Marvelous Mazes

最新推荐文章于 2019-10-06 20:28:00 发布

转载最新推荐文章于 2019-10-06 20:28:00 发布 · 49 阅读

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原文链接：http://www.cnblogs.com/superxuezhazha/p/5293433.html

本文介绍了一种迷宫绘制程序的设计思路与实现方法，通过解析输入文件中的特定字符与数字组合来绘制由字母和特殊符号构成的迷宫图案。文章提供了一个C++实现示例，该程序能够处理复杂的迷宫描述并正确输出。

F - Marvelous Mazes

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status Practice UVA 445

Description

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<bits/stdc++.h>
using namespace std;
int main()
{
    char s[150];
    int sum=0;
    while(gets(s))
    {
        int len=strlen(s);
        for(int i=0;i<len;i++)
        {
            if(s[i]>='0'&&s[i]<='9')
            {
                sum+=s[i]-'0';
            }
            else if(s[i]=='!')
            {
                cout<<endl;
            }
            else if(s[i]=='b')
            {
                for(int j=0;j<sum;j++)
                    cout<<' ';
                sum=0;
            }
            else
            {
                for(int j=0;j<sum;j++)
                {
                    cout<<s[i];
                }
                sum=0;
            }

        }
        cout<<endl;

    }
}

View Code

这里读取说的比较高级，但是用一个gets就解决了，因为它可以读取\n,水

转载于:https://www.cnblogs.com/superxuezhazha/p/5293433.html