[ACTF新生赛2020]frequency

[ACTF新生赛2020]frequency

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属性的详细信息里发现后半段信息

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连在一块进行解码

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根据题目提示应该是频率统计,可以用网站或者脚本实现

https://www.lddgo.net/string/frequency-statistics

_TrehXObmVWKp0Z2IPG0Wn4VVlnHEjVQPOWPdDyzAKo

显然,flag是按照字频从大到小排列的({和}除外)

flag{plokmijnuhbygvrdxeszwq}

脚本

import re

file = open('18.txt')  #根据路径自行调整
line = file.readlines()
file.seek(0,0)
file.close()

result = {}
for i in range(97,123):
   count = 0
   for j in line:
      find_line = re.findall(chr(i),j)
      count += len(find_line)
   result[chr(i)] = count
res = sorted(result.items(),key=lambda item:item[1],reverse=True)

num = 1
for x in res:
      print('频数第{0}: '.format(num),x)
      num += 1

得到结果

python 1.py
频数第1:  ('a', 198)
频数第2:  ('c', 194)
频数第3:  ('t', 188)
频数第4:  ('f', 180)
频数第5:  ('p', 174)
频数第6:  ('l', 170)
频数第7:  ('o', 164)
频数第8:  ('k', 159)
频数第9:  ('m', 152)
频数第10:  ('i', 145)
频数第11:  ('j', 139)
频数第12:  ('n', 133)
频数第13:  ('u', 128)
频数第14:  ('h', 116)
频数第15:  ('b', 109)
频数第16:  ('y', 106)
频数第17:  ('g', 100)
频数第18:  ('v', 92)
频数第19:  ('r', 87)
频数第20:  ('d', 79)
频数第21:  ('x', 73)
频数第22:  ('e', 68)
频数第23:  ('s', 61)
频数第24:  ('z', 54)
频数第25:  ('w', 48)
频数第26:  ('q', 41)

不过使用脚本会有一点点小问题,那就是{和}没有给统计上,需要手动加上

posted on 2023-11-08 10:54  跳河离去的鱼  阅读(103)  评论(0编辑  收藏  举报