NNSCTF 这是BASE??

https://www.nssctf.cn/problem/691

拿到题目可以看到一个数组与编码后的字符，简单查看可以发现其符合base表的特征

image.png

那接下来就简单了，python写一个脚本把键值对遍历一下就好

a = ""
base64_tab = {0: 'J', 1: 'K', 2: 'L', 3: 'M', 4: 'N', 5: 'O', 6: 'x', 7: 'y', 8: 'U', 9: 'V', 10: 'z', 11: 'A', 12: 'B', 13: 'C', 14: 'D', 15: 'E', 16: 'F', 17: 'G', 18: 'H', 19: '7', 20: '8', 21: '9', 22: 'P', 23: 'Q', 24: 'I', 25: 'a', 26: 'b', 27: 'c', 28: 'd', 29: 'e', 30: 'f', 31: 'g', 32: 'h', 33: 'i', 34: 'j', 35: 'k', 36: 'l', 37: 'm', 38: 'W', 39: 'X', 40: 'Y', 41: 'Z', 42: '0', 43: '1', 44: '2', 45: '3', 46: '4', 47: '5', 48: '6', 49: 'R', 50: 'S', 51: 'T', 52: 'n', 53: 'o', 54: 'p', 55: 'q', 56: 'r', 57: 's', 58: 't', 59: 'u', 60: 'v', 61: 'w', 62: '+', 63: '/', 64: '='}
for x in base64_tab.values():
    a += x

print(a)

然后使用CyberChef解码

也可以用python一步到位

import base64
dict={0: 'J', 1: 'K', 2: 'L', 3: 'M', 4: 'N', 5: 'O', 6: 'x', 7: 'y', 8: 'U', 9: 'V', 10: 'z', 11: 'A', 12: 'B', 13: 'C', 14: 'D', 15: 'E', 16: 'F', 17: 'G', 18: 'H', 19: '7', 20: '8', 21: '9', 22: 'P', 23: 'Q', 24: 'I', 25: 'a', 26: 'b', 27: 'c', 28: 'd', 29: 'e', 30: 'f', 31: 'g', 32: 'h', 33: 'i', 34: 'j', 35: 'k', 36: 'l', 37: 'm', 38: 'W', 39: 'X', 40: 'Y', 41: 'Z', 42: '0', 43: '1', 44: '2', 45: '3', 46: '4', 47: '5', 48: '6', 49: 'R', 50: 'S', 51: 'T', 52: 'n', 53: 'o', 54: 'p', 55: 'q', 56: 'r', 57: 's', 58: 't', 59: 'u', 60: 'v', 61: 'w', 62: '+', 63: '/', 64: '='}

#k1 = "JKLMNOxyUVzABCDEFGH789PQIabcdefghijklmWXYZ0123456RSTnopqrstuvw+/="
k1 = ""

for i in dict:
    k1 += dict[i]
    print(k1)

k2 = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/="
en = "FlZNfnF6Qol6e9w17WwQQoGYBQCgIkGTa9w3IQKw"
b = ''
for i in en:
    a = k1.find(i)
    print(a)
    b+=k2[a]
    print(b)

print(base64.b64decode(b))

我当初没接触这种题目的时候猜测的之间替换表然后查询，而是将自定义编码后的参数转化为标准编码，再使用标准base64解码