having center
// coordinates are c1, c2
void printRectPattern(int c1, int c2, int n)
{
// Iterate in the range[0, n-1]
for (int i = 0; i < n; i++) {
// Iterate in the range[0, n-1]
for (int j = 0; j < n; j++) {
cout << (max(abs(c1 - i), abs(c2 - j))) << " ";
}
cout << endl;
}
}
// Driver Code
int main()
{
// Given Input
int c1 = 2;
int c2 = 2;
int n = 5;
// Function Call
printRectPattern(c1, c2, n);
// This code is contributed by Potta Lokesh
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
import java.io.*;
class GFG{
// Function to print the matrix filled
// with rectangle pattern having center
// coordinates are c1, c2
static void printRectPattern(int c1, int c2, int n)
{
// Iterate in the range[0, n-1]
for(int i = 0; i < n; i++)
{
// Iterate in the range[0, n-1]
for(int j = 0; j < n; j++)
{
System.out.print((Math.max(Math.abs(c1 - i),
Math.abs(c2 - j))) + " ");
}
System.out.println();
}
}
// Driver code
public static void main(String[] args)
{
// Given Input
int c1 = 2;
int c2 = 2;
int n = 5;
// Function Call
printRectPattern(c1, c2, n);
}
}
// This code is contributed by sanjoy_62
Python 3
# Python3 program for the above approach
# Function to print the matrix filled
# with rectangle pattern having center
# coordinates are c1, c2
def printRectPattern(c1, c2, n):
# Iterate in the range[0, n-1]
for i in range(n):
# Iterate in the range[0, n-1]
for j in range(n):
print(max(abs(c1 - i), abs(c2 - j)), end = " ")
print("")
# Driver Code
# Given Input
c1 = 2
c2 = 2
n = 5
# Function Call
printRectPattern(c1, c2, n)
C
// C# program for the above approach
using System;
class GFG{
// Function to print the matrix filled
// with rectangle pattern having center
// coordinates are c1, c2
static void printRectPattern(int c1, int c2, int n)
{
// Iterate in the range[0, n-1]
for(int i = 0; i < n; i++)
{
// Iterate in the range[0, n-1]
for(int j = 0; j < n; j++)
{
Console.Write((Math.Max(Math.Abs(c1 - i),
Math.Abs(c2 - j))) + " ");
}
Console.WriteLine();
}
}
// Driver Code
public static void Main(String[] args)
{
// Given Input
int c1 = 2;
int c2 = 2;
int n = 5;
// Function Call
printRectPattern(c1, c2, n);
}
}
// This code is contributed by target_2
java 描述语言
<script>
// Javascript program for the above approach
// Function to print the matrix filled
// with rectangle pattern having center
// coordinates are c1, c2
function printRectPattern(c1, c2, n) {
// Iterate in the range[0, n-1]
for (let i = 0; i < n; i++)
{
// Iterate in the range[0, n-1]
for (let j = 0; j < n; j++) {
document.write(Math.max(Math.abs(c1 - i), Math.abs(c2 - j)) + " ");
}
document.write("<br>");
}
}
// Driver Code
// Given Input
let c1 = 2;
let c2 = 2;
let n = 5;
// Function Call
printRectPattern(c1, c2, n);
// This code is contributed by gfgking
</script>
Output:
2 2 2 2 2
2 1 1 1 2
2 1 0 1 2
2 1 1 1 2
2 2 2 2 2
时间复杂度: O(N ^2) 辅助空间: O(1)
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