寒假刷题之6——迷宫

 Marvelous Mazes 

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

Input and Output

Your program will getthe information for the mazes from the input file. This filewill contain lines of characters which your program mustinterpret to draw a maze. Each row of the maze will be describedby a series of numbers and characters, where the numbers before acharacter tell how many times that character will be used. If there are multiple digits in a number before a character, thenthe number of times to repeat the character is the sum of thedigits before that character.

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

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

There is no limit tothe 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

这题又是一道水题。。。但我把题目看错，花了不少时间，还跑到别人的博客上留言指出，糗大了。。。

我以为字母前面的数字一定是字母重复的次数，考虑非一位数数字的情况考虑了很久。。。

竟然没看到这句“If there are multiple digits in a number before a character, thenthe number of times to repeat the character is the sum of thedigits before that character.”

不过终于是ac了，代码如下：

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


int main()
{
	char ch;
	int count = 0, i;
	
	while ((ch = getchar()) != EOF) {
		if (isdigit(ch)) {
			count += ch - '0';
			continue;
		} 
		else if (ch == '!' || ch == '\n')
			printf("\n");
		else {
			if (ch == 'b')
				ch = ' ';
			for (i=0; i<count; ++i) 
				printf("%c", ch);
			count = 0;
		}
	}
	
	return 0;
}

以后要把题目看清楚了，不知道我已经栽在这种错误上几次了。。。