Bugku——Crypto 1

把猪困在猪圈里

下载附件:是一个文本
文本内容:
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base 64转图片

image
猪圈密码
image
flag{thisispigpassword}

你喜欢下棋吗

棋盘密码在线工具:
image
flag.txt文本内容:

一种5bit的编码 bugku里面的内容为小写 bugku{11111 11001 00011 00111 01001 11011 10110 11111 10000 01110 11011 10110 11111 01001 00001}
5bit编码在线工具解码:

image
bugku{baud0tc0de}

EN-气泡

  • 气泡 >> bubble编码
  • bubble编码特征:x开头,x结尾,某些特定位是元音字母,而且出现频率特别高,而且基本单元是5元组
    网上找的脚本:
点击查看代码

from bubblepy import BubbleBabble

#解三次码得flag bubble编码特征是x开头x结尾,中间都是5位一组其中某个元音字母出现比较频繁

from bubblepy import BubbleBabble

#导入包bubblepy

str='xivak-notuk-cupad-tarek-zesuk-zupid-taryk-zesak-cined-tetuk-nasuk-zoryd-tirak-zysek-zaryd-tyrik-nisyk-nenad-tituk-nysil-hepyd-tovak-zutik-cepyd-toral-husol-henud-titak-hesak-nyrud-tarik-netak-zapad-tupek-hysek-zuned-tytyk-zisuk-hyped-tymik-hysel-hepad-tomak-zysil-nunad-tytak-nirik-copud-tevok-zasyk-nypud-tyruk-niryk-henyd-tityk-zyral-nyred-taryk-zesek-corid-tipek-zysek-nunad-tytal-hitul-hepod-tovik-zurek-hupyd-tavil-hesuk-zined-tetuk-zatel-hopod-tevul-haruk-cupod-tavuk-zesol-ninid-tetok-nasyl-hopid-teryl-nusol-heped-tovuk-hasil-nenod-titek-zyryl-hiped-tivyk-cosok-zorud-tirel-hyrel-hinid-tetok-hirek-zyped-tyrel-hitul-nyrad-tarak-hotok-cuvux'

#str是待解密字符

#print(str)

Str=BubbleBabble()

str1=Str.decode(str).decode()

str2=Str.decode(str1).decode()

str3=Str.decode(str2).decode()

print(str3)

# !/usr/bin/env python
# -*- coding:utf-8 -*-
# @Time  : 2024/7/20 上午10:08
# @Author: yangjian
# @File  : 气泡编码.py

bugku{th1s_1s_A_Bubb13}

你以为是md5吗

文本内容:

bci177a7a9c7udf69c248647b4dfc6fd84o
我认为不是MD5……结果就是MD5,将不属于MD5的范围的字符去掉就可以了
bc177a7a9c7df69c248647b4dfc6fd84
flag{666666666666}

小山丘的秘密

将压缩包解压:
一张图和一个文本:
image
文本内容:
image

flag.txt 中给出A=1,一般的希尔密码是A=0,B=1, C=2 …所以字母表需要修改,现在的字母表为Z=0 , A=1 , B=2 …
那张图片是密码,所以密码为:123014560不是对应的字母:abczadefz,这里弄了好久
image
就得到了flag:
bugku{whatahill}

Math&English

下载附件,文本内容:

你会数学和英语吗? 3×7= 11×3= sin(Π/6)+sin(Π/6)= 2×11= 3×cos(2Π)= 22×2= 6×9= 2.5×2= tan(Π) + 1 = 5×7= 7×9-9= ln(e) + 2 = (6×6)-1= 4×10 + cos(0) = 26×2 = ln(e) × 10 +3 =
写出答案:

image
对应着元音表,就得到flag
image

bugku{yuanyinpwd}

easy_hash

脚本:

import hashlib
import string
md5_list = {}
for c in string.printable:
m = hashlib.md5(c.encode())
md5_list[str(m.hexdigest())] = c
with open('output', 'r') as file:
for line in file:
line = line.split('\n')[0]
if line in md5_list:
print(md5_list[line], end='')

还有一种解法:

8fa14cdd754f91cc6554c9e71929cce7 f
2db95e8e1a9267b7a1188556b2013b33 l
0cc175b9c0f1b6a831c399e269772661 a
b2f5ff47436671b6e533d8dc3614845d g
f95b70fdc3088560732a5ac135644506 {
61e9c06ea9a85a5088a499df6458d276 W
e1671797c52e15f763380b45e841ec32 e
c4ca4238a0b923820dcc509a6f75849b 1
4a8a08f09d37b73795649038408b5f33 c
cfcd208495d565ef66e7dff9f98764da 0
6f8f57715090da2632453988d9a1501b m
e1671797c52e15f763380b45e841ec32 e
b14a7b8059d9c055954c92674ce60032 _
e358efa489f58062f10dd7316b65649e t
cfcd208495d565ef66e7dff9f98764da 0
b14a7b8059d9c055954c92674ce60032 _
e358efa489f58062f10dd7316b65649e t
2510c39011c5be704182423e3a695e91 h
e1671797c52e15f763380b45e841ec32 e
b14a7b8059d9c055954c92674ce60032 _
f1290186a5d0b1ceab27f4e77c0c5d68 w
cfcd208495d565ef66e7dff9f98764da 0
4b43b0aee35624cd95b910189b3dc231 r
c4ca4238a0b923820dcc509a6f75849b 1
8277e0910d750195b448797616e091ad d
b14a7b8059d9c055954c92674ce60032 _
cfcd208495d565ef66e7dff9f98764da 0
8fa14cdd754f91cc6554c9e71929cce7 f
b14a7b8059d9c055954c92674ce60032 _
6f8f57715090da2632453988d9a1501b m
8277e0910d750195b448797616e091ad d
e4da3b7fbbce2345d7772b0674a318d5 5
cbb184dd8e05c9709e5dcaedaa0495cf }

flag{We1c0me_t0_the_w0r1d_0f_md5}

MaybeEasyRSA

task.py文件内容:

from Crypto.Util.number import *
from secret import flag
from sympy import nextprime
flag=b''
r = getRandomNBitInteger(64)
p1 = r5 + r4 - r3 + r2 - r + 2024
q1 = r5 - r4 + r3 - r2 + r + 2024
p =nextprime(p1)
q =nextprime(q1)
n = p*q
def enc(flag, n):
m = bytes_to_long(flag)
return pow(m, 65537, n)
c = enc(flag, n)
print(n)
print(c)
n=1058756604181685917958185571746711428601045466594619627088399455470892502310139608978164573773765910533535849969889860275644658086339362201840191861497991951344100284004205747279828567835277683
c=643011921358774538378881505518732362708757638007688649119354348647468190640688857686213431516881297805187671774564672497176686191162897918672339254715366498963369868410476737543157157149810569

用脚本:

from gmpy2 import *
from Crypto.Util.number import *
from sympy import nextprime
n = 1058756604181685917958185571746711428601045466594619627088399455470892502310139608978164573773765910533535849969889860275644658086339362201840191861497991951344100284004205747279828567835277683
c = 643011921358774538378881505518732362708757638007688649119354348647468190640688857686213431516881297805187671774564672497176686191162897918672339254715366498963369868410476737543157157149810569
e = 65537
r_prox = iroot(n, 10)[0]
for r in range(r_prox - 2000, r_prox + 2000):
p1 = r ** 5 + r ** 4 - r ** 3 + r ** 2 - r + 2024
q1 = r ** 5 - r ** 4 + r ** 3 - r ** 2 + r + 2024
p = nextprime(p1)
q = nextprime(q1)
if p * q == n:
d = invert(e, (p - 1) * (q - 1))
print(long_to_bytes(pow(c, d, n)))
break

bugku{01a25ea3fd6349c6e635a1d0196e75fb}

easy_crypto

全是0和1,转为摩斯密码:

0010 0100 01 110 1111011 11 11111 010 000 0 001101 1010 111 100 0 001101 01111 000 001101 00 10 1 0 010 0 000 1 01111 10 11110 101011 1111101
..-. .-.. .- --. ----.-- -- ----- .-. ... . ..--.- -.-. --- -.. . ..--.- .---- ... ..--.- .. -. - . .-. . ... - .---- -. ----. -.-.-- -----.-

flag{m0rse_code_1s_interest1n9!}

黄道十二宫

下载附件:是一张图片
image
用工具:AZdecrypt
image
flag{alphananke}

一段新闻

题目:
image

推荐一个在线加解密的网址http://www.atoolbox.net/Category.php?Id=27)
找到隐藏字符加密,进行解密操作
image
社会主义编码
image
flag{why_d0nt_you_try_t0_understand_socia1ism?}

7+1+0

文本内容:
4nXna/V7t2LpdLI44mn0fQ==
wp里的脚本:

import base64
c = '4nXna/V7t2LpdLI44mn0fQ=='
decode1_str = base64.b64decode(c)
decode2_str = ''
for i in range(len(decode1_str)):
decode2_str += chr(decode1_str[i])
print(decode2_str)
flag = ''
//ord将字符按ascii码转换为数字,chr将数字转换为字符
for i in range(len(decode2_str)):
if i % 2 == 0:
flag += chr(ord(decode2_str[i])-128)
else:
flag += chr(ord(decode2_str[i]))
print(flag)

bugku{7bit28bit}

这是个盲兔子,竟然在唱歌!

题目就告诉我们解密方式:
盲文——>兔子——>音乐符号
image

道友不来算一算凶吉?

题目:

半仙我夜观天象,掐指一算,卜出卦象如下,不知道的有无道友可解此卦。
密文:升益艮归妹井萃旅离旅困未济屯未济中孚未济升困噬嗑鼎震巽噬嗑解节井萃离未济蒙归妹大畜无妄解兑临睽升睽未济无妄遁涣归妹
嗯?为什么还有a和b呢?
a=5
b=7

脚本:

s='升益艮归妹井萃旅离旅困未济屯未济中孚未济升困噬嗑鼎震巽噬嗑解节井萃离未济蒙归妹大畜无妄解兑临睽升睽未济无妄遁涣归妹'
dic={'坤': '000000', '剥': '000001', '比': '000010', '观': '000011', '豫': '000100', '晋': '000101', '萃': '000110', '否': '000111', '谦': '001000', '艮': '001001', '蹇': '001010', '渐': '001011', '小过': '001100', '旅': '001101', '咸': '001110', '遁': '001111', '师': '010000', '蒙': '010001', '坎': '010010', '涣': '010011', '解': '010100', '未济': '010101', '困': '010110', '讼': '010111', '升': '011000', '蛊': '011001', '井': '011010', '巽': '011011', '恒': '011100', '鼎': '011101', '大过': '011110', '姤': '011111', '复': '100000', '颐': '100001', '屯': '100010', '益': '100011', '震': '100100', '噬嗑': '100101', '随': '100110', '无妄': '100111', '明夷': '101000', '贲': '101001', '既济': '101010', '家人': '101011', '丰': '101100', '离': '101101', '革': '101110', '同人': '101111', '临': '110000', '损': '110001', '节': '110010', '中孚': '110011', '归妹': '110100', '睽': '110101', '兑': '110110', '履': '110111', '泰': '111000', '大畜': '111001', '需': '111010', '小畜': '111011', '大壮': '111100', '大有': '111101', '夬': '111110', '乾': '111111'}
li=[]
k=0
for i in range(len(s)):
if k 1:
k=0
continue
try:
li.append(dic[s[i]])
except:
t=''
t=t+s[i]+s[i+1]
li.append(dic[t])
k=1
ss=''.join(li)
print(ss)
a=1
b=0
enc=""
for i in ss:
if a%8
0:
b=b2+int(i)
enc=enc+chr(b)
b=0
a=1
else:
b=b
2+int(i)
a=a+1
import base64
print(enc)
x=base64.b64decode(enc).decode()
def decrypt4(enc):
temp=''
offset=5
for i in range(len(enc)):
temp+=chr(ord(enc[i])+offset+i)
return temp
print(x)
def decrypt5(flag):
for a in range(1,200):
enc = ''
for i in flag:
for k in range(200):
if (ord(i) - 97 - 7+26*k)%a==0:
enc+= chr((ord(i) - 97 - 7 + 26 * k) // a + 97)
break
print(enc)
x=decrypt4(x)
print(x)
x=decrypt5(x)
flag在输出的结果中

flag{shaodayouxiduoduyijing}

给你私钥吧

将压缩包解压后得到4个文件
image
脚本如下:

from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_OAEP
from gmpy2 import invert
import base64
f = open("pubkey.pem","rb").read()
pub = RSA.importKey(f)
n,e = pub.n,pub.e
//分解n得到pq
p=157790417717035275943197904823645145281147085252905247447260034051878691034747684303715336348507267921249655103263347914128144476912685213431110454636244692224328066884510063590700506729345331153483633231327359450199822698241355428609085077662488946173655043172957247264543259611018596088670385591091710018977
q=167343506005974003380506069679607737381940204686173214188860057004909006055220516074283090160430833007424970980655748310232878462615469792561310560310363430669700009093597847018287568821792168143170329382585883857083334915378884054389878477389765792275111293420203613159303898365894897865177093362621517279751
d = int(invert(e, (p - 1) * (q - 1)))
private_key = RSA.construct((n, e, d, p, q))
de = PKCS1_OAEP.new(private_key)
c = open('flag.enc').read()
c = base64.b64decode(c)
flag = de.decrypt(c)
print(flag)

bugku{tw0_Tig3rs_l0V3_d4nc1ng~ei!}

posted @   XIZen  阅读(44)  评论(0编辑  收藏  举报
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