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MOBILE

Mobile Analysis

这个题目就是简单的mobile分析

import base64
p1 = 'bGFtkaXNwjSVNDQ3'
p1 = p1[10:]+p1[3:9]+p1[0:3]+p1[9:10]
print(p1,end='')
table = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/='
tran_table1 = "g6dfYnaH9qFiIoby5RlZB0WAxJVzhLOD2p4w=uT/rmcGjkMC1X7ePsSKNQ3v8t+UE"
tran_table2 = "6lTD9AC/5YQIinxbw3cs4tUXfoF8egj1EdMyLWkV=NRB02uqvrZhmJPpOza7KG+HS"
b = "J0tpzHRuhTQpLauS"
for i in range(len(b)):
    print(table[tran_table1.find(b[i])],end='')

c = "otG28PYN8CtG"
for i in range(len(c)):
    print(table[tran_table2.find(c[i])],end='')
end="SVNDQ3tkaXNwbGFjZV9hbHRlcm5hdGl2ZV9tb2JpbGV9"
print('')
print(base64.b64decode(end))

通过这个题目对base64和简单的mobile又有了了解。

Easy Mobile

Flag分多段加密

a=[5,44,0x60] 
b=[27,82] 
for i in range(len(a)): 
for j in range(len(b)): 
print(chr(a[i]^b[j]),end=' ') 
print()
#'W7{2'

a = ['W','7','2','{'] 
b = [110,0x7e,0xb9,0xc0] 
for i in range(4): 
for j in range(4): 
print(hex(b[i]-ord(a[j])),end=' ') 
print() //得到' {W72Eb7L'

ZjlDZQ

Base64解码 f9Ce

加减异或

a = [74, 62, 28, 0x20, 60] 
for i in range(len(a)): 
print(chr((a[i]^5)+20),end='')
'cO-9M'

table = '0123456789abcdefghijklmn' 
tran_table = 'njfb7326aeimlhd951048cgk' 
box = [] 
for i in range(len(table)): 
box.append(tran_table.find(table[i])) 
print(box) 
f = '' 
s = '=K5blEMshGi=QLwCzVS1LUPy' 
for i in range(len(s)): 
f += s[box[i]] 
print(f)
# SVME1
zslLChbUwG5PLiKyQ==

Md5解密得到

'87Ed-'

然后是两个base64和一个DES，进行解密得到 T46O4f

p = [ 0x65, 0x27, 0x22, 0x20, 0x26, 0x70, 0x76, 0x26, 0x28, 0x4C, 
0x3B, 0x68, 0x49, 0x6A, 0x6B, 0x5A, 0x25, 0x37, 0x53, 0x36, 
0x36, 0x4F, 0x60, 0x2C, 0x44] 
for i in range(len(p)): 
print(chr(i^p[24-i]),end='')
#'D-bL23U0-SaaEe5C87dc2540}'
ISCC{W72Eb7Lf9CecO-9M87Ed-T46O4fD-bL23U0-SaaEe5C87dc2540}
通过这个题目收获了加减异或算法和md5加密，des加密，flag的多段加密解密。

好玩的？是新语言哦

随机数生成加异或,这个题目难度不大

ISCC{19fb4dd02ea64ca8c96c868f6c5434e4}

通过这个题目学会了随机数的加减异或。题目难度较低。