(一)makeshift
题目:
from secret import FLAG
flag = FLAG[::-1]
new_flag = ''
for i in range(0, len(flag), 3):
new_flag += flag[i+1]
new_flag += flag[i+2]
new_flag += flag[i]
print(new_flag)
思路:
属于很简单的流加密
首先,flag = FLAG[::-1],步长为-1,从后向前取值,对字符串进行了反转,效果如下:
然后,直接看作将flag三位为一组,进行同一个模式的乱序,m+下来,对应顺序为HTB,从len(c)-3开始,步长为-3遍历字符串,直到索引为-3,那么取出的c中THB索引为i、i+1、i+2,对应赋值即可。
脚本:
c = "!?}De!e3d_5n_nipaOw_3eTR3bt4{_THB"
m = ''
for i in range(len(c)-3, -3, -3):
m += c[i+1]
m += c[i]
m += c[i+2]
print(m)
(二)dynastic
题目:
from secret import FLAG
from random import randint
def to_identity_map(a):
return ord(a) - 0x41
def from_identity_map(a):
return chr(a % 26 + 0x41)
def encrypt(m):
c = ''
for i in range(len(m)):
ch = m[i]
if not ch.isalpha():
ech = ch
else:
chi = to_identity_map(ch)
ech = from_identity_map(chi + i)
c += ech
return c
with open('output.txt', 'w') as f:
f.write('Make sure you wrap the decrypted text with the HTB flag format :-]\n')
f.write(encrypt(FLAG))
ech = chr((ord(mch)-0x41 + i)%26 +0x41)
ord(ech) -0x41 = (ord(mch) - 0x41 + i)mod26
思路:
简单的流加密
对字符串中的字母进行了简单的移位,保留特殊字符,写个等式求逆
c = chr((ord(m) - 0x41 + i)%26 + 0x41)
ord(c) = (ord(m) - 0x41 + i)%26 + 0x41
ord(c) - 0x41 = (ord(m) - 0x41 + i)%26
ord(c) - 0x41 + k*26 = ord(m) - 0x41 +i
感觉这里不会溢出,所以不取模
ord(c) - 0x41 + k*26 - i = ord(m) - 0x41
感觉会溢出,所以取模
char((ord(c) - 0x41 -i)%26 + 0x41) = m
脚本:
def decrypt(c):
m = ''
for i in range(len(c)):
ch = c[i]
if not ch.isalpha():
mch = ch
else:
mch = chr((ord(ch)-