网站怎么优化自己免费以品牌推广为目的的广告网络平台

高手进阶区（1）

接下来就是攻防世界的高手进阶区了，这里的题目讲道理对于我这样的新手还是挺难的，尽力而为吧~（tcl）

题目1：你猜猜
题目来源： ISCC-2017
题目描述：我们刚刚拦截了，敌军的文件传输获取一份机密文件，请君速速破解。
题目附件：
504B03040A0001080000626D0A49F4B5091F1E0000001200000008000000666C61672E7478746C9F170D35D0A45826A03E161FB96870EDDFC7C89A11862F9199B4CD78E7504B01023F000A0001080000626D0A49F4B5091F1E00000012000000080024000000000000002000000000000000666C61672E7478740A0020000000000001001800AF150210CAF2D1015CAEAA05CAF2D1015CAEAA05CAF2D101504B050600000000010001005A000000440000000000

解题：
题目中的提示性语句就是“一份机密文件”，可以想到也许可以从文件的格式入手。504B0304是zip文件的文件头，因此我们可以使用winhex将其以zip文件的格式保存。

保存后解压时会有一个密码，试一下123456成功了，得到结果：daczcasdqwdcsdzasd
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题目2：enc
题目来源： 暂无
题目描述：Fady不是很理解加密与编码的区别 所以他使用编码而不是加密，给他的朋友传了一些秘密的消息。
题目附件：

ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO
ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE
ONE ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE
ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE ONE ONE
ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE
ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE
ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO
ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO
ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO
ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO
ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ONE
ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ONE ONE
ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO
ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ONE
ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO
ONE ZERO ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE ONE
ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO
ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO
ONE ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ZERO ZERO
ONE ONE ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE
ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE
ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE
ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO
ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE
ZERO ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ONE ZERO
ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ONE
ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO
ZERO ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO
ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE
ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE
ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO
ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ONE
ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE
ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE ONE ONE ZERO ONE
ZERO ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE ZERO
ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE
ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO
ONE ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE
ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO
ONE ZERO ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE
ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO
ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ONE
ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ONE
ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ONE ONE
ZERO ZERO ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO
ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO
ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE
ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO
ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ONE
ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE
ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE
ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE ZERO
ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE
ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO
ONE ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE
ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO
ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO
ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO
ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO
ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE
ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE
ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO
ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ONE
ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO
ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ONE ONE
ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE
ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO
ZERO ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO
ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO
ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE
ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE
ONE ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO
ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO
ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO
ZERO ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ZERO
ZERO ONE ZERO ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO
ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO
ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ONE ZERO
ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE
ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO
ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO
ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ONE
ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE ONE ONE ZERO ONE
ZERO ZERO ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO
ZERO ONE ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO ZERO ONE ONE ONE
ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ONE ZERO
ONE ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ZERO ZERO ZERO ONE ZERO ONE
ONE ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ZERO
ONE ZERO ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ZERO ZERO ZERO ZERO
ZERO ONE ONE ONE ZERO ONE ZERO ONE ZERO ONE ZERO ZERO ONE ONE ZERO
ZERO ZERO ONE ONE ZERO ONE ZERO ZERO ONE ZERO ZERO ONE ONE ZERO ZERO
ZERO ZERO ZERO ONE ONE ZERO ZERO ONE ONE ONE ZERO ONE ZERO ZERO ONE
ONE ZERO ZERO ZERO ONE ZERO ONE ZERO ZERO ZERO ONE ZERO ZERO ONE ONE
ONE ONE ZERO ONE ZERO ZERO ONE ONE ONE ONE ZERO ONE

解题：
整个文件里只有zero和one，显然是01编码，不禁感叹他的这位盆友为什么要用这么**的方法将他们替换成英文。将其替换成01编码后，我们可以想到常用的编码方法就是8位二进制编码可以共同编码一个字符，因此我们将它们8位分为一组，找到对应的字符，结果如下：
Li0gLi0uLiAuIC0uLi0gLS4tLiAtIC4uLS4gLSAuLi4uIC4tLS0tIC4uLi4uIC0tLSAuLS0tLSAuLi4gLS0tIC4uLi4uIC4uLSAuLS0uIC4uLi0tIC4tLiAtLS0gLi4uLi4gLiAtLi0uIC4tLiAuLi4tLSAtIC0tLSAtIC0uLi0gLQ==
好明显的base64编码~结果为：
.- .-… . -…- -.-. - …-. - … .---- … — .---- … — … …- .–. …-- .-. — … . -.-. .-. …-- - — - -…- -
又又又是莫斯密码，解码后就可以得到结果了：ALEXCTFTH15O1SO5UP3RO5ECR3TOTXT

import base64
#import morse_talk as mtalk
with open('zero_one', 'r') as f:data = f.read()
data = data.replace("ZERO","0").replace("ONE","1").replace(' ','').replace('\n','')
word=''
for i in range(0, len(data), 8):word+=(chr(int(data[i:i+8], 2)))
word=base64.b64decode(word).decode(encoding='UTF-8')
s = word.split(" ")
print(s)
dict = {'.-': 'A','-...': 'B','-.-.': 'C','-..':'D','.':'E','..-.':'F','--.': 'G','....': 'H','..': 'I','.---':'J','-.-': 'K','.-..': 'L','--': 'M','-.': 'N','---': 'O','.--.': 'P','--.-': 'Q','.-.': 'R','...': 'S','-': 'T','..-': 'U','...-': 'V','.--': 'W','-..-': 'X','-.--': 'Y','--..': 'Z','.----': '1','..---': '2','...--': '3','....-': '4','.....': '5','-....': '6','--...': '7','---..': '8','----.': '9','-----': '0','..--..': '?','-..-.': '/','-.--.-': '()','-....-': '-','.-.-.-': '.'}
for item in s:print (dict[item],end='')

————————————————————————————————————

题目3：告诉你个秘密
题目来源： ISCC-2017
题目描述：暂无
题目附件：
636A56355279427363446C4A49454A7154534230526D6843
56445A31614342354E326C4B4946467A5769426961453067

解题：
给出的题目是两行16位数，所以就试着用ASCII码转换了一下，得到了cjV5RyBscDlJIEJqTSB0RmhCVDZ1aCB5N2lKIFFzWiBiaE0g
然后用base64解码，得到了r5yG lp9I BjM tFhB T6uh y7iJ QsZ bhM
这神奇的密码想了我好久，最后竟然是！！键盘！！
每4个键包围了一个字符，最终结果是：TONGYUAN
————————————————————————————————————
题目4：Easy-one
题目来源： Hack-you-2014
题目描述：破解密文，解密msg002.enc文件
题目附件：
encryptor.c

#include <stdlib.h>
#include <stdio.h>
#include <string.h>int main(int argc, char **argv) {if (argc != 3) {printf("USAGE: %s INPUT OUTPUT\n", argv[0]);return 0;}FILE* input  = fopen(argv[1], "rb");FILE* output = fopen(argv[2], "wb");if (!input || !output) {printf("Error\n");return 0;}char k[] = "CENSORED";char c, p, t = 0;int i = 0;while ((p = fgetc(input)) != EOF) {c = (p + (k[i % strlen(k)] ^ t) + i*i) & 0xff;t = p;i++;fputc(c, output);}return 0;
}

msg001

  Hi! This is only test message

msg001.enc

灄@佇紦矘缑N1i_5徙?辏犆?R

msg002.enc

獨僂鐙摪陦(?p\}㈩欸1^シ廛T偾?咘芩鬯?h?}荊tKk蝌)??侂?w壙7㎜Z６縌*猙灗襤侊鎟2$萇?旞噓T?场S?&碡涉翾汉€enN疓璃暇侁鴊D諞褲`'^街?[}qh9吽頵槃
N?z?B??P内?暣?0;}瞌宏-儝錻2奌兜)豿?l飢=?鸦<'硌??锑B璦!a珉腷鈻紭%粵f.厂嚘嶑??鯠z姄{D?RO?'M梮饂柫鹥務Ns潮瞤?務<?r弞?桞y*???~m鴵桘bd嬣桖裪(椋31?|脮Q+
藟/6,覘I?!?魘蓪湹V?╟uc壂毈汾臔i1?'=}斚R]??

解题：
显然encryptor.c中是一段加密的密码，而msg001中是一段明文文字 ，msg001.enc中是其加密后的文字，我们要找的flag应该是msg002.enc对应的解密后的明文。
这道题看上去就是一道写逆代码的题目，但是里面有一个坑，我们尝试后发现msg001加密后并非是msg001.enc中的内容，这是因为它的加密代码中key是错误的，所以我们需要找出正确的key。
找出key的代码如下：

#include <stdlib.h>
#include <stdio.h>
#include <string.h>int main(int argc, char **argv) {FILE* input  = fopen("msg001.enc", "rb");if (!input ) {//如果有一个文件无法打开 printf("Error\n");return 0;}char c, p, t = 0;int i=0;char w[] = "Hi! This is only test message\n";  //原来input中的值unsigned int j = 0;while ((p = fgetc(input)) != EOF) {for (j=31;j<125;j++) {//在找k的真正值，只有当能够逆向回w时，真正找到k c = (p - (j ^ t) - i*i) & 0xff;if (c == w[i]) {printf("%c",j);t = c;i++;break;}}}return 0;
}

得到密钥为：VeryLongKeyYouWillNeverGuess
然后使用密钥去得到解密，代码如下：

#include <stdlib.h>#include <stdio.h>#include <string.h>int main(int argc, char **argv) {FILE* input  = fopen("msg002.enc", "rb");FILE* output = fopen("msg002.txt", "wb");if (!input || !output) {printf("Error\n");return 0;}char c, p, t = 0;int i = 0;char k[] = "VeryLongKeyYouWillNeverGuess";i = 0;c, p, t = 0;int g = 0;while ((p = fgetc(input)) != EOF) {//printf("%c",p);c = (p - (k[i % strlen(k)] ^ t) - i*i) & 0xff;printf("Decrypting %x i=%d t=%d k=%d -> %d\n",p,i,t,(k[i % strlen(k)] ^ t),c);t = c;//c是改变之前的c i++;printf("%c",c);fputc(c, output);g++;if (g>450) {break;}}return 0;}

得到的结果为：
The known-plaintext attack (KPA) is an attack model for cryptanalysis where the attacker has samples of both the plaintext (called a crib), and its encrypted version (ciphertext). These can be used to reveal further secret information such as secret keys and code books. The term “crib” originated at Bletchley Park, the British World War II decryption operation.
The flag is CTF{6d5eba48508efb13dc87220879306619}