atcoder ABC 357-C题详解

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​atcoder ABC 357-C题详解

Problem Statement

For a non-negative integer K, we define a level-Kcarpet as follows:

A level-0 carpet is a 1×1 grid consisting of a single black cell.
For K>0, a level-K carpet is a 3K×3Kgrid. When this grid is divided into nine 3K−1×3K−1 blocks:
The central block consists entirely of white cells.
The other eight blocks are level-(K−1) carpets.
You are given a non-negative integer N.
Print a level-N carpet according to the specified format.

Constraints

0≤N≤6
N is an integer.

Input

The input is given from Standard Input in the following format:

Output

Print 3N lines.
The i-th line (1≤i≤3N) should contain a string Si​ of length 3N consisting of . and #.
The j-th character of Si​ (1≤j≤3N) should be # if the cell at the i-th row from the top and j-th column from the left of a level-N carpet is black, and . if it is white.

Sample Input 1

1

Sample Output 1

#.#

A level-1 carpet is a 3×3 grid as follows:

When output according to the specified format, it looks like the sample output.

Sample Input 2

2

Sample Output 2

#########
#.##.##.#
#########
###…###
#.#…#.#
###…###
#########
#.##.##.#
#########
A level-2 carpet is a 9×9 grid.

思路分析：

​本题可以用递归来实现，其中要把八个小正方形全部遍历到，需要用八个坐标。把白色的地毯赋值为1。

code：

#include <iostream>
#include <cmath>
using namespace std;
int a[2000][2000];//0:黑,1:白
void dfs(int x,int y,int k){if(k==0){a[x][y]=0;return ;}dfs(x,y,k-1);dfs(x+pow(3,k-1),y,k-1);dfs(x+pow(3,k-1)*2,y,k-1);dfs(x,y+pow(3,k-1),k-1);dfs(x,y+pow(3,k-1)*2,k-1);dfs(x+pow(3,k-1),y+pow(3,k-1)*2,k-1);dfs(x+pow(3,k-1)*2,y+pow(3,k-1),k-1);dfs(x+pow(3,k-1)*2,y+pow(3,k-1)*2,k-1);for(int i=x+pow(3,k-1);i<x+pow(3,k-1)*2;i++){for(int j=y+pow(3,k-1);j<y+pow(3,k-1)*2;j++){a[i][j]=1;}}
}
int main(){int n;cin>>n;dfs(0,0,n);for(int i=0;i<pow(3,n);i++){for(int j=0;j<pow(3,n);j++){if(a[i][j]) cout<<'.';else cout<<'#';}cout<<endl;}
}

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