UVA 445 Marvelous Mazes

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

题意：

将字符串转换成非凡的迷宫，‘！’对应回车，‘b’对应空格，其他字母或字符前面的数字累加的和就是输出这个东西的个数。

疏忽大意，作了好久！

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cctype>
using namespace std;
int main(){
    char ch[150];
    int i, j;
    while(gets(ch) != NULL){
        int s = 0;
        for(i = 0; i < strlen(ch); i++){
            if(isdigit(ch[i])) s += ch[i] - '0';  //是数字的话，累加
            else{
                if(ch[i] == '!') printf("\n");
                else{
                    while(s--) {
                        if(ch[i] == 'b') printf(" ");
                        else printf("%c", ch[i]);
                    }
                    s = 0;
                }
            }
        }
        printf("\n");
    }
    return 0;
}