2018 HEBTUCTF Sudoku&Viginere

題目給了一個word 一個數(shù)獨然后 Viginere

rry55t1r13
15_1t_3a5y
分別嘗試提交，最后確定flag為：HEBTUCTF{15_1t_3a5y}

柵欄密碼

最總flag{76d6207ceb064719cdf7b8d6168fefda}

4進制

查看題目

四進制很明顯有三個前面的零省略了補上，轉16，再轉文本試一下

看到666心里就有譜了

最終flag{Fourbase123}

Ook

查看題目

ook編碼在線解密

最終flag{1c470f09af4c86b7}

2018 AFCTF Morse

查看題目
摩斯密碼解密

提交不對
十六進制轉文本試一下

最終 afctf{1s’t_s0_345y}

CRC32 BOOM！

查看題目

先crc32碰撞 在明文攻擊
得到密碼為bugku_newctf
解壓出flag.jpg。結果無法打開。
用文本打開，可以看到flag就在文件尾部：

最終flag{Crcrcrcrc_32_BOOM}

2018 HEBTUCTF 社會主義接班人

查看題目

在線解密就完了
最終HEBTUCTF{ejvovdasfjfvmrfmsdemxj}

查看題目

柵欄解密得到

柵欄中的base

查看題目

把這些 base16 base32 base64挨個轉
再柵欄

最終flag{0939_F2A_BACD0}

2018 AFCTF 可憐的RSA

得到兩個文件（中間那個公鑰的也可以直接分解得到ne，然后分解n得到pq也是可以的）
不說了直接上腳本

from Crypto.PublicKey import RSA from gmpy2 import * from Crypto.PublicKey import RSA from Crypto.Cipher import PKCS1_OAEP from base64 import b64decodefrom Crypto.PublicKey import RSA f = open('D:\\pycharm\\venv\\mima\\gongyao\\public.key', 'rb').read() pub = RSA.importKey(f) n = pub.n e = pub.e p = 3133337 q = 25478326064937419292200172136399497719081842914528228316455906211693118321971399936004729134841162974144246271486439695786036588117424611881955950996219646807378822278285638261582099108339438949573034101215141156156408742843820048066830863814362379885720395082318462850002901605689761876319151147352730090957556940842144299887394678743607766937828094478336401159449035878306853716216548374273462386508307367713112073004011383418967894930554067582453248981022011922883374442736848045920676341361871231787163441467533076890081721882179369168787287724769642665399992556052144845878600126283968890273067575342061776244939 d = int(invert(e, (p-1)*(q-1))) from Crypto.PublicKey import RSA from Crypto.Cipher import PKCS1_OAEP from base64 import b64decode key_info = RSA.construct((n, e, d, p, q)) key = RSA.importKey(key_info.exportKey()) key = PKCS1_OAEP.new(key) f = open('D:\\pycharm\\venv\\mima\\gongyao\\flag.enc', 'r').read() c= b64decode(f) flag = key.decrypt(c) print(flag)

運行即可得到flagb'afctf{R54_|5_$0_B0rin9}'

山東省大學生網絡技術大賽-baby

查看題目

# -*- coding: utf-8 -*-from Crypto.PublicKey import RSA import libnum import uuidflag = "flag{***************}" rsa = RSA.generate(4096,e=3) p = rsa.p d = rsa.d e = rsa.e N = rsa.n m = libnum.s2n(flag) c = pow(m, e, N) print "[+]c:",c print "[+]N:",N ''' [+]c: 3442467842482561323703237574537907554035337622762971103210557480050349359873041624336261782731509068910003360547049942482415036862904844600484976674423604861710166033558576921438068555951948966099658902606725292551952345193132973996288566246138708754810511646811362017769063041425115712305629748341207792305694590742066971202523405301561233341991037374101265623265332070787449332991792097090044761973705909217137119649091313457206589803479797894924402017273543719924849592070328396276760381501612934039653 [+]N: 691316677109436623113422493782665795857921917893759942123087462879884062720557906429183155859597756890896192044003240821906332575292476160072039505771794531255542244123516929671277306361467074545720823735806308003091983427678300287709469582282466572230066580195227278214776280213722215953097747453437289734469454712426107967188109548966907237877840316009828476200388327329144783877033491238709954473809991152727333616022406517443130542713167206421787038596312975153165848625721911080561242646092299016802662913017071685740548699163836007474224715426587609549372289181977830092677128368806113131459831182390520942892670696447128631485606579943885812260640805756035377584155135770155915782120025116486061540105139339655722904721294629149025033066823599823964444620779259106176913478839370100891213072100063101232635183636552360952762838656307300621195248059253614745118852163569388418086291748805100175008658387803878200034840215506516715640621165661642177371863874586069524022258642915100615596032443145034847031564356671559179212705466145609698475546210994748949121359853094247990533075004393534565421776468785821261291309463205314057882016266066365636018084499158806717036972590848458891019171583268920180691221168453612029698510271 '''

第一眼看到e=3大致就知道是什么了后面給你nc。確定是低指數(shù)攻擊。
腳本如下

from gmpy2 import iroot import libnumn = 691316677109436623113422493782665795857921917893759942123087462879884062720557906429183155859597756890896192044003240821906332575292476160072039505771794531255542244123516929671277306361467074545720823735806308003091983427678300287709469582282466572230066580195227278214776280213722215953097747453437289734469454712426107967188109548966907237877840316009828476200388327329144783877033491238709954473809991152727333616022406517443130542713167206421787038596312975153165848625721911080561242646092299016802662913017071685740548699163836007474224715426587609549372289181977830092677128368806113131459831182390520942892670696447128631485606579943885812260640805756035377584155135770155915782120025116486061540105139339655722904721294629149025033066823599823964444620779259106176913478839370100891213072100063101232635183636552360952762838656307300621195248059253614745118852163569388418086291748805100175008658387803878200034840215506516715640621165661642177371863874586069524022258642915100615596032443145034847031564356671559179212705466145609698475546210994748949121359853094247990533075004393534565421776468785821261291309463205314057882016266066365636018084499158806717036972590848458891019171583268920180691221168453612029698510271 c = 3442467842482561323703237574537907554035337622762971103210557480050349359873041624336261782731509068910003360547049942482415036862904844600484976674423604861710166033558576921438068555951948966099658902606725292551952345193132973996288566246138708754810511646811362017769063041425115712305629748341207792305694590742066971202523405301561233341991037374101265623265332070787449332991792097090044761973705909217137119649091313457206589803479797894924402017273543719924849592070328396276760381501612934039653 m = 0 while 1:res = iroot(c + m * n, 3)if (res[1] == True):print(libnum.n2s(int(res[0])))breakm = m + 1

運行得到b'flag{4c466c3d0949118a3ca3319b43fe792bef9e94a19c8f666d2ec6c890034d88ba}'

2018 AFCTF MagicNum

查看題目

double類型數(shù)據(jù)，轉化成內部存儲形式，然后再解碼成字節(jié)碼

#include <stdio.h> #include <stdlib.h>int main() {double x;while(scanf("%llf",&x)){unsigned char *p;char buff[9];int i;p=(unsigned char *)&x;for (i=0; i<8; i++){itoa(*p++,buff,2);printf("%08s",buff);}printf("\n");}return 0; }

去掉=號結束

easy_base

查看題目


你沒猜錯就是base64 40連解
flag是啥我也不說了自己慢慢連解去吧。。

鍵盤之爭

但是按照自己的鍵盤看了半天。。。
沖以前的對抗賽找了原題是一個dvorak鍵盤然后就開始咱們的移形換位

dic = {r"'": "q",r",": "w",r".": "e","p": "4","y": "t","f": "y","g": "u","c": "i","r": "o","l": "p",r"/": r"[",r"=": r"]",r'"': 'Q',r"<": "W",r">": "E","P": "R","Y": "T","F": "Y","G": "U","C": "I","R": "O","L": "P",r"?": r"{",r"+": r"}","a": "a","A": "A","o": "s","O": "S","e": "d","E": "D","u": "f","U": "F","i": "g","I": "G","d": "h","D": "H","h": "j","H": "J","t": "k","T": "K","n": "l","N": "L","s": ";","S": ":",r"-": r"'",r'_': r'"',r";": "z",r":": "Z","q": "x","Q": "X","j": "c","J": "C","k": "v","K": "V","x": "b","X": "B","b": "n","B": "N","m": "m","M": "M","w": r",","W": r"<","v": r".","V": r">","z": r"/","Z": r"?",r'!': "!",r"@": r"@",r"#": r"#",r"$": r"$",r"%": r"%",r"^": r"^",r"&": r"&",r"*": r"*",r"(": r"(",r")": r")",r"[": r"-",r"]": r"=",r"{": r"_",r"}": r"+"}s = r'ypau_kjg;"g;"ypau+'for i in s:print(" ".join([key for key, value in dic.items() if value == i]), end='')

輸出flag{this_is_flag}（建議收藏這個鍵盤換位py）
MD5加密就行了 最終flag為flag{951c712ac2c3e57053c43d80c0a9e543}

總結

以上是生活随笔為你收集整理的BMZCTF 密码的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。