每天10道Crypto Day3
1.救世捷径
2.坏蛋是雷宾
3.[ACTF新生赛2020]crypto-classic0
题目
hint.txt :哼，压缩包的密码？这是小Z童鞋的生日吧==
cipher.txt: Ygvdmq[lYate[elghqvakl}
以及一个howtoencrypt.zip的加密压缩包
题目提示压缩包密码是生日，那么通过爆破 8位数的年月日。
打开加密文件得到代码
#include解密python脚本
ciper = 'Ygvdmq[lYate[elghqvakl}'
for i in range(0,23):
flag = ord(ciper[i])^0x7
flag+=3
print(chr(flag),end='')
flag{my_naive_encrytion}
4.[NPUCTF2020]这是什么觅
字母代表星期的首字母，其中 S 和 T 都出现两次，所以 S1 代表 SAT，S2 代表 SUN ，每组最后一个数字即代表第几行，F1 W1 S22 S21 T12 S11 W1 S13对应得到 3 1 12 5 14 4 1 18，对照字母表 calendar。
flag{calendar}
5.[AFCTF2018]Single
直接词频分析爆破
flag{Oh_U_found_it_nice_tRy}
6.[WUSTCTF2020]B@se
看的师傅的博客
class base64:
def __init__(self,alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"):
self.alphabet = alphabet
def _EnInsideManage(self,strlist):
strflag = ""
temp = ord(strlist[0]) >> 2
strflag += self.alphabet[temp]
temp = ((ord(strlist[0])&3)<<4)|(ord(strlist[1])>>4)
strflag += self.alphabet[temp]
temp = ((ord(strlist[1])&15)<<2)|(ord(strlist[2])>>6)
strflag += self.alphabet[temp]
temp = (ord(strlist[2])&63)
strflag += self.alphabet[temp]
return strflag
def enbase64(self,charString):
encode = ""
for i in range(len(charString)//3):
encode += self._EnInsideManage(charString[i*3:i*3+3])
if len(charString)%3!=0:
if len(charString)%3 == 1:
encode += self._EnInsideManage(charString[-1:]+chr(0)+chr(0))[:2]+"=="
if len(charString)%3 == 2:
encode += self._EnInsideManage(charString[-2:]+chr(0))[:3]+'='
return encode
def TenToBin(self,tenum):
binstr = ""
for i in range(5,-1,-1):
if 1 == (tenum//(2**i)):
binstr += '1'
tenum = tenum%(2**i)
else:
binstr += '0'
return binstr
def BinToStr(self,strbin):
"Turn the binary string to a ASCII string"
strten = ""
for i in range(len(strbin)//8):
num = 0
test = strbin[i*8:i*8+8]
for j in range(8):
num += int(test[j])*(2**(7-j))
strten += chr(num)
return strten
def debase64(self,base64string):
binstr = ""
for i in base64string:
binstr += self.TenToBin(self.alphabet.find(i))
return self.BinToStr(binstr)
from itertools import combinations, permutations
for i in list(permutations(['3','4','j','u'], 4)):
try:
password = 'JASGBWcQPRXEFLbCDIlmnHUVKTYZdMovwipatNOefghq56rs{}kxyz012789+/'.format(''.join(i))
print(password)
newobj = base64(alphabet=password)
print(newobj.debase64("MyLkTaP3FaA7KOWjTmKkVjWjVzKjdeNvTnAjoH9iZOIvTeHbvD=="))
except:
pass
{base64_1s_v3ry_e@sy_and_fuN}
7.[ACTF新生赛2020]crypto-classic1
键盘加密 和 维吉尼亚密码
获取的维吉尼亚密文
SRLU{LZPL_S_UASHKXUPD_NXYTFTJT}
通过爆破获取 密钥key ,我们可以猜到 SRLU的明文是ACTF
所以
#破解key python3
s='ABCDEFGHIJKLMNOPQRSTUVWXYZ'
s1='ACTF'
s2='SRLU'
key =''
for i in range(len(s1)):
key+=s[(s.find(s2[i])-s.find(s1[i]))%26]
print(key)
得到 key = sp
flag{what_a_classical_vigenere}
8.EasyProgram
参考文章
key = 'whoami'
s = []
t = []
j = 0
for i in range(256):
s.append(i)
for i in range(256):
t.append(key[i % len(key)])
for i in range(256):
j=(j+s[i]+ord(t[i])) % 256
s[i],s[j]=s[j],s[i]
i=0
j=0
f = [0x00,0xBA,0x8F,0x11,0x2B,0x22,0x9F,0x51,0xA1,0x2F,0xAB,0xB7,0x4B,0xD7,0x3F,0xEF,0xE1,0xB5,0x13,0xBE,0xC4,0xD4,0x5D,0x03,0xD9,0x00,0x7A,0xCA,0x1D,0x51,0xA4,0x73,0xB5,0xEF,0x3D,0x9B,0x31,0xB3]
flag = ''
for m in range(38):
i = (i + 1)% 256
j = (j + s[i])% 256
s[i],s[j]=s[j],s[i]
x = (s[i]+(s[j] % 256)) % 256
flag += chr(f[m] ^ s[x])
print(flag)
flag{f238yu28323uf28u2yef2ud8uf289euf}
9.[BJDCTF2020]Polybius
参考文章 (波利比奥斯方阵密码)
得到 flag{flagispolybius}
10.[WUSTCTF2020]大数计算
关于宇宙终极问题的答案 x,y,z
P1 = 1
for i in range(1,2021):
P1 *=i
'''
P2 = (520**1314 + 2333**666)
P3 = 80538738812075974 + 80435758145817515 + 12602123297335631
P4 = (22**2+36)*1314
print(P1)
print('-----------------')
print(P2)
print('--------------')
print(P3)
print('------------------')
print(P4)
'''
a =38609695
b =67358675
c =17357662
d =683280
print(hex(a)[2:]+'-'+hex(b)[2:]+'-'+hex(c)[2:]+'-'+hex(d)[2:])
flag{24d231f-403cfd3-108db5e-a6d10}