1.通过nc端口可以获得一个加密算法的五个参数n,e,d,c2,r。几次nc发现参数n,e,d不发生变化，而c2和r在不断的变化，所以猜测这是一个RSA加密，而c2和r都是密文，进行尝试，发现c2解出的明文与r^-1相乘可以获得一段有意义的明文，即为答案
import gmpy2 def shuchu(mingwenstr): if mingwenstr[len(mingwenstr)-1]=='L': mingwenstr=mingwenstr[2:len(mingwenstr)-1] else: mingwenstr=mingwenstr[2:len(mingwenstr)] if not len(mingwenstr)%2==0: mingwenstr='0'+mingwenstr i=len(mingwenstr) mingwen="" while i>=1: str1=mingwenstr[i-2:i] mingwen=chr(int(str1,16))+mingwen i=i-2 return mingwen e=3 d=58070556439793090526057208709744281266699522732880809417222160656726252113813697356285226658626010881834410509800517446002936497512727830847585298932899297251384197149372765798336891928961982606512715939584602138570176466492576873015396850289610395524868997323113173151554667479227153672743044388610706469956522886976551988118566524382547827999574700187204163033785878760087039158299804693857061230328733771956786333627783002819502573976524444691753704136431366061704756184802834459959184533250175265286723943071655257551597505042735965819319292478912706745310604757468335858175002424726406132378973591389079242021955 n=87105834659689635789085813064616421900049284099321214125833240985089378170720546034427839987939016322751615764700776169004404746269091746271377948399348945877076295724059148697505337893442973909769073909376903207855264699738865309523095275434415593287303495984669759727332001218840730509114566582916059704935385321982580346749147006925780956372612590055880897804341000627901507033770398466328696077102639753825819349512854490295124049540409694800789160103174699469939773219086406181469636327731496893645403727898138438731456170483734919419215754658944354898505598862246290878791817393880153989112477526189234162033361 c2=30132543736268406128006366464667229771073658466016673121647580163529342379328161036460170843634938824748741593475228015576606281766122693295164938051152259925237031266538565360075085380831091576758642909966414706080267848654347459452041009557966508871753912315759169663197469559281085213023761241462601694456397469906847104846936615928238362311208495235132581926980537708257685299831667264927106739017142385174176373835139386078500160518744072262902657518141254914887340207566012259036468547203920094710762375729853416491332936827048049647501996480349776493220949016766098297834731142453582162651375789941891914976509 r=40362150765525455398305794281797305490609797222633763924968255583201392938971882099474127231258279012753683149475392582835066322472023630708986909192229345380946916114425191164277988260735513951657449996156426068500820728361370965468934689968116991853678252892154230050077402502313809894311821473925115876231283367951031098705719436798093988664714173078719633447013891858952353430120322426205074873650730720829400817256332721447324365618558304550240380256207038022701869134112893112523282671510529083738640331200124328981859414011412619244224887134339783731254380294415758009647671108465065236784576156455594956835456 p3=pow(c2,d,n) rinv=gmpy2.invert(r,n) p4=(p3*rinv)%n p4=hex(p4) print shuchu(p4) 2.题目为x=chr（random.randint（0,0xff））+chr（random.randint（0,0xff）+chr（random.randint（0,0x1f）+ chr（random.randint（0,0x1f）） hashlib.sha256（x）.hexdigest（）[0:8]==‘ caade32b ‘，尝试直接按题目代码进行暴力破解，发现无法得到结果，仔细思考明白在ascii码中只有32-128对应字符可以键入，其余字母是无法作为flag键入的，于是修改程序暴力破解，获得答案
import random import hashlib for a in xrange(32,128): for b in xrange(32,128): for c in xrange(32,128): for d in xrange(32,128): x=chr(a)+chr(b)+chr(c)+chr(d) if hashlib.sha256(x).hexdigest()[0:8]=='caade32b': print x 3.题目为Alice与Bob正在通信，Alice发给Bob两段密文如下： Li4tLS0gLi4tLS0g4oCmLS0gLi4tLS0g4oCmLS0gLi0tLS0gLi4tLS0gLi4tLS0g4oCmLS0g4oCmLS0g4oCmLS0gLi0tLS0gLi4tLS0gLi4tLS0g4oCmLi0gLi0tLS0g4oCmLS0g4oCmLS0= usphcakdacvayunmtnqhp Eve截获了这两段密文，请你帮助他将密文破译出来。我们在题目中发现两端密文“Li4tLS0gLi4tLS0g4oCmLS0gLi4tLS0g4oCmLS0gLi0tLS0gLi4tLS0gLi4tLS0g4oCmLS0g4oCmLS0g4oCmLS0gLi0tLS0gLi4tLS0gLi4tLS0g4oCmLi0gLi0tLS0g4oCmLS0g4oCmLS0=”和“usphcakdacvayunmtnqhp”，很明显第一段明文使用base64加密，在线解密可得“..--- ..--- …-- ..--- …-- .---- ..--- ..--- …-- …-- …-- .---- ..--- ..--- ….- .---- …-- …--“，这是一段摩斯密码，解密可得“223231223331224133”，使用手机键盘即可发现其对应的是bedbfdbgf，这时一个密文时3*3，一个密文是3*7，猜测这里进行了hill加密，直接使用hill解密无果，尝试对第二段密文进行修改，将‘usphcakdacvayunmtnqhp’编程3*7的矩阵[usphcak,dacvayu,nmtnqhp]，在对第一三行做反转，可得[kcahpsu,dacvayu,phqntmn]，逆向栅栏可得字符串ppdkhaaqccnvhtapmysnuu，在对其使用hill密码解密可得gotothemoviethisnight。
4.题目为Eve窃听了Alice和Bob的通信，得到了Alice和Bob通信的公钥是（4846403,97）。以及下列一系列数据，请帮助Eve解密出这些数据的明文？ 1668222、313433、3647056、2851410、21188、3766877、3904324、1869820、3632941、2414731，可知这只是一个简单RSA解密，解密获得答案即可：
import gmpy2 n=4846403 e=97 c=[1668222,313433,3647056,2851410,21188,3766877,3904324,1869820,3632941,2414731] p=2153 q=2251 d=int(gmpy2.invert(e,(p-1)*(q-1))) m=[] for i in c: mingwen=pow(i,d,n) m.append(mingwen) print m 5.题目：
Alice与Bob正在进行加密通话。Eve从中进行了窃听，截获了Bob发给Alice的口令明文：Alice：d*@*68707600*K。Eve尝试用该用户和口令去登录Bob的服务器，发现该口令不正确。于是Eve入侵了Bob的电脑，在Bob的服务器上找到了Alice对应的口令的摘要，下载打开后如下所示：4*de5f*d23l9fbdbll9fl3*8797**de75d7**94*。Eve尝试破译该口令时，发现该摘要和口令存在很多问题。你能帮助Eve破译出Alice的口令吗？有口令可以猜到这里大概是hash加密，而密文有四十位可以让我们想到这里应该是sha-1加密，这时暴力破解即可：
import hashlib import re s="4.de5f.d2319fbdb119f13.8797..de75d7..94." for i in xrange(32,128): for j in xrange(32,128): for k in xrange(32,128): mingwen="d"+chr(i)+"@"+chr(j)+"68707600"+chr(k)+"K" temp=hashlib.sha1(mingwen).hexdigest() if re.match(s,temp): print mingwen