uva445 - Marvelous Mazes

本文深入探讨如何通过编程解析并实现复杂迷宫的绘制,包括输入文件解析、字符重复处理及输出绘制逻辑,旨在为程序员提供迷宫绘制程序的设计思路。

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Your mission, if you decide to accept it, is to create a maze drawing program. A maze will consist of the alphabetic characters A-Z, * (asterisk), and spaces.

Input and Output

Your program will get the information for the mazes from the input file. This file will contain lines of characters which your program must interpret to draw a maze. Each row of the maze will be described by a series of numbers and characters, where the numbers before a character tell how many times that character will be used. If there are multiple digits in a number before a character, then the number of times to repeat the character is the sum of the digits before that character.

The lowercase letter "b" will be used in the input file to represent spaces in the maze. The descriptions for different rows in the maze will be separated by an exclamation point (!) or by an end of line.

Descriptions for different mazes will be separated by a blank line in both input and output. The input file will be terminated by an end of file.

There is no limit to the number of rows in a maze or the number of mazes in a file, though no row will contain more than 132 characters.

Happy mazing!

Sample Input

1T1b5T!1T2b1T1b2T!1T1b1T2b2T!1T3b1T1b1T!3T3b1T!1T3b1T1b1T!5T1*1T
 
11X21b1X
4X1b1X

Sample Output

T TTTTT
T  T TT
T T  TT
T   T T
TTT   T
T   T T
TTTTT*T
 
XX   X
XXXX X

第二次用的fgets,这道题看输入输出就能理解对于英语不好的我来说万幸啊,用的fgets读入了一个空行所以也输出了一个空行,开始还以为要换行错了。

#include <stdio.h>
#include <string.h>
#define max 150
void main ()
{char s[151];
 int i,j,k,l,num;
 while (fgets(s,max,stdin))
 {l=strlen(s);
  k=0;
  while (k<l-2)
  {num=0;
   while ((s[k]>='0')&&(s[k]<='9'))
   {num=num+s[k]-'0';++k;}
   if (s[k]!='!')
   {
   for (i=1;i<=num;i++)
   if (s[k]=='b') printf(" ");
             else printf("%c",s[k]);
   }
   else printf("\n");
   ++k;
  }
  printf("\n");
 }
}

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