Codeforces Beta Round #36 / 36B Fractal(分形&模拟)

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本文提供了一道名为B.Fractal的编程题解析，该题要求根据给定的模型绘制出经过特定步骤迭代后的分形图案。文章详细介绍了问题背景、输入输出格式，并给出了完整的代码实现。

Ever since Kalevitch, a famous Berland abstractionist, heard of fractals, he made them the main topic of his canvases. Every morning the artist takes a piece of graph paper and starts with making a model of his future canvas. He takes a square as big as n × n squares and paints some of them black. Then he takes a clean square piece of paper and paints the fractal using the following algorithm:

Step 1. The paper is divided into n² identical squares and some of them are painted black according to the model.

Step 2. Every square that remains white is divided into n² smaller squares and some of them are painted black according to the model.

Every following step repeats step 2.

$\text{[math]}$

Unfortunately, this tiresome work demands too much time from the painting genius. Kalevitch has been dreaming of making the process automatic to move to making 3D or even 4D fractals.

Input

The first line contains integers n and k (2 ≤ n ≤ 3, 1 ≤ k ≤ 5), where k is the amount of steps of the algorithm. Each of the following nlines contains n symbols that determine the model. Symbol «.» stands for a white square, whereas «*» stands for a black one. It is guaranteed that the model has at least one white square.

Output

Output a matrix n^k × n^k which is what a picture should look like after k steps of the algorithm.

   Sample test(s)
 

     input
   
2 3
.*
..

     output
   
.*******
..******
.*.*****
....****
.***.***
..**..**
.*.*.*.*
........

     input
   
3 2
.*.
***
.*.

     output
   
.*.***.*.
*********
.*.***.*.
*********
*********
*********
.*.***.*.
*********
.*.***.*.

复制一开始给出的图形，然后迭代复制。

完整代码：

/*30ms,200KB*/

#include<cstdio>

char pix[3][3];
char expix[2][243][243];

int main()
{
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	int n, k;
	bool change = false;///一定要初始化!!
	scanf("%d%d", &n, &k);
	int exn = n;
	getchar();
	for (int i = 0; i < n; ++i)
	{
		for (int j = 0; j < n; ++j)
		{
			scanf("%c", &pix[i][j]);
			expix[0][i][j] = pix[i][j];
			expix[1][i][j] = pix[i][j];
		}
		getchar();
	}
	///////////////////////////////
	while (--k)
	{
		change = !change;
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < n; ++j)  ///按最初的n*n复制形状
				if (pix[i][j] == '.')
					for (int ii = i * exn; ii < (i + 1)*exn; ++ii)
						for (int jj = j * exn; jj < (j + 1)*exn; ++jj)
							expix[change][ii][jj] = expix[!change][ii - i * exn][jj - j * exn];
				else
					for (int ii = i * exn; ii < (i + 1)*exn ; ++ii)
						for (int jj = j * exn; jj < (j + 1)*exn ; ++jj)
							expix[change][ii][jj] = '*';
		exn *= n;
	}
	for (int i = 0; i < exn; ++i)
	{
		for (int j = 0; j < exn; ++j)
			printf("%c", expix[change][i][j]);
		printf("\n");
	}
	return 0;
}