WeChall_Enlightment (Encoding)

解题：

　　刚开始一看这题就蒙逼了，完全没思路，过了几天后再仔细去想想，应该是二进制的ascii码，但是原来的三张图虽然都是8的倍数，但完全转换不成有用的东西，题目的意思能否找到光，百度了一下关于三原色的知识，有如下图。　　

有点思路了，三种颜色刚好对应数字的颜色，于是便想，将三个颜色的二进制数相同的位置统计1的数量，只要数量大于0，这点便取1，然后写了个python程序生成处理后的二进制串。

a = ''
b = ''
c = ''

for i in range(len(a)):
    if int(a[i])+int(b[i])+int(c[i]) > 0:
        print('',end = '')
    else:
        print('',end = '')
        

得到的结果为：

01000001011010000110000100100001001000000100100101110100001000000111001101100101011001010110110101110011001000000111100101101111011101010010000001100111011011110111010000100000011100110110111101101101011001010111010001101000011010010110111001100111001000000110100101101110011101000110010101110010011001010111001101110100011010010110111001100111001000010000110100001010010101110110010101101100011011000010000001110100011011110010000001100111011011110010000001110100011011110010000001110100011010000110010100100000011011100110010101111000011101000010000001110011011101000110000101100111011001010010110000100000011001110110111100100000011101000110100001100101011100100110010100111010000011010000101000110000001100010011000000110000001100010011000100110000001100000011000000110001001100010011000000110001001100000011000000110001001100000011000100110001001100000011000000110001001100010011000100110000001100010011000100110000001100010011000000110000001100000000110100001010001100000011000000110001001100010011000000110001001100010011000100110000001100010011000000110001001100010011000100110001001100010011000000110001001100000011000000110001001100010011000000110000001100000011000100110001001100000011000000110001001100000011000100001101000010100011000000110001001100010011000100110000001100010011000100110000001100000011000100110001001100000011000000110001001100000011000100110000001100010011000100110000001100010011000100110000001100000011000000110000001100010011000100110000001100000011000100110000000011010000101000110000001100000011000100110000001100010011000100110001001100000011000000110001001100010011000100110000001100000011000000110000001100000011000100110001001100000011000100110000001100000011000000110000001100010011000100110001001100000011000000110000001100000000110100001010

在将串数放入JPocketKnife v4.06a软件将串按8位分组，然后用acsii解码，卧槽，hhh。

得到如下信息：

Aha! It seems you got something interesting!
Well to go to the next stage, go there:
01001100011010010110011101101000
00110111010111110100110001100101
01110110011001010110110000110010
00101110011100000110100001110000

再将解出的二进制串做相同处理。

得到：Ligh7_Level2.php

于是打开http://www.wechall.net/challenge/anto/enlightment/Ligh7_Level2.php

卧槽，居然还藏着第二个页面，这次字符串的颜色变了，再去找相关资料。

心里想着应该是一样的套路，看图的话，应该是三个加起来<3，用python输出后发现结果不对，试了下 and ，结果也不对，又试了下^ 符号，结果看起来挺正常的，用JPocketKnife v4.06a软件处理后，果然出来了结果。

a = ''
b = ''
c = ''

for i in range(len(a)):
    if int(a[i])^int(b[i])^int(c[i]):
        print('',end = '')
    else:
        print('',end = '')

结果：

01010100011100100110100101110000011011000110010100101101010110000010110101001111010100100010110000100000011100100110100101100111011010000111010000111111000011010000101001000111011100100110010101100001011101000010000100100000010010000110010101110010011001010010000001101001011100110010000001110111011010000110000101110100001000000111100101101111011101010010000001110011011010000110111101110101011011000110010000100000011000100110010100100000011011000110111101101111011010110110100101101110011001110010000001100110011011110111001000101110001011100010111000001101000010100010001001000111011010010110110101101101011001010101111101000100011000010101111101001100011010010110011101101000011101000010001000001101000010100110010101101110011101000110010101110010001000000111010001101000011010010111001100100000011000010111001100100000011100000110000101110011011100110111011101101111011100100110010000100001

处理后得到：

Triple-X-OR, right?
Great! Here is what you should be looking for...
"Gimme_Da_Light"

hhh，问题解决了，至于为什么是异或，我也想不出原因= =！