生成印度地图的代码(带说明)

给定生成印度地图的混淆代码，请说明其工作方式。

以下代码在执行后会生成印度地图-

#include "stdio.h"
int main()
{
    int a, b, c;
    for (b=c=10; a="Hello!Welcome to GeeksForGeeks.\
      TFy!QJu ROo TNn(ROo)SLq SLq ULo+\
      UHs UJq TNn*RPn/QPbEWS_JSWQAIJO^\
      NBELPeHBFHT}TnALVlBLOFAkHFOuFETp\
      HCStHAUFAgcEAelclcn^r^r\\tZvYxXy\
      T|S~Pn SPm SOn TNn ULo0ULo#ULo-W\
      Hq!WFs XDt!" [b+++21]; )
  
      for (; a-- > 64 ; )
         putchar ( ++c=='Z' ? c = c/ 9:33^b&1);
    return 0;
}

上面的代码是混淆代码(即人类难以理解的代码)的典型示例。

它是如何工作的?

基本上，字符串是印度地图的行程编码。字符串中的交替字符存储连续绘制空格的次数，以及连续绘制感叹号的次数。

这是对该程序不同元素的分析–

编码的字符串

"Hello!Welcome to GeeksForGeeks."
"TFy!QJu ROo TNn(ROo)SLq SLq ULo+UHs UJq TNn*RPn/QPbEWS_JSWQAIJO^NBELPeHBFHT}TnALVlBL"
"OFAkHFOuFETpHCStHAUFAgcEAelclcn^r^r\\tZvYxXyT|S~Pn SPm SOn TNn ULo0ULo#ULo-WHq!WFs XDt!";

注意[b +++ 21]在编码字符串的末尾。由于b +++ 21等于(b ++ + 21)，其结果为31(10 + 21)，因此该字符串的前31个字符将被忽略，并且不做任何贡献。其余的编码字符串包含用于绘制地图的指令。各个字符确定要连续绘制多少个空格或感叹号。

外循环

这个循环越过字符串中的字符。每次迭代将b的值加1，然后将字符串的下一个字符分配给a。

内循环

此循环绘制单个字符，并在到达行尾时绘制换行符。考虑一下这个putchar语句

putchar(++c=='Z' ? c = c/9 : 33^b&1);

由于“Z”代表ASCII 90号，第90/9会给我们10这是一个字符。十进制33是ASCII，表示“！”。切换33的低阶位可得到32，即空格的ASCII码。这引起 ！如果b为奇数，则打印；如果b为偶数，则打印空白。

下面是上面代码的不太混淆的版本–

C

// C program to print map of India
#include 
  
int main()
{
    int a = 10, b = 0, c = 10;
  
    // The encoded string after removing first 31 characters
    // Its individual characters determine how many spaces
    // or exclamation marks to draw consecutively.
    char* str = "TFy!QJu ROo TNn(ROo)SLq SLq ULo+UHs UJq "
                "TNn*RPn/QPbEWS_JSWQAIJO^NBELPeHBFHT}TnALVlBL"
                "OFAkHFOuFETpHCStHAUFAgcEAelclcn^r^r\\tZvYxXyT|S~Pn SPm "
                "SOn TNn ULo0ULo#ULo-WHq!WFs XDt!";
  
    while (a != 0)
    {
        // read each character of encoded string
        a = str[b++];
        while (a-- > 64)
        {
            if (++c == 90) // 'Z' is 90 in ascii
            {
                // reset c to 10 when the end of line is reached
                c = 10;        // '\n' is 10 in ascii
  
                // print newline
                putchar('\n'); // or putchar(c);
            }
            else
            {
                // draw the appropriate character
                // depending on whether b is even or odd
                if (b % 2 == 0)
                    putchar('!');
                else
                    putchar(' ');
            }
        }
    }
  
    return 0;
}

Java

// Java program to print map of India
class GFG
{
    public static void main(String[] args) 
    {
        int a =10, b = 0, c = 10;
          
        // The encoded string after removing first 31 characters
        // Its individual characters determine how many spaces
        // or exclamation marks to draw consecutively.
        String s1="TFy!QJu ROo TNn(ROo)SLq SLq ULo+UHs UJq TNn*RPn/QP,\n"
        + "bEWS_JSWQAIJO^NBELPeHBFHT}TnALVlBLOFAkHFOuFETpHCStHAUFAgcEAelc,\n"
        + "lcn^r^r\\tZvYxXyT|S~Pn SPm SOn TNn ULo0ULo#ULo-WHq!WFs XDt!";
          
        // read each character of encoded string
        a=s1.charAt(b);
          
        while (a != 0)
        {
            if (b < 170)
            {
                a = s1.charAt(b);
                b++;
                while (a-- > 64)
                {
                      
                    if (++c=='Z')
                    {
                          
                        c/=9; 
                          
                        System.out.print((char)(c));
                    }
                    else
                        System.out.print((char)(33 ^ (b & 0x01)));
                }
            }
            else
                break;
        }
    }
}

Python3

# Python3 program to print map of India 
a = 10
b = 0
c = 10
  
# The encoded string after removing first
# 31 characters. Its individual characters 
# determine how many spaces or exclamation 
# marks to draw consecutively. 
s = ("TFy!QJu ROo TNn(ROo)SLq SLq ULo+UHs"
     " UJq TNn*RPn/QPbEWS_JSWQAIJO^NBELPe"
     "HBFHT}TnALVlBLOFAkHFOuFETpHCStHAUFA"
     "gcEAelclcn^r^r\\tZvYxXyT|S~Pn SPm S"
     "On TNn ULo0ULo#ULo-WHq!WFs XDt!")
  
# Read each character of encoded string 
a = ord(s[b]) 
  
while a != 0: 
    if b < 170: 
        a = ord(s[b]) 
        b += 1
          
        while a > 64: 
            a -= 1
            c += 1
              
            if c == 90: 
                c = c // 9
                print(end = chr(c)) 
            else: 
                print(chr(33 ^ (b & 0X01)), end = '') 
    else: 
        break
  
# The code is contributed by aayush_chouhan

C#

// C# program to print map of India
using System;
  
class GFG
{
    public static void Main() 
    {
        int a = 10, b = 0, c = 10;
          
        // The encoded string after removing first 31 characters
        // Its individual characters determine how many spaces
        // or exclamation marks to draw consecutively.
        string s1 = "TFy!QJu ROo TNn(ROo)SLq SLq ULo+UHs UJq TNn*RPn/QP,\n"
        + "bEWS_JSWQAIJO^NBELPeHBFHT}TnALVlBLOFAkHFOuFETpHCStHAUFAgcEAelc,\n"
        + "lcn^r^r\\tZvYxXyT|S~Pn SPm SOn TNn ULo0ULo#ULo-WHq!WFs XDt!";
          
        // read each character of encoded string
        a = s1[b];
          
        while (a != 0)
        {
            if (b < 170)
            {
                a = s1[b];
                b++;
                while (a-- > 64)
                {
                      
                    if (++c == 'Z')
                    {
                          
                        c/=9; 
                          
                    Console.Write((char)(c));
                    }
                    else
                    Console.Write((char)(33 ^ (b & 0x01)));
                }
            }
            else
                break;
        }
    }
}
  
//This code is contributed by vt_m.

PHP

 64)
            {
                  
                if (++$c==90)
                {
                      
                    $c=floor($c/9); 
                      
                    echo chr($c);
                }
                else
                    printf(chr(33 ^ ($b & 0x01)));
            }
        }
        else
            break;
    }
  
// note: ord() function convert the 
// characters into its Ascii value
  
//this code is contributed by mits 
?>

输出：

参考： http : //stackoverflow.com/questions/3533348/how-does-this-code-generate-the-map-of-india

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