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這是一道凱撒密碼的題目，題目裡提示先五移位，在八移位就可以得到

完成，這裡使用的軟體米斯特安全工具或者使用線上的工具也可以

單表替換密碼

Lw!

Gyzvecy ke WvyVKT!

W'zz by reso dsbdkwksky tzjq teo kly ujr. Teo keujr, gy joy dksurwmq bjdwv vorakeqojalr jmu wkd jaazwvjkwemd. Vorakeqojalr ljd j zemq lwdkeor, jzklesql gwkl kly juxymk et vecaskyod wk ljd qekkym oyjzzr vecazwvjkyu. Decy dwcazy ezu vwalyod joy kly Vjydjo vwalyo, kly Xwqymyoy vwalyo, kly dsbdkwkskwem vwalyo, glwvl wd klwd emy, jmu de em. Jzcedk jzz et klydy vwalyod joy yjdwzr boeiym keujr gwkl kly lyza et vecaskyod. Decy myg ymvorakwem cykleud joy JYD, kly vsooymk dkjmujou teo ymvorakwem, jzemq gwkl ODJ. Vorakeqojalr wd j xjdk twyzu jmu wd xyor wmkyoydkwmq klesql. De iwvi bjvi, oyju sa em decy veez vwalyod jmu ljxy tsm!

El jmu teo reso oyveoud cr mjcy wd WvyVKT{jzgjrd_zwdkym_ke_reso_dsbdkwksky_tzjqd}.
 

大多數時候可以使用線上網站進行詞頻分析，如https://quipqiup.com/，這種特別原理的題目的則是沒什麼太大的技術

提交已知資訊，加大成功率

Hi! Welcome to IceCTF! I'll be your substitute flag for the day. For today, we are studying basic cryptography and its applications. Cryptography has a long history, although with the ad?ent of computers it has gotten really complicated. Some simple old ciphers are the Caesar cipher, the ?igenere cipher, the substitution cipher, which is this one, and so on. Almost all of these ciphers are easily bro?en today with the help of computers. Some new encryption methods are AES, the current standard for encryption, along with RSA. Cryptography is a ?ast field and is ?ery interesting though. So ?ic? bac?, read up on some cool ciphers and ha?e fun! Oh and for your records my name is IceCTF{always_listen_to_your_substitute_flags}.
 

下面的是一個替換指令碼

ule = {'T':'T','a':'a','b':'b','c':'c','d':'d','e':'e','f':'f','g':'g',
        'h':'h','i':'i','j':'j','k':'k','l':'l','m':'m','n':'n','o':'o',
        'p':'p','q':'q','r':'r','s':'s','t':'t','u':'u','v':'v','w':'w',
        'x':'x','y':'y','z':'z'}

rule['a'] = 'd',
rule['b'] = 'm'
rule['d'] = 'e'
rule['e'] = 'n'
rule['g'] = 'f'
rule['h'] = 'o'
rule['i'] = 'x'
rule['k'] = 'p'
rule['p'] = 'i'
rule['q'] = 'r'
rule['r'] = 'a'
rule['t'] = 's'
rule['u'] = 'b'
rule['w'] = 't'
rule['y'] = 'l'
rule['z'] = 'u'
rule['m'] = 'h'
rule['j'] = 'g'
rule['x'] = 'c'
rule['l'] = 'y'
rule['s'] = 't'
rule['f'] = 'w'
rule['v'] = 'k'
print rule 

with open('encryption.txt', 'r') as f:
    data=f.read()
print data

new_data =''
for i in xrange(0,len(data)):
    if data[i].isalpha():
        new_data += rule[data[i]]
    else:
        new_data += data[i]
print new_data

with open('result.txt', 'w') as r:
    r.write(new_data)
 

RSA 最簡單模式

為什麼這麼說，因為

N=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

e=0x1

c=0x4963654354467b66616c6c735f61706172745f736f5f656173696c795f616e645f7265617373656d626c65645f736f5f63727564656c797d

e=1，相當於密文就是明文，測試k=0時直接16進位制解密，得到

IceCTF{falls_apart_so_easily_and_reassembled_so_crudely}

這是RSA中最簡單的情況，也是非常特殊的情況，有關的原理大家要回憶起上文講的 想到原文和密文的關係

n**e ≡ c mod N

訊息解密，利用金鑰 d 進行解密得到明文n：

c**d ≡ n (mod N)

倒數第二簡單的RSA

為啥有倒數第二了因為這個就是原理題，當初是讓我們瞭解原理的

//題目的已知資訊
N=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

e=0x10001

phi=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

d=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

c=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

根據原理，我們解密需要的是密文n 私鑰d 大素數N            n**d mod N就是我們要的明文，然而題中都給了，我們只需要簡單跑一下指令碼

#!/usr/bin/python2.7  3的跑不了 因為字串沒有decode屬性方法
# -*- coding: utf-8 -*-
import math   # 匯入 math 模組

def normaldecode(c,d,N):
 # 普通解密需要密文c 私鑰d 大素數N c**d mod N
    #pow（a,b,c）計算a的b次冪 如果有c則modc    [2,-1]去掉0x字元
    return hex(pow(c, d, N))[2:-1].decode('hex')  # 再解密返回

關於這個rsa再多說兩句，一個是要熟悉python的模 餘 次方等運算的函式，另一個就是gmpy2庫，利用gmpy2破解rsa是常用的，裡面有很多有關加密解密函式，用於計算私鑰 解密之類的

Classicl三連

Ld hcrakewcfaxr, f hofjjlhfo hlaxuc lj f krau ev hlaxuc kxfk zfj tjui xljkeclhfoor gtk dez xfj vfooud, vec kxu pejk afck, ldke iljtju. Ld hedkcfjk ke peiucd hcrakewcfaxlh foweclkxpj, pejk hofjjlhfo hlaxucj hfd gu acfhklhfoor hepatkui fdi jeoyui gr xfdi. Xezuyuc, OrmkO3vydJCoe2qyNLmcN2qlpJXnM3SxM2Xke3q9 kxur fcu foje tjtfoor yucr jlpaou ke gcufn zlkx peiucd kuhxdeoewr. Kxu kucp ldhotiuj kxu jlpaou jrjkupj tjui jldhu Wcuun fdi Cepfd klpuj, kxu uofgecfku Cudfljjfdhu hlaxucj, Zecoi Zfc LL hcrakewcfaxr jthx fj kxu Udlwpf pfhxldu fdi guredi. F btlhn gcezd veq mtpa eyuc kxu ofsr iew.

看樣子是替換密碼，上https://quipqiup.com/

In cryptography, a classical cipher is a type of cipher that was used historically but now has fallen, for the most part, into disuse. In contrast to modern cryptographic algorithms, most classical ciphers can be practically computed and solved by hand. However, LyjtL3fvnSRlo2xvKIjrK2ximSHkJ3ZhJ2Hto3x9 they are also usually very simple to break with modern technology. The term includes the simple systems used since Greek and Roman times, the elaborate Renaissance ciphers, World War II cryptography such as the Enigma machine and beyond. A quick brown fox jump over the lazy dog.

有一串莫名奇妙的東西，LyjtL3fvnSRlo2xvKIjrK2ximSHkJ3ZhJ2Hto3x9 應該是移位，即凱撒加密，

然而解碼沒有特別的提示，base全位置爆破試試，其中一個就是flag

RSA  n爆破解密

n is 966808932627497190635859236054960349099463975227350564265384373280336699853387254070662881265937565163000758606154308757944030571837175048514574473061401566330836334647176655282619268592560172726526643074499534129878217409046045533656897050117438496357231575999185527675071002803951800635220029015932007465117818739948903750200830856115668691007706836952244842719419452946259275251773298338162389930518838272704908887016474007051397194588396039111216708866214614779627566959335170676055025850932631053641576566165694121420546081043285806783239296799795655191121966377590175780618944910532816988143056757054052679968538901460893571204904394975714081055455240523895653305315517745729334114549756695334171142876080477105070409544777981602152762154610738540163796164295222810243309051503090866674634440359226192530724635477051576515179864461174911975667162597286769079380660782647952944808596310476973939156187472076952935728249061137481887589103973591082872988641958270285169650803792395556363304056290077801453980822097583574309682935697260204862756923865556397686696854239564541407185709940107806536773160263764483443859425726953142964148216209968437587044617613518058779287167853349364533716458676066734216877566181514607693882375533
e is 65537
c is 168502910088858295634315070244377409556567637139736308082186369003227771936407321783557795624279162162305200436446903976385948677897665466290852769877562167487142385308027341639816401055081820497002018908896202860342391029082581621987305533097386652183849657065952062433988387640990383623264405525144003500286531262674315900537001845043225363148359766771033899680111076181672797077410584747509581932045540801777738548872747597899965366950827505529432483779821158152928899947837196391555666165486441878183288008753561108995715961920472927844877569855940505148843530998878113722830427807926679324241141182238903567682042410145345551889442158895157875798990903715105782682083886461661307063583447696168828687126956147955886493383805513557604179029050981678755054945607866353195793654108403939242723861651919152369923904002966873994811826391080318146260416978499377182540684409790357257490816203138499369634490897553227763563553981246891677613446390134477832143175248992161641698011195968792105201847976082322786623390242470226740685822218140263182024226228692159380557661591633072091945077334191987860262448385123599459647228562137369178069072804498049463136233856337817385977990145571042231795332995523988174895432819872832170029690848

n不是很大的時候，我們可以暴力分解，網站

p q c e都知道了，那麼我們就知道d，就可以解密了

#!/usr/bin/env python
# coding:utf-8

__author__ = 'zhengjim'

import gmpy2

p = gmpy2.mpz(31093551302922880999883020803665536616272147022877428745314830867519351013248914244880101094365815998050115415308439610066700139164376274980650005150267949853671653233491784289493988946869396093730966325659249796545878080119206283512342980854475734097108975670778836003822789405498941374798016753689377992355122774401780930185598458240894362246194248623911382284169677595864501475308194644140602272961699230282993020507668939980205079239221924230430230318076991507619960330144745307022538024878444458717587446601559546292026245318907293584609320115374632235270795633933755350928537598242214216674496409625928997877221)
q = gmpy2.mpz(31093551302922880999883020803665536616272147022877428745314830867519351013248914244880101094365815998050115415308439610066700139164376274980650005150267949853671653233491784289493988946869396093730966325659249796545878080119206283512342980854475734097108975670778836003822789405498941374798016753689377992355122774401780930185598458240894362246194248623911382284169677595864501475308194644140602272961699230282993020507668939980205079239221924230430230318076991507619960330144745307022538024878444458717587446601559546292026245318907293584609320115374632235270795633933755350928537598242214216674496409625928797450473)
e = gmpy2.mpz(65537)
#計算d
phi_n = (p - 1) * (q - 1)   # 計算r = phin
d = gmpy2.invert(e, phi_n)   # 計算私鑰d ed≡1(modr)  e和r互質。並求得e關於r的模反元素
print "private key:"
print d

#求密文
c = gmpy2.mpz(168502910088858295634315070244377409556567637139736308082186369003227771936407321783557795624279162162305200436446903976385948677897665466290852769877562167487142385308027341639816401055081820497002018908896202860342391029082581621987305533097386652183849657065952062433988387640990383623264405525144003500286531262674315900537001845043225363148359766771033899680111076181672797077410584747509581932045540801777738548872747597899965366950827505529432483779821158152928899947837196391555666165486441878183288008753561108995715961920472927844877569855940505148843530998878113722830427807926679324241141182238903567682042410145345551889442158895157875798990903715105782682083886461661307063583447696168828687126956147955886493383805513557604179029050981678755054945607866353195793654108403939242723861651919152369923904002966873994811826391080318146260416978499377182540684409790357257490816203138499369634490897553227763563553981246891677613446390134477832143175248992161641698011195968792105201847976082322786623390242470226740685822218140263182024226228692159380557661591633072091945077334191987860262448385123599459647228562137369178069072804498049463136233856337817385977990145571042231795332995523988174895432819872832170029690848)
print "plaintext:"

print "plaintext:"
M  =  pow(c,d,p*q)   # 從c**d mod N(N=p*q)
print '[10進位制]'+str(M)
flag = str(hex(M))[2:]  # 這裡擷取2開始 是為了去除0x兩個字元
print '[16進位制]'+flag
print '[ASCII碼]'+flag.decode('hex')