[BJDCTF2020]这是base??

发表于 2024-06-16

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: '='}

chipertext:
FlZNfnF6Qol6e9w17WwQQoGYBQCgIkGTa9w3IQKw

很简单的思路

就是将密文对照着字典重新编辑一遍

也就是找到每个字母对应的数字，将其替换为base64表的一般字符

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: '='}

a = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/='    #标准表

c='FlZNfnF6Qol6e9w17WwQQoGYBQCgIkGTa9w3IQKw'

ds=''  #把dict转换成字符串方便处理
for i in range(65):
    ds+=dict[i]

l=[]
for i in range(len(c)):
    l.append(ds.index(c[i]))  #无论换不换表，base64变换本身产生的6位二进制数对应的十进制数是不变的，这里就是找到密文c的每个字符在dict表中键值

#print(l)  #l中存的是索引值（下标数字）

m1=''
for ll in l:
    m1+=a[ll]  #找到l中所存的每个数字在标准的base64加密表中所对应的字符
print(m1)  #m1是标准base64表编码结果

m2=base64.b64decode(m1)  #直接调用函数恢复出明文
print(m2)

得到flag