最后一位被整除 oracle,【CTF WriteUp】2020第四届强网杯部分Crypto题解
寫在前邊
強網杯還是難。。去年正賽賽題一道都不會，只能靠臨時補充的強網先鋒題目拿分的情景歷歷在目。今年也沒好哪去，只能寫一點是一點吧。
modestudy
這道題是一道六合一塊密碼大雜燴，考察基礎知識與變換，六道小的題目全做完以后拿到flag。其中4、5、6三道題需要大量交互，因此答案不變，可以單次做完后保留答案一起提交。1、2、3三道題當場變換即可。
Stage 1
[$] challenge 1
[+] cookie:session=6b1f33a78c5b9c17;admin=0;checksum=552ebbeb9276a8cd9f741b7d29f8c9e1eb455757207ac420659e7eab56ddee25
[+] checksum=aes128cbc.encrypt(session=6b1f33a78c5b9c17;admin=0)
[+] only admin can see flag
[-] cookie:(待輸入)
本題采用CBC模式加密。我們還是借用紅黑聯盟那個CBC解密的圖說事
如圖所示，需要將admin=0改為admin=1，只要把上一段密文的對應位異或’0’再異或’1’即可。這樣雖然會破壞上一段明文，但是無所謂，admin=1構造出來了
session=6b1f33a7 552ebbeb9276a8cd9f741b7d29f8c9e1
8c5b9c17;admin=0 eb455757207ac420659e7eab56ddee25
把第一行最后一位異或1即可過關
[$] challenge 1
[+] cookie:session=6043fb92858e4080;admin=0;checksum=e1b54fe73c59fc5d07542002d90b7ef53e730401cc8736d4312e6ca55f4758a6
[+] checksum=aes128cbc.encrypt(session=6043fb92858e4080;admin=0)
[+] only admin can see flag
[-] cookie:session=6043fb92858e4080;admin=0;checksum=e1b54fe73c59fc5d07542002d90b7ef43e730401cc8736d4312e6ca55f4758a6
[+] decrypt(checksum):竴痬€萠窰賦?CH?58e4080;admin=1
[+] passed
Stage 2
[$] challenge 2
[+] sha256(iv)=11f595abc9d7b986d24fce986d1f9ddfb0d83f2f978db20b862ea730e570bff0
[+] 1. server's job: print aes_cbc_dec(key,iv,your_input_c).encode('hex')
[+] 2. your job: guess iv
[-] your choice:(待輸入)
標準的CBC選擇密文攻擊。還用上邊那張圖。設我們輸入為c0 c1，得到輸出為m0 m1，則
dec(c0) = m0 ^ iv
dec(c1) = m1 ^ c0
我們讓c1 = c0，可以看到
dec(c0) = m0 ^ iv
dec(c1) = dec(c0) = m1 ^ c0 = m0 ^ iv
iv = m0 ^ m1 ^ c0
所以輸入兩段一樣的16字符內容，拿這段內容和兩段解密出來的明文異或即為iv
[$] challenge 2
[+] sha256(iv)=11f595abc9d7b986d24fce986d1f9ddfb0d83f2f978db20b862ea730e570bff0
[+] 1. server's job: print aes_cbc_dec(key,iv,your_input_c).encode('hex')
[+] 2. your job: guess iv
[-] your choice:1
[-] c:80808080808080808080808080808080
[+] fee1313801108cd3dc7dfb547a3126d82893f3d8e1541cc11799a1546e4fef0e
[+] 1. server's job: print aes_cbc_dec(key,iv,your_input_c).encode('hex')
[+] 2. your job: guess iv
[-] your choice:2
[-] iv(encode hex):ee42fad0d874a822f3d462302c4ef1e6
[+] passed
Stage 3
[$] challenge 3
[+] cookie=session:0884ce7c;timedl=1;admin=0;guess_cookie_ma=1;guess_mp_ab=1;guess_cookie_mb=0;hell_pad=233
[+] 128bit_ecb_encrypt(cookie):0632b3e7adb2f6d5ae4a92f553f2f4a4c9f95b099cfd8e3408137d134eb51d147045f278246a831fabdbdccde099b4b694b07e699ffae6a82c6cfbc4454816b1c78e5d1be8b67e235fbbfd2a75e73f32bac806808bc2e102db8f2c159b250415
[+] only admin can see the flag
[-] input your encrypted cookie(encode hex):(待輸入)
ECB模式的加密每塊之間相互獨立，加密的六個明文段依次為：
session:0884ce7c
;timedl=1;admin=
0;guess_cookie_m
a=1;guess_mp_ab=
1;guess_cookie_m
b=0;hell_pad=233
我們只需要把原始的第五段密文覆蓋掉原始的第三段密文，即可做出admin=1，通過
[$] challenge 3
[+] cookie=session:31e820c8;timedl=1;admin=0;guess_cookie_ma=1;guess_mp_ab=1;guess_cookie_mb=0;hell_pad=233
[+] 128bit_ecb_encrypt(cookie):1ab7386c76b1c8749b05c6fc3b9ef740ef5a0fc8370638514069f522736af63aefa2c39ca7bc127d2b02d5481a01562da88bf83f7efc639d8fcd953572936ef92397f151ae10e4e8180a0a576ba8327df98eb4a8787959889cb63476eca247a8
[+] only admin can see the flag
[-] input your encrypted cookie(encode hex):1ab7386c76b1c8749b05c6fc3b9ef740ef5a0fc8370638514069f522736af63a2397f151ae10e4e8180a0a576ba8327da88bf83f7efc639d8fcd953572936ef92397f151ae10e4e8180a0a576ba8327df98eb4a8787959889cb63476eca247a8
[+] decrypted:session:31e820c8;timedl=1;admin=1;guess_cookie_ma=1;guess_mp_ab=1;guess_cookie_mb=0;hell_pad=233
[+] passed
Stage 4
[$] challenge 4
[+] sha256(secret)=d014cbddd2cbb0fa2404c519c166bc85c03ee3445d643f451a5f5d6244e7e34d
[+] assert len(secret)==16
[+] 1. server's job: print aes_ecb(key,input+secret+'x00'*((16-(len(input+secret) % 16)) % 16))
[+] 2. your job: guess secret
[-] your choice:(待輸入)
從這里開始就不是拿到題直接能夠解決的了，需要計算出結果保存，
由于ECB模式加密每個塊互不相關，所以這里可以按位爆破，如：
(1)首先記錄’a’ * 15 + 任一字符加密后的的256個值
(2)輸入’a’ * 15，看’a’ * 15+secret首位是上述哪個
(3)用’a’ * 14 + secret首位代替’a’ * 15，重復以上步驟，直到secret全部爆出
(代碼最后一起給)
Stage 5
[$] challenge 5
[+] sha256(secret)=4c766c8749526dd1a14fdf37619d2fbebbbc2478e25e27e63e591ae0aafb305d
[+] assert len(secret)==16
[+] myblockencrypt_ecb(secret).encode("hex")=4c21cc2bc7941b224ed45bd02ee11b60
[+] In this challenge, you need to try something.
[+] 1. server's job: print myblockencrypt_ecb(your_input)
[+] 2. your job: guess secret
[-] your choice:(待輸入)
這里果然是 need try something。我們嘗試加密幾個看看
[+] In this challenge, you need to try something.
[+] 1. server's job: print myblockencrypt_ecb(your_input)
[+] 2. your job: guess secret
[-] your choice:1
[-] input(encode hex):80808080808080808080808080808080
[+] myblockencrypt_ecb(your_input).encode("hex"):35093509350935093509350935093509
[+] 1. server's job: print myblockencrypt_ecb(your_input)
[+] 2. your job: guess secret
[-] your choice:1
[-] input(encode hex):66666666666666666666666666666666
[+] myblockencrypt_ecb(your_input).encode("hex"):022c022c022c022c022c022c022c022c
可以看出，這個自定義的myblockencrypt_ecb實際上就是每16位一段進行加密且互不影響。鑒于數次訪問的加密結果完全相同，我們姑且認為這個加密是個固定的映射，因此嘗試從0000到ffff的所有輸入，然后去對密文就可以了。由于交互時間限制，一次連接中爆不完全部65536個值，所以需要邊爆邊記錄，最后拿密文去對就好，這題參數不會變。
Stage 6
[$] challenge 6
[+] padding(m):m+chr(16-(len(m)%16))*(16-(len(m)%16))
[+] unpadding(m):m[:-ord(m[-1])]
[+] len(secret):16
[+] iv+aes128_cbc(key,iv,padding(secret)):31313131313131313131313131313131891f3925b171f363bfc1ab0f4c03c9172cdfbd56dbd225d2accaf2b6fcbd0780
[+] 1. server's job: decrypt and unpadding
[+] 2. your job: guess secret
[-] your choice:(待輸入)
本題解密并不返回結果，只返回能不能成功解密，因此要利用本題的填充模式。我們已知如下信息：
dec(c0) = iv ^ m0
dec(c1) = c0 ^ m1
iv = 1111111111111111
c0、c1告知
m1 = 'x10'*16
要求m0，就必須使用唯一與其相關的c0；利用填充規則，就必須將c0置于最后一段。考察這樣的解密輸入：
第一段：'x00' * 15 + ?
第二段：c0
對于?取值0~255依次進行測試。根據CBC解密流程，如果當?=a時能夠解密成功，說明此時解密的第二段明文末位為x01，即m0(secret)的末位為 a ^ iv ^ 0x01
接著用a ^ 0x01 ^ 0x02來代替a作為固定末位，此時解密的最后一位固定為x02。繼續調整輸入第一段的倒數第二位，使得解密結果末尾為x02x02，此時解密成功，可求出m0的倒數第二位。以此類推可以求出m0
(這玩意兒是叫 padding-oracle 來著？忘了)
完整代碼
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *
import hashlib
from Crypto.Util.number import bytes_to_long, long_to_bytes
team_token = "隊伍token"
p = remote("106.14.66.172", 7777)
def strxor(a, b):
return ''.join(chr(ord(a[i])^ord(b[i%len(b)])) for i in range(len(a)))
def mysha256(text):
mysha = hashlib.sha256()
mysha.update(text)
hashresult = mysha.digest()
bits=''.join(bin(ord(j))[2:].zfill(8) for j in hashresult)
return bits
def passPoW(text):
print text
count = 0
while True:
bits = mysha256(text + "a" + str(count))
if(bits.startswith("00000")):
break
count += 1
return "a" + str(count)
def solve_step_4(level, known):
print "Level %s" % str(level)
if(level == 16):
return known
records = []
for i in range(256):
p.recvuntil("[-] your choice:")
p.sendline("1")
p.recvuntil("[-] input(encode hex):")
tmpstr = ('a'*(15-level) + known + chr(i)).encode('hex')
p.sendline(tmpstr)
text = p.recvuntil("[+] 2. your job: guess secret")
s = text.split("[+] ")[2].strip()
s = s[15:47]
print "Process: %s/256 %s %s" % (str(i), tmpstr, s)
records.append(s)
p.recvuntil("[-] your choice:")
p.sendline("1")
p.recvuntil("[-] input(encode hex):")
tmpstr = ('a'*(15-level)).encode('hex')
p.sendline(tmpstr)
text = p.recvuntil("[+] 2. your job: guess secret")
s = text.split("[+] ")[2].strip()
s = s[15:47]
newknown = known + chr(records.index(s))
print "==========================="
print newknown.encode('hex')
print "==========================="
return solve_step_4(level+1, newknown)
def pad4(text):
return '0'*(4-len(text))+text
def solve_step_5(cipher):
f = open("step5.txt",'w+')
for i in range(0, 65536, 8):
print "Processing: %s/65536" % str(i)
tmpstr = ""
for j in range(8):
tmpstr += pad4(long_to_bytes(i+j).encode('hex'))
p.recvuntil("[-] your choice:")
p.sendline("1")
p.recvuntil("[-] input(encode hex):")
p.sendline(tmpstr)
text = p.recvuntil("[+] 2. your job: guess secret")
cipher = text.split("[+] ")[1]
cipher = cipher[45:]
for j in range(8):
f.write("'"+tmpstr[4*j:4*j+4]+"' ")
f.write("=== ")
for j in range(8):
f.write("""+cipher[4*j:4*j+4]+"" ")
f.write("
")
return ""
def solve_step_6(c0, level, known):
tmpknown = strxor(known, chr(level))
newknown = ""
if(level>=17):
return known
for i in range(256):
print "Process: %s/256" % str(i)
tmpcipher = '1'*16+'x00'*(16-level)+chr(i)+tmpknown+c0
p.recvuntil("[-] your choice:")
p.sendline("1")
p.recvuntil("[-] input your iv+c (encode hex):")
p.sendline(tmpcipher.encode('hex'))
text = p.recvuntil("[+] 2. your job: guess secret")
if text.find("[+] unpadding success") >= 0:
newknown = chr(i^level) + known
print "==============================="
print newknown.encode('hex')
print "==============================="
break
return solve_step_6(c0, level+1, newknown)
print p.recvline().strip()
# pass the PoW
tmpline = p.recvline()
print tmpline.strip()
powtext = passPoW(tmpline[11:19])
print p.recvuntil("[-] ?=")
print "SEND: %s" % powtext
p.sendline(powtext)
print p.recvuntil("[+] teamtoken=")
print "SEND: %s" % team_token
p.sendline(team_token)
print p.recvuntil("[-] your choice:")
# Step 1
print "SEND: 1"
p.sendline("1")
text = p.recvuntil("[-] cookie:")
print text
s = text.split("[+] ")[1].strip()
result = s[:79]+chr(ord(s[79:81].decode('hex'))^0x1).encode('hex')+s[81:]
print "SEND: %s" % result[7:]
p.sendline(result[7:])
print p.recvuntil("[-] your choice:")
# Step 2
print "SEND: 2"
p.sendline("2")
print p.recvuntil("[-] your choice:")
print "SEND: 1"
p.sendline("1")
print p.recvuntil("[-] c:")
print "SEND: 80808080808080808080808080808080"
p.sendline("80808080808080808080808080808080")
text = p.recvuntil("[-] your choice:")
print text
s = text.split("[+] ")[1].strip()
c0 = "8080808080808080"
s1, s2 = s[:32].decode('hex'), s[32:].decode('hex')
result = strxor(strxor(c0, s2), s1).encode('hex')
print "SEND: 2"
p.sendline("2")
print p.recvuntil("[-] iv(encode hex):")
print "SEND: %s" % result
p.sendline(result)
print p.recvuntil("[-] your choice:")
# Step 3
print "SEND: 3"
p.sendline("3")
text = p.recvuntil("[-] input your encrypted cookie(encode hex):")
print text
s = text.split("[+] ")[2].strip()
s = s[27:]
result = s[:64]+s[128:160]+s[96:]
print "SEND: %s" % result
p.sendline(result)
print p.recvuntil("[-] your choice:")
# # Step 4
# print "SEND: 4"
# p.sendline("4")
# text = p.recvuntil("[+] 2. your job: guess secret")
# print text
# result = solve_step_4(0, "")
# print p.recvuntil("[-] your choice:")
# print "SEND: 2"
# p.sendline("2")
# print p.recvuntil("[-] secret(encode hex):")
# print "SEND: %s" % result
# p.sendline(result)
# print p.recvuntil("[-] your choice:")
# Step 4(finished)
result = "98c2ae1ef3ff5aee4999172ec6bd32f3"
print "SEND: 4"
p.sendline("4")
print p.recvuntil("[-] your choice:")
print "SEND: 2"
p.sendline("2")
print p.recvuntil("[-] secret(encode hex):")
print "SEND: %s" % result
p.sendline(result)
print p.recvuntil("[-] your choice:")
# # Step 5
# print "SEND: 5"
# p.sendline("5")
# text = p.recvuntil("[+] 2. your job: guess secret")
# print text
# ciphertext = text.split("[+] ")[3].strip()
# ciphertext = ciphertext[41:]
# result = solve_step_5(ciphertext)
# print "SEND: %s" % result
# p.sendline(result)
# print p.recvuntil("[-] your choice:")
# Step 5(finished)
result = "6122db344a73a14e8247fb2856fbbf94"
print "SEND: 5"
p.sendline("5")
print p.recvuntil("[-] your choice:")
print "SEND: 2"
p.sendline("2")
print p.recvuntil("[-] secret(encode hex):")
print "SEND: %s" % result
p.sendline(result)
print p.recvuntil("[-] your choice:")
# # Step 6
# print "SEND: 6"
# p.sendline("6")
# text = p.recvuntil("[+] 2. your job: guess secret")
# s = text.split("[+] ")[4].strip()
# s = s[70:]
# result = solve_step_6(s[:32].decode('hex'), 1, "")
# print result.encode('hex')
# result = strxor(result, '1')
# print result.encode('hex')
# Step 6(finished)
result = "0a1c9e1464925d23d4a3068313b407ee".decode('hex')
result = strxor(result, '1')
print "SEND: 6"
p.sendline("6")
print p.recvuntil("[-] your choice:")
print "SEND: 2"
p.sendline("2")
print p.recvuntil("[-] secret(encode hex):")
print "SEND: %s" % result.encode('hex')
p.sendline(result.encode('hex'))
print p.recvuntil("[-] your choice:")
p.interactive()
強網先鋒-baby_crt
(好久沒見過這類純數學變換的題目了，里邊部分變換步驟可能有些跳步，熟悉數論的應該能看懂，不熟悉的請先去補充一下基礎知識，或嘗試自行證明)
首先我們知道
S = (Cp * Sp + Cq * Sq) % (n * t1 * t2)
所以
S % t1 = (Cp * Sp + Cq * Sq) % t1
S % t1 = (q * t2 * inv(q * t2, p * t1)) * pow(m + k, dp, p * t1) + 一個t1的倍數 % t1
S % t1 = (a * p * t1 + 1) * pow(m + k, dp, p * t1) % t1
S % t1 = pow(m + k, dp, p * t1) % t1
S & t1 = pow(m + k, dp, t1)
于是
c1 = (m - pow(S, et1, t1) + 1) % t1
c1 = (m - pow(m + k, dp*et1, t1) + 1) % t1
我們知道
et1 = inv(d, t1-1)
d * et1 = x * (t1-1) + 1
dp = d % ((p-1)*(t1-1))
d = dp + y * (p-1) * (t1-1)
所以 (dp + y * (p-1) * (t1-1)) * et1 = x * (t1-1) + 1
dp * et1 = 1 (mod t1-1)
t1是質數，m+k大概率與t1互質，根據費馬小定理
c1 = (m - pow(m + k, dp*et1, t1) + 1) % t1
c1 = (m - (m + k) + 1) % t1
c1 = (1-k) % t1
同理可以推出
c2 = 1 % t2 = 1
于是
sig = pow(S, c1 * c2, n)
變成了
sig = pow(S, c1, n)
這個c1是(1-k) % t1，而k是65536以內隨機的質數，因此不能直接求。接下來我們想辦法處理這個sig。處理sig的關鍵點在S?；氐絊的式子
S = (Cp * Sp + Cq * Sq) % (p * q * t1 * t2)
S % p = (Cp * Sp + Cq * Sq) % p
S % p = (q * t2 * inv(q * t2, p * t1)) * pow(m + k, dp, p * t1) + 一個p的倍數
S % p = (a * p * t1 + 1) * pow(m + k, dp, p * t1) % p
S % p = pow(m + k, dp, p)
同理推出
S % q = pow(m, dq, q)
根據中國剩余定理，可以求出S % n的通解：
S % n = pow(m+k,dp,p)*q*inv(q,p)+pow(m,dq,q)*p*inv(p,q)
于是有
sig = pow(S, c1, n)
sig = (pow(m+k,dp,p)*q*inv(q,p) + pow(m,dq,q)*p*inv(p,q)) ^ c1 % n
這里是形如(a+b) ^ n的形式，用二項式定理展開，注意到前者能被q整除，后者能被p整除，所以展開后只剩下一頭一尾不能被n整除，其余全部消掉。即：
sig = (pow(m+k,dp,p)*q*inv(q,p))^c1 + (pow(m,dq,q)*p*inv(p,q))^c1 % n
sig % q = (pow(m,dq,q)*p*inv(p,q))^c1 % q
sig % q = pow(m,dq*c1,q)
sig^e % q = pow(m,dq*c1*e,q)
sig^e % q = pow(m,c1,q)
此處最后一步是因為
dq = d % ((q-1)*(t2-1))
d = dq + a*(q-1)*(t2-1)
e*d = e*dq + e**(q-1)*(t2-1)
b*(p-1)*(q-1)+1 = e*dq + e**(q-1)*(t2-1)
1 = e*dq (mod q-1)
pow(m, e*dq, q) = m
于是我們可以通過爆破c1，計算sig ^ e - m ^ c1與n的最大公約數得到q。完整代碼如下：
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import gmpy2
n = 26318358382258215770827770763384603359524444566146134039272065206657135513496897321983920652242182112479484135343436206815722605756557098241887233837248519031879444740922789351356138322947108346833956405647578838873425658405513192437479359531790697924285889505666769580176431360506227506064132034621123828090480606055877425480739950809109048177976884825589023444901953529913585288143291544181183810227553891973915960951526154469344587083295640034876874318610991153058462811369615555470571469517472865469502025030548451296909857667669963720366290084062470583318590585472209798523021029182199921435625983186101089395997
m = 26275493320706026144196966398886196833815170413807705805287763413013100962831703774640332765503838087434904835657988276064660304427802961609185997964665440867416900711128517859267504657627160598700248689738045243142111489179673375819308779535247214660694211698799461044354352200950309392321861021920968200334344131893259850468214901266208090469265809729514249143938043521579678234754670097056281556861805568096657415974805578299196440362791907408888958917063668867208257370099324084840742435785960681801625180611324948953657666742195051492610613830629731633827861546693629268844700581558851830936504144170791124745540
sig = 20152941369122888414130075002845764046912727471716839854671280255845798928738103824595339885345405419943354215456598381228519131902698373225795339649300359363119754605698321052334731477127433796964107633109608706030111197156701607379086766944096066649323367976786383015106681896479446835419143225832320978530554399851074180762308322092339721839566642144908864530466017614731679525392259796511789624080228587080621454084957169193343724515867468178242402356741884890739873250658960438450287159439457730127074563991513030091456771906853781028159857466498315359846665211412644316716082898396009119848634426989676119219246
e = 65537
tmpsig = pow(sig, e, n)
for c1 in range(65536):
tmpq = (pow(m, c1, n)-tmpsig)%n
if(gmpy2.gcd(tmpq, n)>1):
q = gmpy2.gcd(tmpq, n)
p = n // q
print "p = " + str(p)
# p = 149580444233086025790179573414856711556292635219028492250676309233306926698347672114881938858364663009604584293251087505211665595084567645013671843742930965283973691725982783226085450276850737892819830479107029360437303876681309102929919702367620122904659518881851295355796994990602745486313662851309278333637
強網先鋒-紅方輔助
查看TCP流量原始數據可得
根據代碼，每5行實際上是一次交互。交互內容與格式為：
1. Client發送字母G(x47)表示開始；
2. Server返回btime；
3. Client發送boffset；
4. Client發送數據，數據格式為：
0~3 數據包編號
4~7 數據包長度
8 當前的sn(x30 or x31 or x32)
9 當前的salt值
5. Server返回下一個數據包編號值
利用btime和offset可以算出t，根據t和salt就可以算出原始明文。最后是一個點陣圖，豎著讀出flag，最后加上QWB{}提交。代碼略
(一定要代碼嘛。。好吧)
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import struct
lines = open("tcp.txt",'r').readlines()
for i in range(0, len(lines), 5):
btime = lines[i+1].strip().decode('hex')
boffset = struct.unpack("
fn = chr(int(lines[i+3][16:18], 16))
salt = int(lines[i+3][18:20], 16)
t = struct.unpack("
t -= boffset
t = struct.pack("
m = ""
count = 0
c = lines[i+3][20:].strip().decode('hex')
for j in range(len(c)):
if(fn == '0'):
m += chr(((ord(c[j]) + salt) & 0xff) ^ ord(t[count]))
elif(fn == '1'):
m += chr(((ord(c[j]) - salt) & 0xff) ^ ord(t[count]))
elif(fn == '2'):
m += chr(((ord(c[j]) ^ salt) & 0xff) ^ ord(t[count]))
count = (count + 1) % 4
print m.strip()
強網先鋒-bank
一個模擬區塊鏈交易的題目，題目核心內容是要偽造交易記錄。根據提示，每一條交易記錄都是
enc(發送方) + enc(接收方) + enc(金額)
雖然我們不知道key，但是明顯我們不需要知道，從view records的交易記錄里直接搬運就可以了。解題所需要的三個要素如下獲取：
enc(我)：請給Alice轉1塊錢，得到的交易記錄前32位就是；
enc(別人)：同上，交易記錄第32~64位就是；
enc(金額)：把交易記錄里邊的大額都讀一遍就夠了。
由于本題不允許使用重復記錄，所以或者構造Alice給我轉不同金額的記錄，或者構造不同的人給我轉相同金額的記錄，不影響解題。解題代碼如下：
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from pwn import *
import hashlib
import string
def mysha256(text):
mysha = hashlib.sha256()
mysha.update(text)
return mysha.hexdigest()
def passPoW(suffix, target):
for a in string.printable:
for b in string.printable:
for c in string.printable:
tmp = a+b+c+suffix
if mysha256(tmp)==target:
return a+b+c
teamtoken = "隊伍token"
p = remote("39.101.134.52", 8005)
text = p.recvuntil("Give me XXX:")
print text
suffix = text.split("+")[1][:17]
target = text.split(" == ")[1][:64]
result = passPoW(suffix, target)
p.sendline(result)
print p.recvuntil("teamtoken:").strip()
p.sendline(teamtoken)
print p.recvuntil("give me your name:").strip()
p.sendline("chainer")
print p.recvuntil("you can choose: transact, view records, provide a record, get flag, hint").strip()
p.sendline("transact")
print p.recvuntil("please give me the trader and the amount(for example:Alice 1)").strip()
p.sendline("Alice 1")
text = p.recvuntil("you can choose: transact, view records, provide a record, get flag, hint")
print text.strip()
s = text.split('
')[1]
nameme = s[2:34]
nameAlice = s[34:66]
p.sendline("view records")
text = p.recvuntil("you can choose: transact, view records, provide a record, get flag, hint")
print text.strip()
newtrans = []
for i in range(2,12):
s = text.split('
')[i]
newtrans.append(nameAlice + nameme + s[64:96])
print newtrans
for i in range(10):
p.sendline("provide a record")
print p.recvuntil("My system is secure if you can give me other records, the receiver can also get the money.").strip()
p.sendline(newtrans[i])
print p.recvuntil("you can choose: transact, view records, provide a record, get flag, hint").strip()
p.sendline("get flag")
p.interactive()
最后說兩句
國密那個題看著就不想做怎么破。。一點動力都沒有
總結
以上是生活随笔為你收集整理的最后一位被整除 oracle,【CTF WriteUp】2020第四届强网杯部分Crypto题解的全部內容，希望文章能夠幫你解決所遇到的問題。
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