HGAME2021-week2-writeup

week2-writeup-6vv+

Reverse

fake_debugger beta

先输入个hgame{aaaaaaaaaaaaaa}试试

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eax: 127
ebx: 23
ecx: 0
zf: 0				h
--------------INFO--------------
 
eax: 127
ebx: 127
ecx: 0
zf: 1				
--------------INFO--------------
 
eax: 74
ebx: 45
ecx: 1
zf: 0				g
--------------INFO--------------
 
eax: 74
ebx: 74
ecx: 1
zf: 1
--------------INFO--------------
 
eax: 34
ebx: 67
ecx: 2
zf: 0				a
--------------INFO--------------
 
eax: 34
ebx: 34
ecx: 2
zf: 1
--------------INFO--------------
 
eax: 52
ebx: 89
ecx: 3
zf: 0				m
--------------INFO--------------
 
eax: 52
ebx: 52
ecx: 3
zf: 1
--------------INFO--------------
 
eax: 104
ebx: 13
ecx: 4
zf: 0				e
--------------INFO--------------
 
eax: 104
ebx: 104
ecx: 4
zf: 1
--------------INFO--------------
 
eax: 99
ebx: 24
ecx: 5
zf: 0				{
--------------INFO--------------
 
eax: 99
ebx: 99
ecx: 5
zf: 1
--------------INFO--------------
 
eax: 66
ebx: 35
ecx: 6
zf: 0				a

可以看到，输入不同字符会改变eax的值，而正确的字符的ascii码为eax异或ebx，然后正确的eax值是zf=1时ebx的值，因此只需要计算每一轮zf分别等于0和1时ebx的值的异或，ecx表示循环次数，可以用来判断第几位字符

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eax: 65
ebx: 35
ecx: 6
zf: 0
--------------INFO--------------
 
eax: 65
ebx: 122
ecx: 6
zf: 1				122 xor 35 = 89 (Y)
--------------INFO--------------

以此一位一位地类推，得到flag

hgame{You_Kn0w_debuGg3r}

就是有点费键盘

crypto

whitegiveRSA

公钥与私钥的产生：

(1)进行加密之前，首先找出2个不同的大质数p和q

(2)计算n=p*q

(3)根据欧拉函数，求得φ(n)=φ(p)φ(q)=(p−1)(q−1)

(4)找出一个公钥e，e要满足: 1<e<φ(n) 的整数，且使e和φ(N)互质。

(5)根据e*d除以φ(n)余数为1，找到私钥d。

(6)所以,公钥就是(n,e) 私钥就是(n,d)

消息加密:

m^e除以n求余数即为c(密文)

img

消息解密:

c^d除以n求余数即为m(明文)

img

binascii
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import gmpy2
n=882564595536224140639625987659416029426239230804614613279163
#这里我用yafu分解了n
p=857504083339712752489993810777
q=1029224947942998075080348647219
e=65537
c=747831491353896780365654517748216624798517769637260742155527
phi=(p-1)*(q-1)
d=gmpy2.invert(e,phi)
m=pow(c,d,n)
print(hex(m))
print(binascii.unhexlify(hex(m)[2:].strip("L")))

gcd or more?

e=2，用上一题的脚本显示找不到d，百度到了rabin加密

https://en.wikipedia.org/wiki/Rabin_cryptosystem

然后题目直接给出了p和q，不用手动分解n，就直接套公式解，从四个输出中找可能存在的真正明文

exp

binascii
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import gmpy2
import libnum
p = 85228565021128901853314934583129083441989045225022541298550570449389839609019
q = 111614714641364911312915294479850549131835378046002423977989457843071188836271
n = p * q
#cipher = pow(s2n(FLAG), 2, n)
#print(cipher)
c=7665003682830666456193894491015989641647854826647177873141984107202099081475984827806007287830472899616818080907276606744467453445908923054975393623509539
e=2


mp = pow(c, (p + 1) / 4, p)
mq = pow(c, (q + 1) / 4, q)
yp = gmpy2.invert(p, q)
yq = gmpy2.invert(q, p)
r = (yp * p * mq + yq * q * mp) % n
rr = n - r
s = (yp * p * mq - yq * q * mp) % n
ss = n - s
print libnum.n2s(r)
print libnum.n2s(rr)
print libnum.n2s(s)
print libnum.n2s(ss)

hgame{3xgcd~is~really~difficult}(狗头）

signin

#竟是数学题

c=a^p*m（mod p）

c=(a^p%p)*(m%p)%p

c=(a%p)*(m%p)%p #费马小定理

c=a*(m%p)%p

c*a^(p-2)=m%p #费马小定理

exp

libnum import *
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a = 143476170144915086020651461328049339656692126954818003650099852868145841319041470073020587125213976925474341076505642227853407676058773113885865826640680023147360447444464935801614525784110903480871654141802517672864060026452463495645509064414259143561988416176609991854421929607534242834743124652491022090289
p = 144990456446230144694263694418769858495021475043392279183925650174062004178559844401086360607928511714506442495515664176646230769554023179040226957152981442993492885354384799122222074255748702871095896167656675843624069307328779207379633792655160544271545569917511963865120135589875939937494010889658831130797
c = 120094382885627426802043931936833471575878884523974767764562747592243608117847482125722859363444069094268578813034406859087905808108331619918394213908184448430084247326673068817903131008945590119114470183031722078599931222887555377046173518111618239410979061284925061871538324577871419928874075180494037057601
p2=p-2
am=pow(a,p2,p)
cm=c*am
m=cm%p
print(hex(m))
m=n2s(m)
print(m)

misc

Telegraph：1601 6639 3459 3134 0892

这名字很可疑，百度一下，我就知道

好吧，那咱打开au，显示频谱

打开滤波器，过滤850Hz

oh我的上帝，这熟悉的点杠隔

然后把莫斯电码写下来

-.–/—/..-/.-./..-./.-../.-/–./../…/—…/….-/–./—–/—–/-../…/—–/-./–./-…/..-/-/-./—–/-/….-/–./—–/—–/-../–/.-/-./—–/…–/—-./…–/.—-/—–/-.-/..

解密得

yourflagis:4g00ds0ngbutn0t4g00dman039310ki

转化大写包上hgame{}就能交

Hallucigenia

放进stegsolve康康有小惊喜

扫一扫

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gmBCrkRORUkAAAAA+jrgsWajaq0BeC3IQhCEIQhCKZw1MxTzSlNKnmJpivW9IHVPrTjvkkuI3sP7bWAEdIHWCbDsGsRkZ9IUJC9AhfZFbpqrmZBtI+ZvptWC/KCPrL0gFeRPOcI2WyqjndfUWlNj+dgWpe1qSTEcdurXzMRAc5EihsEflmIN8RzuguWq61JWRQpSI51/KHHT/6/ztPZJ33SSKbieTa1C5koONbLcf9aYmsVh7RW6p3SpASnUSb3JuSvpUBKxscbyBjiOpOTq8jcdRsx5/IndXw3VgJV6iO1+6jl4gjVpWouViO6ih9ZmybSPkhaqyNUxVXpV5cYU+Xx5sQTfKystDLipmqaMhxIcgvplLqF/LWZzIS5PvwbqOvrSlNHVEYchCEIQISICSZJijwu50rRQHDyUpaF0y///p6FEDCCDFsuW7YFoVEFEST0BAACLgLOrAAAAAggUAAAAtAAAAFJESEkNAAAAChoKDUdOUIk=

b64解出来看见行末有个GNP.根据题目应该是要翻转过来

60 42 AE 44 4E 45 49 00 00 00 00 FA 3A E0 B1
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00 01 02 03 04 05 06 07  08 09 0A 0B 0C 0D 0E 0F 
--------------------------------------------------------------------
82 60 42 AE 44 4E 45 49  00 00 00 00 FA 3A E0 B1   | .`B.DNEI.....:..
66 A3 6A AD 01 78 2D C8  42 10 84 21 08 42 29 9C   | f.j..x-.B..!.B).
35 33 14 F3 4A 53 4A 9E  62 69 8A F5 BD 20 75 4F   | 53..JSJ.bi... uO
AD 38 EF 92 4B 88 DE C3  FB 6D 60 04 74 81 D6 09   | .8..K....m`.t...
B0 EC 1A C4 64 67 D2 14  24 2F 40 85 F6 45 6E 9A   | ....dg..$/@..En.
AB 99 90 6D 23 E6 6F A6  D5 82 FC A0 8F AC BD 20   | ...m#.o........ 
15 E4 4F 39 C2 36 5B 2A  A3 9D D7 D4 5A 53 63 F9   | ..O9.6[*....ZSc.
D8 16 A5 ED 6A 49 31 1C  76 EA D7 CC C4 40 73 91   | ....jI1.v....@s.
22 86 C1 1F 96 62 0D F1  1C EE 82 E5 AA EB 52 56   | "....b........RV
45 0A 52 23 9D 7F 28 71  D3 FF AF F3 B4 F6 49 DF   | E.R#.(q......I.
74 92 29 B8 9E 4D AD 42  E6 4A 0E 35 B2 DC 7F D6   | t.)..M.B.J.5...
98 9A C5 61 ED 15 BA A7  74 A9 01 29 D4 49 BD C9   | ...a....t..).I..
B9 2B E9 50 12 B1 B1 C6  F2 06 38 8E A4 E4 EA F2   | .+.P......8.....
37 1D 46 CC 79 FC 89 DD  5F 0D D5 80 95 7A 88 ED   | 7.F.y..._....z..
7E EA 39 78 82 35 69 5A  8B 95 88 EE A2 87 D6 66   | ~.9x.5iZ.......f
C9 B4 8F 92 16 AA C8 D5  31 55 7A 55 E5 C6 14 F9   | ........1UzU....
7C 79 B1 04 DF 2B 2B 2D  0C B8 A9 9A A6 8C 87 12   | |y...++-........
1C 82 FA 65 2E A1 7F 2D  66 73 21 2E 4F BF 06 EA   | ...e..-fs!.O...
3A FA D2 94 D1 D5 11 87  21 08 42 10 21 22 02 49   | :.......!.B.!".I
92 62 8F 0B B9 D2 B4 50  1C 3C 94 A5 A1 74 CB FF   | .b.....P.<...t..
FF A7 A1 44 0C 20 83 16  CB 96 ED 81 68 54 41 44   | ...D. ......hTAD
49 3D 01 00 00 8B 80 B3  AB 00 00 00 02 08 14 00   | I=..............
00 00 B4 00 00 00 52 44  48 49 0D 00 00 00 0A 1A   | ......RDHI......
0A 0D 47 4E 50 89                                 | ..GNP.

反转后保存成png文件，是一个长得很别致的flag

翻转一下就好

ctf-hgame2021
#HGAME-ctf
HGAME2021-week2-writeup
http://example.com/2021/03/02/HGAME2021-week2-writeup-0/
作者
Magnesium
发布于
2021年3月2日
许可协议