攻防世界部分MISC做题记录
不太难，又有点东西，就一块说了。
很普通的数独
题目来源：ISCC-2017
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这题还挺考验眼力的。仔细看1，5，21是二维码的定位点。但是顺序不对，需要将图片1，5，21重命名成：5，21，1。然后把这25个图片按5×5排列，然后把有数字的格记为数字1（代表黑色），没有的记为0（代表白色），再把得到的数字用python画出二维码。
111111101010101000101000001111110000101111111 100000101100111101010011101100011001001000001 101110101110011111010011111101000101001011101 101110101101100010001010000011110001101011101 101110100011100100001111101111111011101011101 100000101100100000011000100001110100001000001 111111101010101010101010101010101011101111111 000000000011001101001000110100110011100000000 110011100100100001111111100100101000000101111 101001001011111111101110101011110101101001100 100000111100100100000110001101001101010001010 001100010011010001010011000100000010110010000 010110101010001111110100011101001110101101111 100011000100011100111011101101100101101110001 001100110100000000010010000111100101101011010 101000001011010111110011011111101001110100011 110111110111011001101100010100001110000100000 110101000010101000011101101101110101101001100 010011111110001011111010001000011011101101100 011001011001010101100011110101001100001010010 010111111111101011111111101101101111111111100 011110001100000100001000101000100100100011110 111110101110011100111010110100110100101010010 110010001011101011101000111100000011100010000 101011111011100111101111111100001010111110010 110100011000111000100111101101111101000100010 111101111110001001000011010110001111110111110 011001010101000110010100010001000101101010001 011101110101101101100100001101101000111101001 110110001001101100010101101111110100101100110 000011100111000000000100001010101111100010010 111010010011110011101110010100001011111010010 101001100010111111110100000100001010101010100 000010011001001101110101001111100101111101101 000010111101110001101011000001000101110100110 011110011010100010100000011011000001110010000 100110100100001101111111101100101110111110011 000000001111110101101000101011100100100011010 111111100011111011011010101101110011101011110 100000101110101101101000111110010001100010001 101110101011100001111111101101001000111111011 101110100110111101101000001001101100011101101 101110100000011101100001101010110010010010001 100000101011001011111011001011000011010110000 111111101010101001111011110101101110000101101
from PIL import Image import re x = 45 #宽 y = 45 #长 #长*宽是行数 im = Image.new("RGB",(x,y)) #创建图片 file = open(r'2.txt') #打开rgb值图片 black = (255,255,255) white = (0,0,0) #通过一个个rgb点生成图片 for i in range(0,x): line = file.readline()#获取一行 for j in range(0,y): if(line[j]=='1'): im.putpixel((i,j),black) else: im.putpixel((i,j),white) im.show()
ps:生成好的二维码记得锐化一下，要反色！！要么识别不了。
多解几次得到flag。
很普通的Disco
题目来源：ISCC-2017
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放Audacity里频谱图和波形没什么发现。。后来仔细找了下，在这首歌最开始的部分有一段可疑的杂音，疑似二进制：
1100110 1101100 1100001 1100111 1111011 1010111 0110000 1010111 0101010 1100110 1110101 1101110 1101110 1111001 1111101
至于为什么是7位一组。。105能被7整除，每个前面加个0，进行解码得到flag：
number = ["01100110","01101100","01100001","01100111","01111011","01010111","00110000","01010111","00101010","01100110","01110101","01101110","01101110","01111001","01111101"]
for i in number:
print(chr(int("0"+i,2)),end="")
miscmisc
题目来源：2019湖湘杯
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前面还比较简单的，binwalk然后是什么明文攻击。不赘述了。解压得到压缩文件：
其中doc里有隐藏文字，我以为是解压密码挨个试了下都不行。。
word1里用steg神器可以发现可疑字母e^za：
。。后面是查看别人的WP了，显然是t m 不想让人做上呗。最后的密码是 pass内容+world里每行字符串的最后一个字符。
z^ea4zaa3azf8
解压得到flag
