Hust oj 2112 Print（递归）_[递归]汉诺塔hustoj-优快云博客

本文链接：https://blog.youkuaiyun.com/Sara_YF/article/details/52074629

本文介绍了一个使用递归方法实现的算法，该算法可以根据输入的整数打印出特定的字符图案。通过递归调用和字符填充的方式，可以生成不同大小和形状的图形，适用于初学者理解和实践递归思想。

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Time Limit: 1000 MS

Memory Limit: 32768 K

Total Submit: 118(77 users)

Total Accepted: 88(75 users)

Rating:

Special Judge: No

Description

Give you a number, print characters like this:

1:
*

2:
...
.*.
...

3:
*****
*...*
*.*.*
*...*
*****

Input

The input contains multiple test cases. Each case consist of a integer n.

1<=n<=50.

Output

For each test case, print the character table like the examples.

Sample Input

Sample Output

*
...
.*.
...
*****
*...*
*.*.*
*...*
*****
.......
.*****.
.*...*.
.*.*.*.
.*...*.
.*****.
.......
*********
*.......*
*.*****.*
*.*...*.*
*.*.*.*.*
*.*...*.*
*.*****.*
*.......*
*********

递归模拟

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int N;

const int maxn = 1005;
char a[maxn][maxn];
void Printf(int n,int ans)
{
    if(n == 0) return ;
    else if(n % 2 == 1)
    {
        for(int i=1;i<=2*n-1;i++)
        {
            a[ans+1][i+ans] = a[2*N-1-ans][i+ans] = '*';
            a[i+ans][ans+1] = a[i+ans][2*N-1-ans] = '*';
        }
    }
    else
    {
        for(int i=1;i<=2*n-1;i++)
        {
            a[ans+1][i+ans] = a[2*N-1-ans][i+ans] = '.';
            a[i+ans][ans+1] = a[i+ans][2*N-1-ans] = '.';
        }
    }
    Printf(n-1,ans+1);
}
int main()
{
    while(~scanf("%d",&N))
    {
       Printf(N,0);
        for(int i=1;i<=2*N-1;i++)
        {
            for(int j=1;j<=2*N-1;j++)
            {
                printf("%c",a[i][j]);
            }
            printf("\n");
        }
    }
}