http://www.shiyanbar.com/ctf/1980
image.png
将源码下载下来，进行解读
#!/usr/bin/env python
import sys
alphaL = "abcdefghijklnmopqrstuvqxyz"
alphaU = "ABCDEFGHIJKLMNOPQRSTUVQXYZ"
num = "0123456789"
keychars = num+alphaL+alphaU
if len(sys.argv) != 3: #如果用户输入的参数不是三个的话，就输出本身文件路径，且结束程序
print "Usage: %s SECRET_KEY PLAINTEXT"%(sys.argv[0])
sys.exit()
key = sys.argv[1] #key接收第二个参数
if not key.isalnum(): #如果第二个参数不是全数字与字母组合，则打印不合法，并结束程序
print "Your key is invalid, it may only be alphanumeric characters"
sys.exit()
plaintext = sys.argv[2] #接收第三个参数
ciphertext = ""
for i in range(len(plaintext)):
rotate_amount = keychars.index(key[i%len(key)]) #检测第二个参数的每一位，并返回与keychars匹配的位置
if plaintext[i] in alphaL: #遍历第三个参数的每一位，转为ASCII码，分别按照不同的加密方式进行加密，再转为字符型
enc_char = ord('a') + (ord(plaintext[i])-ord('a')+rotate_amount)%26
elif plaintext[i] in alphaU:
enc_char = ord('A') + (ord(plaintext[i])-ord('A')+rotate_amount)%26
elif plaintext[i] in num:
enc_char = ord('0') + (ord(plaintext[i])-ord('0')+rotate_amount)%10
else:
enc_char = ord(plaintext[i])
ciphertext = ciphertext + chr(enc_char)
print "Encryption complete, ENC(%s,%s) = %s"%(plaintext,key,ciphertext)
按照原题的意思，key=T0pS3cre7key,ciphertext=Bot kmws mikferuigmzf rmfrxrwqe abs perudsf! Nvm kda ut ab8bv_w4ue0_ab8v_DDU，所以我们需要反求plaintext的值
解题思路：
1.将字符串转为ASCII码
2.len(key)=12
image.png
可以看到rotate_amount值一共有12个，会一直循环下去。所以解码配置rotate_amount值时只配置12个。
解码代码：
#!/usr/bin/env python
import sys
alphaL = "abcdefghijklnmopqrstuvqxyz"
alphaU = "ABCDEFGHIJKLMNOPQRSTUVQXYZ"
num = "0123456789"
keychars = num+alphaL+alphaU
key='T0pS3cre7key'
ciphertext = "Bot kmws mikferuigmzf rmfrxrwqe abs perudsf! Nvm kda ut ab8bv_w4ue0_ab8v_DDU"
rotate_amount=[]
for i in key:
rotate_amount.append(keychars.index(i))
print rotate_amount
plain_text=''
for j in range(len(ciphertext)):
if ciphertext[j] in alphaL:
for i in range(26):
if (rotate_amount[j%12]+i)%26==(ord(ciphertext[j])-ord('a')):
plain_text=plain_text+chr(ord('a')+i)
elif ciphertext[j] in alphaU:
for i in range(26):
if (rotate_amount[j%12]+i)%26==(ord(ciphertext[j])-ord('A')):
plain_text=plain_text+chr(ord('A')+i)
elif ciphertext[j] in num:
for i in range(10):
if (rotate_amount[j%12]+i)%10==(ord(ciphertext[j])-ord('0')):
plain_text=plain_text+chr(ord('0')+i)
else:
plain_text=plain_text+ciphertext[j]
print(plain_text)
image.png
得到flag：th4ts_w0rs3_th4n_DES
难点在于取余数的逆运算，且得琢磨一会儿
