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  • 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
    
    
    C++语言: Codee#23813
    01 char str[max_size][164];
    02 int main()
    03 {
    04
    05     int n;                //empty string
    06     for( n = 0; fgets(str[n], sizeof(str[n]), stdin); ++n)
    07         ;
    08     for(int j = 0; j < n; ++j)
    09     {
    10         int len = strlen(str[j]);
    11         int cnt = 0;
    12         for(int k = 0; k < len - 1; ++k)
    13             if(isdigit(str[j][k]))
    14                 cnt = cnt + str[j][k] - '0';
    15             else if('!' == str[j][k])
    16                 fprintf(fout, "\n");
    17             else
    18             {
    19                 for(int l = 0; l < cnt; ++l)//blank      
    20                     fprintf(fout, "%c", str[j][k] == 'b' ? ' ' : str[j][k]);
    21                 cnt = 0;
    22             }
    23         fprintf(fout, "\n");
    24     }
    25
    26     return 0;
    27 }
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  • 原文地址:https://www.cnblogs.com/invisible/p/2237308.html
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