BUUCTF_Crypto_[WUSTCTF2020]B@se

密文：MyLkTaP3FaA7KOWjTmKkVjWjVzKjdeNvTnAjoH9iZOIvTeHbvD==
JASGBWcQPRXEFLbCDIlmnHUVKTYZdMovwipatNOefghq56rs****kxyz012789+/
题目给了一串密文和base64的表,这个一样是自定义的表,但是缺了一点,

爆破出来就好了,脚本:

import base64
import string
file=open('C:/Users/25287/Desktop/1.txt','w')
str1='MyLkTaP3FaA7KOWjTmKkVjWjVzKjdeNvTnAjoH9iZOIvTeHbvD=='
string2='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
for i in range(65,122):
    for j in range(65,122):
        for k in range(65,122):
            for l in range(65,122):
                string1='JASGBWcQPRXEFLbCDIlmnHUVKTYZdMovwipatNOefghq56rs'+chr(i)+chr(j)+chr(k)+chr(l)+'kxyz012789+/'
                file.write(str(base64.b64decode(str1.translate(str.maketrans(string1,string2)))))
            string1=''

我把输出存在了文件里,方便查找,在文件的末尾位置得到flag:

flag{base64_1s_v3ry_e@sy_and_fuN}

标签：BUUCTF,122,base64,Crypto,chr,65,range,se,string1
来源： https://www.cnblogs.com/1ucky/p/15840972.html