mcfx's blog - 2020年2月
/2020/02/
个人博客，（曾经）基本只放题解，现在随便写点啥了吧（HackTM CTF Quals 2020 Writeup
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2020-02-06T03:32:00+08:00由于寒假比较闲，所以找点比赛打。由于需要上交 wp，所以是英文的。
# Crypto
## RSA is easy #1
Since $$N$$ is known, we can compute the encrypted value of each printable character. Then match them with the given encrypted flag, we can decrypt it.
```python
e=65537
n=...
s=[...(encrypted flag)]
from gmpy2 import *
f={}
for j in range(32,127):
f[j**e%n]=j
r=''
for i in s:
r+=chr(f[i])
print r
```
## RSA is easy #2
We doesn't know $$N$$, but we know that $$c=n^e\ mod\ N$$. So $$N$$ is a factor of $$n^e-c$$. Then for two $$c$$, we can check all $$n$$, and calculate their $$gcd$$. Then just follow #1.
```python
from gmpy2 import *
s=open('c').read()
s1='Encrypted flag:'
s=s[s.find(s1)+len(s1):]
s=eval(s.strip())
a=s[0]
b=s[1]
for i in range(97,97+26):
for j in range(97,97+26):
t=gcd(i**65537-a,j**65537-b)
if t>100:
print '|',t
print i
```
## Prison Break
Maintain the difference of every two adjacent numbers.
```cpp
#include
typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef double lf;
typedef long double llf;
typedef std::pair pii;
#define xx first
#define yy second
template inline T max(T a,T b){return a>b?a:b;}
template inline T min(T a,T b){return a0?a:-a;}
template inline bool repr(T &a,T b){return ab?a=b,1:0;}
template inline T gcd(T a,T b){T t;if(a> >>.>*./N>.NN/N>N . /N >N>/>/>N/
WN V*\\NVW\WVN* NVNN WN\ \ . *.NN . N NWW\\.W\ W\\ NW N\W
```
```python
from Crypto.Cipher import AES
c='059fd04bca4152a5938262220f822ed6997f9b4d9334db02ea1223c231d4c73bfbac61e7f4bf1c48001dca2fe3a75c975b0284486398c019259f4fee7dda8fec'.decode('hex')
iv = '42042042042042042042042042042042'.decode('hex')
def chk(key):
aes = AES.new(key,AES.MODE_CBC, iv)
r=aes.decrypt(c)
for i in r:
if ord(i)126:
return False
print(r)
return True
def chkn(x):
r=''
for i in range(32):
r+=chr(x&255)
x>>=8
assert x==0
r=r[::-1]
return chk(r)
a,b=open('key.txt').readlines()[:2]
a=a.strip()
b=b.strip()
va={'>':2, '/':1, 'N':3, ' ':0, '.':0}
vb={'N':3, 'W':4, ' ':0, '\\':1, '.':0, 'V':2}
unk=[]
tot=0
for i in range(59):
p=58-i
po=20**i
if a[p]=='*':
assert b[p]=='*'
unk.append((po,i))
else:
tot+=(va[a[p]]*5+vb[b[p]])*po
def dfs(x,cur):
if x==len(unk):
return chkn(cur)
for i in range(20):
dfs(x+1,cur+unk[x][0]*i)
dfs(0,tot)
```
# Forensics
## RR
It seems to be raid5. After some observation, I found the way to decrypt it.
(decrypt script, 2.img is the $$xor$$ of 1.img and 3.img)
```cpp
#include
typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef double lf;
typedef long double llf;
typedef std::pair pii;
#define xx first
#define yy second
template inline T max(T a,T b){return a>b?a:b;}
template inline T min(T a,T b){return a0?a:-a;}
template inline bool repr(T &a,T b){return ab?a=b,1:0;}
template inline T gcd(T a,T b){T t;if(a>3
return s[:xt]+bytes([s[xt]^(13
return s[:xt]+bytes([s[xt]^(1