九度OJ 1161：Repeater（复制器） （递归）

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：1449

解决：508

题目描述：

Harmony is indispensible in our daily life and no one can live without it----may be Facer is the only exception. One day it is rumored that repeat painting will create harmony and then hundreds of people started their endless drawing. Their paintings were based
on a small template and a simple method of duplicating. Though Facer can easily imagine the style of the whole picture, but he cannot find the essential harmony. Now you need to help Facer by showing the picture on computer.

You will be given a template containing only one kind of character and spaces, and the template shows how the endless picture is created----use the characters as basic elements and put them in the right position to form a bigger template, and then repeat and
repeat doing that. Here is an example.

# #

 #      <-template

# #

So the Level 1 picture will be

# #

 #

# #

Level 2 picture will be

# #     # #

 #         #

# #     # #

     # #   

      #    

     # #   

# #    # #

 #        # 

# #    # #

输入：

The input contains multiple test cases.

The first line of each case is an integer N, representing the size of the template is N*N (N could only be 3, 4 or 5).

Next N lines describe the template.

The following line contains an integer Q, which is the Scale Level of the picture.

Input is ended with a case of N=0.

It is guaranteed that the size of one picture will not exceed 3000*3000.

输出：

For each test case, just print the Level Q picture by using the given template.

样例输入：

3
# #
 #
# #
1
3
# #
 #
# #
3
4
 OO
O  O
O  O
 OO
2
0

样例输出：

# #
 #
# #
# #   # #         # #   # #
 #     #           #     #
# #   # #         # #   # #
   # #               # #
    #                 #
   # #               # #
# #   # #         # #   # #
 #     #           #     #
# #   # #         # #   # #
         # #   # #
          #     #
         # #   # #
            # #
             #
            # #
         # #   # #
          #     #
         # #   # #
# #   # #         # #   # #
 #     #           #     #
# #   # #         # #   # #
   # #               # #
    #                 #
   # #               # #
# #   # #         # #   # #
 #     #           #     #
# #   # #         # #   # #
     OO  OO
    O  OO  O
    O  OO  O
     OO  OO
 OO          OO
O  O        O  O
O  O        O  O
 OO          OO
 OO          OO
O  O        O  O
O  O        O  O
 OO          OO
     OO  OO
    O  OO  O
    O  OO  O
     OO  OO     

来源：

2011年北京大学计算机研究生机试真题

思路：

开始以为这个题很难，后来发现起始就是基于模板画图，用递归来做就行。

容易出细节错误。

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>

#define N 3000
#define M 5

char p[N][N+1];
char tem[M][M+1];
int m;

void cp(int x, int y)
{
    int i, j;
    for (i=0; i<m; i++)
    {
        for (j=0; j<m; j++)
        {
            p[i+x][j+y] = tem[i][j];
        }
    }
}

void set(int n, int x, int y)
{
    if (n == 1)
    {
        cp(x, y);
        return;
    }
    int i, j;
    int size = pow(m, n-1);
    for (i=0; i<m; i++)
    {
        for (j=0; j<m; j++)
        {
            if (tem[i][j] != ' ')
                set(n-1, x+i*size, y+j*size);
        }
    }
}

void print(int n)
{
    int size = pow(m, n);
    for (int i=0; i<size; i++)
    {
        for (int j=0; j<size; j++)
        {
            printf("%c", p[i][j]);
        }
        printf("\n");
    }
}

int main(void)
{
    int n, i, j;
    while (scanf("%d", &m) != EOF && m)
    {
        getchar();
        for (i=0; i<N; i++)
        {
            for (j=0; j<N; j++)
                p[i][j] = ' ';
            p[i][N] = '\0';
        }
        for (i=0; i<m; i++)
        {
            gets(tem[i]);
        }
        scanf("%d", &n);
        getchar();
        set(n, 0, 0);
        print(n);
    }

    return 0;
}
/**************************************************************
    Problem: 1161
    User: liangrx06
    Language: C
    Result: Accepted
    Time:260 ms
    Memory:9796 kb
****************************************************************/