UVA - 445 Marvelous Mazes

最新推荐文章于 2019-10-06 20:28:00 发布

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#Marvelous Mazes #Uva

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本文介绍了一个迷宫绘制程序的设计与实现。程序从输入文件中读取特定格式的数据，解析后绘制出由字符组成的迷宫。文章提供了完整的源代码，并解释了如何通过不同的字符和数字组合来构建各种形状。

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.

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

题目大意：

给出一个字符串，由数字、大写字母、* 、和b组成，

当s[i]=='b'时，输出空格；
当是数字时，如s[i]=2,s[i+1]='T',连续输出两个T；
如果出现连续数字，则输出这些连续数字的和个字母，如11A,则输出AA
当s[i]=='!'时，输出换行

#include <string>
#include <iostream>
using namespace std;
const int N = 1000;
int main() {
	string str;
	int n = 0;	
	while( getline(cin,str) ) {
		int len = str.size();
		for(int i = 0; i < len; i++) {
			if( str[i] >= '0' && str[i] <= '9') {
				n += (str[i] - '0');
			}
			else if(str[i] == '!') {
				cout<<endl;
			}
			else {
				if(str[i] == 'b')
					str[i] = ' ';
				for(int j = 0; j < n; j++) {
					cout<<str[i];
				}
				n = 0;
			}
		}
		cout<<endl;
	}
	return 0;
}