445 - Marvelous Mazes

最新推荐文章于 2018-04-19 13:40:41 发布

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本文介绍了一个使用C语言实现的迷宫绘制程序。该程序能够从输入文件读取迷宫描述并将其转换为由字符组成的迷宫图案。文章提供了完整的源代码，并通过样例展示了如何解析输入文件以生成迷宫。

Marvelous Mazes

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

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

int main()
{
    char str[150];
    int i, j;
    int len;
    int sum;

    while (fgets(str, sizeof(str), stdin) != NULL) {
        len = strlen(str);
        sum = 0;
        for (i = 0; i < len; ++i) {
            if (isdigit(str[i])) {
                sum += str[i] - '0';
            }
            else if (str[i] == '!') {
                printf("/n");
            }
            else if (str[i] == 'b') {
                for (j = 0; j < sum; ++j)
                    printf(" ");
                sum = 0;
            }
            else {
                for (j = 0; j < sum; ++j)
       printf("%c", str[i]);
                sum = 0;
            }
        }
        printf("/n");
    }

    return 0;
}