LeftOrRight

題目
Left?Middle?No，I want right！（flag is right？！）

解題
用010Editor打開

JPG應該是以FFD8開頭，將這之前的刪去后得到：

一顆，樹！？二叉樹？

在照片前面，有
0x66 0x30 0x39 0x65 0x35 0x34 0x63 0x31 0x62 0x61 0x64 0x32 0x78 0x33 0x38 0x6d 0x76 0x79 0x67 0x37 0x77 0x7a 0x6c 0x73 0x75 0x68 0x6b 0x69 0x6a 0x6e 0x6f 0x70

照片最后，應該以FFD9結尾，在這之后，還有一些數(shù)：
0x39 0x30 0x35 0x65 0x34 0x63 0x31 0x66 0x61 0x78 0x33 0x32 0x38 0x6d 0x64 0x79 0x76 0x67 0x37 0x77 0x62 0x73 0x75 0x68 0x6b 0x6c 0x69 0x6a 0x7a 0x6e 0x6f 0x70

將這兩段十六進制轉為ASCII碼：

f09e54c1bad2x38mvyg7wzlsuhkijnop
905e4c1fax328mdyvg7wbsuhklijznop

def get_after_deep(pre, mid, a):#已知前中，求后，a就是后序if len(pre) == 1:a.append(pre[0])returnif len(pre) == 0:returnroot = pre[0]root_index = mid.index(root)get_after_deep(pre[1:root_index+1], mid[:root_index], a)get_after_deep(pre[root_index+1:], mid[root_index+1:], a)a.append(root)return adef get_hou():pre=input("請依次輸入前序遍歷、中序遍歷的結果，以換行分割:\n")mid=input()pre_list=list(pre)mid_list=list(mid)a=[]res_list=get_after_deep(pre,mid,a)res="".join(res_list)print("后序遍歷為:",res)get_hou()

運行得到：后序遍歷為: 951c4e03xm82yw7gvdakhusjilponzbf

答案
flag{951c4e03xm82yw7gvdakhusjilponzbf}

RSA4

題目

N = 331310324212000030020214312244232222400142410423413104441140203003243002104333214202031202212403400220031202142322434104143104244241214204444443323000244130122022422310201104411044030113302323014101331214303223312402430402404413033243132101010422240133122211400434023222214231402403403200012221023341333340042343122302113410210110221233241303024431330001303404020104442443120130000334110042432010203401440404010003442001223042211442001413004 c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c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c = 10013444120141130322433204124002242224332334011124210012440241402342100410331131441303242011002101323040403311120421304422222200324402244243322422444414043342130111111330022213203030324422101133032212042042243101434342203204121042113212104212423330331134311311114143200011240002111312122234340003403312040401043021433112031334324322123304112340014030132021432101130211241134422413442312013042141212003102211300321404043012124332013240431242

解題

from gmpy2 import * from Crypto.Util.number import * from functools import reduceN1 = int(str(331310324212000030020214312244232222400142410423413104441140203003243002104333214202031202212403400220031202142322434104143104244241214204444443323000244130122022422310201104411044030113302323014101331214303223312402430402404413033243132101010422240133122211400434023222214231402403403200012221023341333340042343122302113410210110221233241303024431330001303404020104442443120130000334110042432010203401440404010003442001223042211442001413004),5) c1 = int(str(310020004234033304244200421414413320341301002123030311202340222410301423440312412440240244110200112141140201224032402232131204213012303204422003300004011434102141321223311243242010014140422411342304322201241112402132203101131221223004022003120002110230023341143201404311340311134230140231412201333333142402423134333211302102413111111424430032440123340034044314223400401224111323000242234420441240411021023100222003123214343030122032301042243),5) N2 = int(str(302240000040421410144422133334143140011011044322223144412002220243001141141114123223331331304421113021231204322233120121444434210041232214144413244434424302311222143224402302432102242132244032010020113224011121043232143221203424243134044314022212024343100042342002432331144300214212414033414120004344211330224020301223033334324244031204240122301242232011303211220044222411134403012132420311110302442344021122101224411230002203344140143044114),5) c2 = int(str(112200203404013430330214124004404423210041321043000303233141423344144222343401042200334033203124030011440014210112103234440312134032123400444344144233020130110134042102220302002413321102022414130443041144240310121020100310104334204234412411424420321211112232031121330310333414423433343322024400121200333330432223421433344122023012440013041401423202210124024431040013414313121123433424113113414422043330422002314144111134142044333404112240344),5) N3 = int(str(332200324410041111434222123043121331442103233332422341041340412034230003314420311333101344231212130200312041044324431141033004333110021013020140020011222012300020041342040004002220210223122111314112124333211132230332124022423141214031303144444134403024420111423244424030030003340213032121303213343020401304243330001314023030121034113334404440421242240113103203013341231330004332040302440011324004130324034323430143102401440130242321424020323),5) c3 = int(str(10013444120141130322433204124002242224332334011124210012440241402342100410331131441303242011002101323040403311120421304422222200324402244243322422444414043342130111111330022213203030324422101133032212042042243101434342203204121042113212104212423330331134311311114143200011240002111312122234340003403312040401043021433112031334324322123304112340014030132021432101130211241134422413442312013042141212003102211300321404043012124332013240431242),5)N = [N1,N2,N3] c = [c1,c2,c3]def chinese_remainder(modulus, remainders):Sum = 0prod = reduce(lambda a, b: a*b, modulus)for m_i, r_i in zip(modulus, remainders):p = prod // m_iSum += r_i * (inverse(p,m_i)*p)return Sum % prod e = 3 # print(chinese_remainder(N,c)) pow_m_e = chinese_remainder(N,c) # pow_m_e = 17446992834638639179129969961058029457462398677361658450137832328330435503838651797276948890990069700515669656391607670623897280684064423087023742140145529356863469816868212911716782075239982647322703714504545802436551322108638975695013439206776300941300053940942685511792851350404139366581130688518772175108412341696958930756520037 m = iroot(pow_m_e,3)[0] print(long_to_bytes(m))

運行得到：b'noxCTF{D4mn_y0u_h4s74d_wh47_4_b100dy_b4s74rd!}'
答案
flag{D4mn_y0u_h4s74d_wh47_4_b100dy_b4s74rd!}

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