2024Geek WriteUp

前言：

这个算是我打的第一个正式的CTF了，也是第一次打的这么认真，一放题就是做，睡醒就是做，天天都是密码题，天天破防，结局是好的，校第二，队伍总分一万出头，我拿了三千多分，被带飞了，哈哈，唯一的遗憾就是没有把纯密码部分ak，week2，week3,week4都还差一道，其他的混做题等以后学了其他方向再去做！
——2024.12.1

凯撒加密：

签到送分题

YEI{CKRIUSK_ZU_2024_MKKQ_INGRRKTMK}

一眼凯撒，简简单单的送分题

找到SYC即可；

RSA：

看题发现，就是普普通通的入门RSA，很简单(原理请看RSA原理那篇文章)

直接出就好了：
已知n = p * q, p, q, c, e


from Crypto.Util.number import long_to_bytes, getPrime
import gmpy2

p = getPrime(128)
q = getPrime(128)
n = p*q
e = 65537

n = 33108009203593648507706487693709965711774665216872550007309537128959455938833

p = 192173332221883349384646293941837353967

q = 172282016556631997385463935089230918399

c = 5366332878961364744687912786162467698377615956518615197391990327680664213847

phi = (p-1)*(q-1)

d = pow(e,-1, phi)

m = pow(c, d, n)

m = 110789406762543471388626033919925647741

print(long_to_bytes(m))

#SYC{RSA_is_easy}

dp：

看题也就知道是普普通通的dp泄露，那就直接打！
这里注释一下dp= d mod (p - 1)
根据rsa的原理，很容易知道

import  binascii
from gmpy2 import*
from Crypto.Util.number import *

c =  127916287434936224964530288403657504450134210781148845328357237956681373722556447001247137686758965891751380034827824922625307521221598031789165449134994998397717982461775225812413476283147124013667777578827293691666320739053915493782515447112364470583788127477537555786778672970196314874316507098162498135060

n =  157667866005866043809675592336288962106125998780791920007920833145068421861029354497045918471672956655205541928071253023208751202980457919399456984628429198438149779785543371372206661553180051432786094530268099696823142821724314197245158942206348670703497441629288741715352106143317909146546420870645633338871

e =  65537

dp =  2509050304161548479367108202753097217949816106531036020623500808413533337006939302155166063392071003278307018323129989037561756887882853296553118973548769

#这里直接爆破就好，还是很容易爆破出来的，
r = e*dp-1
for i in range(1,e):                  
    if r%i == 0:
        if n%((r//i)+1) == 0:
            p = r//i + 1
#爆破出来了p，那么离flag也很接近了吧
q = n//p
phi = (p-1)*(q-1)
d = gmpy2.invert(e,phi)
m = pow(c,d,n)

print(long_to_bytes(m))

这里简单的解释一下dp泄露爆破的原理：
(1)edp = ed mod (p-1)，有(3)edp-ed = k1(p-1)

又

(2)ed = 1 mod (p-1)*(q-1),有(4)ed-1=k2(p-1)(q-1)
根据(1)(2)可以得出，edp - 1 = 0 mod (p-1)，可以知道edp-1 是p-1的倍数
根据(3)(4)换掉ed可知：
edp=【k2(q-1)+k1】(p-1)+1
然后直接爆破p就好

XOR：

又是看了题就知道是异或，但是并没有那么简单，直接看题吧

这里要知道，python中 ’^‘是异或而不是次方

据题，可知，f1，f2，e2，又因为e2 = e1 ^ f2,那么根据异或小知识，可以知道e1 = e2 ^ f2, enc = e1 ^ f1,
flag = xor(long_to_bytes(enc), key)
但是key不知道，这里根据len(key) == 4,并且flag的题头是‘SYC{’
这里我是直接把enc给long_to_bytes, 根据前四个对应的字符来异或’SYC{‘
从来可以得到key，然后就可以得到flag了

from Crypto.Util.number  import long_to_bytes
from pwn import xor

e2 = 10706859949950921239354880312196039515724907
f1 = 4585958212176920650644941909171976689111990
f2 = 3062959364761961602614252587049328627114908

key = b'><0R'

e1 = e2 ^ f2

enc = e1 ^ f1

flag = xor(long_to_bytes(enc), key)

print(flag)

print(len(long_to_bytes(enc)))

print(long_to_bytes(enc))

#下面的步骤是得到key的过程：
print(chr(ord('m')^ord('S')))

print(chr(ord('e')^ord('Y')))

print(chr(ord('s')^ord('C')))

print(chr(ord(')')^ord('{')))

#SYC{a_part_0f_X0R}

优化：

from Crypto.Util.number import * 
from pwn import xor 
f1 = 4585958212176920650644941909171976689111990 
f2 = 3062959364761961602614252587049328627114908 
e2 = 10706859949950921239354880312196039515724907 
enc = e2^f1^f2 
key = xor(long_to_bytes(enc),b'SYC{')[:4] 
flag = xor(long_to_bytes(enc),key) 
print(flag) 
#SYC{a_part_0f_X0R}

共模攻击：

由题目可知，本题是共模攻击，c = pow (m , $e_{i}$ ,n)
e1 ,e2, c1,c2, n,都已经知道
c1 = pow (m,e1,n), c2 = pow (m,e2,n)
需要求得m，那么这里我根据结论反证一下吧
$m = (c 1^{s 1} * c 2^{s 2}) m o d n$
因为模数相同，将c1,c2带入
可得：(( $m^{e 1 * s 1}$ )*( $m^{e 2 * s 2}$ ))mod n
有： $m^{(e 1 * s 1 + e 2 * s 2)} m o d n$
因为m < n, 所以，上式子等于m
得证，直接写python脚本即可

import gmpy2
from Crypto.Util.number import long_to_bytes

n = 19742875423645690846073637620470497648804310111201409901059297083827103813674034450200432098143959078292346910591785265323563248781526393718834491458926162514713269984791730816121181307827624489725923763353393879316510062227511469438742429290073999388690825732236465647396755899136346150862848924231619666069528077790933176798057396704758072769660663756346237040909579775389576227450505746914753205890194457812893098491264392293949768193694560954874603451253079446652049592976605414438411872223250039782381259212718733455588477129910357095186014496957765297934289263536712574572533650393220492870445376144568199077767

e1 = 911

e2 = 967

c1 = 18676091924461946809127036439355116782539894105245796626898495935702348484076501694838877829307466429933623102626122909782775514926293363853121828819237500456062111805212209491398720528499589486241208820804465599279152640624618194425740368495072591471531868392274503936869225072123214869399971636428177516761675388589238329574042518038702529606188240859751459632643230538522947412931990009143731829484941397093509641320264169403755707495153433568106934850283614529793695266717330769019091782929139589939928210818515744604847453929432990185347112319971445630830477574679898503825626294542336195240055995445217249602983

c2 = 4229417863231092939788858229435938841085459330992709019823280977891432565586698228613770964563920779991584732527715378842621171338649745186081520176123907689669636473919678398014317024138622949923292787095400632018991311254591786179660603414693984024161009444842277220189315861986306573182865656366278782315864366857374874763243428496061153290565891942968876789905670073321426112497113145141539289020571684634406829272902118484670099097148727072718299512735637087933649345419433312872607209633402427461708181971718804026293074540519907755129917132236240606834816534369171888633588190859475764799895410284484045429152

_,r,s = gmpy2.gcdext(e1,e2)

m = pow(c1, r, n) * pow(c2, s, n) % n

print(long_to_bytes(m))

#SYC{U_can_really_attack}

这里用到了扩展欧几里得算法，简单解释一下扩展欧几里得的原理：
前置知识：
裴蜀定理：ax + by = m , 其有解的充要条件为 gcd(a,b) | m
即充要条件是 m 必须能够被 gcd(a,b) 整除
欧几里得：
def gcd(a, b):
return a if b == 0 else gcd(b, a%b)
回归正传：
已经知道ax1 + by1 = gcd(a,b)和bx2+(a%b)y2 = gcd(b,a%b)
且gcd(a,b) == gcd(b,a%b)
ax1 + by1 = bx2 + （a%b）y2
a%b = a - (a/b )* b （说一下，比如 3 % 2 = 1， 3 - （3 / 2 ）* 2 = 1这里一个是3整除2结果为1）
$a x 1 + b y 1 = b x 2 + [a - (a / b) * b] y 2$
$a x 1 + b y 1 = a y 2 + b [x 2 - (a / b) y 2]$
x1 = y2 , $y 1 = x 2 - a / b * y 2$
当b = 0 , ax = gcd(a,0),即ax = a, 所以x = 1, y = 0,然后我们回推递归求解即可！

特解：
ax + by = c (c% gcd(a,b) == 0）
此时一定有解，而且有多个，所以可求通解！我们先从特解开始：
ax + by = c 相当于 ax + by = gcd(a,b) 两侧同时乘了一个 c/gcd(a,b)
所以此条件下的特解x0 = x1 * c/gcd(a,b), y0 = y1 * c/gcd(a,b);
通解：
所以：ax = ax + lcm(a,b) by = by - lcm(a,b)
而： lcm(a,b) = ab/gcd(a,b)
由此： ax = ax + ab/gcd(a,b)
约掉a得： x = x + b/gcd(a,b) 即通解 x = x0 + b/gcd(a,b) *k (k = 0 ,+-1,+-2......)
同理可得： y = y - a/gcd(a,b) 即通解y = y0 - a/gcd(a,b) * k (k = 0 ,+-1,+-2......)

不是套娃：

额，这道题真的有够我破防的，呜呜呜

-..-/../-./..--.-/.---/../..--.-/--../../..--.-/.--/.-/..--.-/-.--/../..--.-/--../../..--.-/--/---/..--.-/-.--/../..--.-/-../..-/.-/..--.-/--../..

直接摩西密码，秒了

得到：XIN_JI_ZI_WA_YI_ZI_MO_YI_DUA_ZI

一直不行，当时甚至中文，德文，日语什么的都来了一遍都不行，直到看到lower，才知道

xin_ji_zi_wa_yi_zi_mo_yi_dua_zi

进入发现Vigenere，提示：commander，然后我用了caesar把字符聚拢了一下，方便看到flag，呜呜呜，出题人真的是个二次元，太感动了

得到sunflower

解开发现居然还有一层啊！！！！！崩溃

看题：

a1fdbce928af7aae

一开始以为是十六进制，但是发现不是啊，太悲伤了，半夜三四点突然想到是md5啊

得到了：HaiKav

真的会破防，就在我以为要出了的时候，里面居然还有，我瞬间绷不住了，我真是又菜又爱玩啊

打开之后得到：NEFICPIC&CRTCTNEYO

emmm，开始以为是位置乱了，移位发现还是不行，但是我仔细一看发现，里面好像有crypto，巧，然后把所有的都提出来，发现NICECTF&NICECRYPTO，wowww~ ~刚刚好，
但其实是栅栏，栅栏密码key = 3,就可以得到NICECTF&NICECRYPTO了，
打开发现还有四层，当时附件还改了，我不知道，后面才看见，太悲伤了，一直出不来

废话不多说：

打开发现是表情包，想到base100，我寻思着有base13吗，然后去看tip，发现了端倪，上面Ba & Ro，ba是指base，ro是指rot，然后rot13，然后base64，最后直接base65536解出，

base100： � �👏👋🐾👚👪👇👱👞👚🐦👡👙👚👏👍🐯👏🐽👐👜🐦👇👱👜👙👤👡👙 🐬👥🐻🐯👏👋👈👚👪👇👘👰👛👮👡👙👎🐭🐯
rot13：8XTGcsPzgc/jbcXV8XFYe/Pzebmjb5nD8XTQcsPaydwjbW68
base64：8KGTpfCmtp/wopKI8KSLr/Cmrozwo5aQ8KGDpfCnlqjwoJ68
base65536 ：𡓥𦶟𢒈𤋯𦮌𣖐𡃥𧖨𠞼：原爞，启动！ -> 原神，启动！
(出题的师傅人是真的好，还来提醒我附件换了，真的挺感动的，在奔溃的边缘得以缓解)

得到：SYC{H0W_P3RF3C+_YU0_AR3!}

imREM:

额，这道题拿到手不知道是什么，然后一直到week4的时候才知道是证书题，

这个public和key都是完整的，openssl，工具也可以一把梭，这个不详细解释了（解出来会得到n,e,c， c可以从key中提取），只解释一下private的做法：

这个private是不完整的，是证书的下半部分，那么我们需要把这部分的证书给手撕开来，得到一些私钥的线索

这个需要先用base64解码，但是不要用工具去解码，因为用工具输出的应该都是utf-8或者unicode具体我也不清楚，反正当时我也是用工具，但就是出不来，后来在python中用base64解码就出了，再转hex即可，

找到02开头81或80，然后就可以判断出dp,dq,inv(q , p)

然后应该就和dp泄露差不多了

from base64 import b64decode
import binascii

s = 2uqP/7tlAPCJrMVD8zoRsRviyUnqO6y7cV7G0Fo1AoGBALpb6vOC9Ya/BEYV8+yL
814K0KQc7LITkwqGrDWoDZAxmgWqVtoxHuhDWr5jF+ttFZBPx4fTezTEjarXXSt3
pB1c7EwnzO0Hy9qJH++g2dcCUCO8uGXBbAACZqCQZAAydSrM5zozWAYMBWcj54j0
LtyOEg4PWPXiTzJ//xxWaH+xAoGBAKFGRIyYD89JvlZA6oz7YnjzsnDlTq01td32
XAuw62dZQHWmg1npC3YtzFlgTyNY2QelObmryyc2vFnxVhTYcDXFLQwrX8X5YV4A
rFAAlyzxpNzYPzDHrdqLD6PhMU+wRuHVPyAtNBsL0N+mgQcsWJJvngSTHg86kJOl
HlNuLBGhAoGAFiG3VR+lubcPvXOVAvkt+c8rF6qcmXlb2Og0hNwDJ2roX98aqOVy
p5AWGPoA1siI4/RPIp1ClfEwKMjraun1ZJs/jKemaQk2hdhWkQ+6QinvUJbA1Lqm
TcRmKa1emY/U6I8ce6N69e7ver1DV4I/ugSahJlZT/JRyF5qj1uVZ/k=

print(b64decode(s).hex())

s = b64decode(s)

print(binascii.hexlify(s))
 
s = daea8fffbb6500f089acc543f33a11b11be2c949ea3bacbb715ec6d05a35

028181 00ba5beaf382f586bf044615f3ec8bf35e0ad0a41cecb213930a86ac35a80d90319a05aa56da311ee8435abe6317eb6d15904fc787d37b34c48daad75d2b77a41d5cec4c27cced07cbda891fefa0d9d7025023bcb865c16c000266a090640032752acce73a3358060c056723e788f42edc8e120e0f58f5e24f327fff1c56687fb1

028181 00a146448c980fcf49be5640ea8cfb6278f3b270e54ead35b5ddf65c0bb0eb67594075a68359e90b762dcc59604f2358d907a539b9abcb2736bc59f15614d87035c52d0c2b5fc5f9615e00ac5000972cf1a4dcd83f30c7adda8b0fa3e1314fb046e1d53f202d341b0bd0dfa681072c58926f9e04931e0f3a9093a51e536e2c11a1

028180 1621b7551fa5b9b70fbd739502f92df9cf2b17aa9c99795bd8e83484dc03276ae85fdf1aa8e572a7901618fa00d6c888e3f44f229d4295f13028c8eb6ae9f5649b3f8ca7a669093685d856910fba4229ef5096c0d4baa64dc46629ad5e998fd4e88f1c7ba37af5eeef7abd4357823fba049a8499594ff251c85e6a8f5b9567f9

exp：

from Crypto.Util.number import *
import gmpy2

n = 19909823107653171578063591352961144331355821517081529017694276790993397379180193511755806958091550033235815834847549265751244192211761569881061742997804509098095297146520946343734939782935970054031389390447526577982809412074573616889663964050032806967234869741452169276495048891650957729859124875343810181695665153129824527065062831773013713479602052423965588148263292672514853749647510938621421264137177666497737099263481269491867593812579958283619882699866915561357437484323854411934718569074869408000191465973545262527827064708529700071967884924505367105288433603597633451764659888020272057850625902629529400734213

e = 65537
  
c = 7919943289368038783724317497692126508862944891824633900678369076067377304246633550201479733917999309653837868431720402842331248180337133419210706758864980771784429471586885607950008535770975052759319391393162975172158181546945556996836823037899332592611500055778040689826681310669586932854911372024839505403827196536681522194462010678759612069086605347172317442348948842018650570415028244207642694056753005420698743763441702332704811711253108369634221705269362120683326305095756923569255799172080587394198712839298803950747051721445378968267999151065222977429033753473443764902682897042586783088412947719949355428108
  
p1 = 0xdaea8fffbb6500f089acc543f33a11b11be2c949ea3bacbb715ec6d05a35

dq = 0x00ba5beaf382f586bf044615f3ec8bf35e0ad0a41cecb213930a86ac35a80d90319a05aa56da311ee8435abe6317eb6d15904fc787d37b34c48daad75d2b77a41d5cec4c27cced07cbda891fefa0d9d7025023bcb865c16c000266a090640032752acce73a3358060c056723e788f42edc8e120e0f58f5e24f327fff1c56687fb1

dp = 0x00a146448c980fcf49be5640ea8cfb6278f3b270e54ead35b5ddf65c0bb0eb67594075a68359e90b762dcc59604f2358d907a539b9abcb2736bc59f15614d87035c52d0c2b5fc5f9615e00ac5000972cf1a4dcd83f30c7adda8b0fa3e1314fb046e1d53f202d341b0bd0dfa681072c58926f9e04931e0f3a9093a51e536e2c11a1

a = 0x1621b7551fa5b9b70fbd739502f92df9cf2b17aa9c99795bd8e83484dc03276ae85fdf1aa8e572a7901618fa00d6c888e3f44f229d4295f13028c8eb6ae9f5649b3f8ca7a669093685d856910fba4229ef5096c0d4baa64dc46629ad5e998fd4e88f1c7ba37af5eeef7abd4357823fba049a8499594ff251c85e6a8f5b9567f9
#转化成dp泄露，这个给了很多条件，我不知道有什么用，证书题其实也不常见
for i in range(1,65538):

    p = ((dp*e-1)//i)+1

    if n % p == 0:

        print(p)

p2= 139366780536806265952989755984946005613683040843550738567578228396467356994133236355995164875843300802280010895144102144241447133593670597630811746145934665872367005294375732111408915384205212738302382734711912565019677628580803708931282556787630816046663643055532580256795845291765806134460998759909294103093

print(isPrime(p2))

print(isPrime(p))

q = n // p2

phi = (q-1)*(p2-1)

d = inverse(e,phi)

m = pow(c,d,n)

print(long_to_bytes(m))

#You_are_r1ght_bu7_Genshin_Impact_1s_@_brand_new_open_wor1d_@dventure_gam3_dev3loped_6y_miHoYo.

#输入压缩包，即可获得flag哦！

#SYC{OVvanSh3nO_Q1D0ng!}

额,这个好像写麻烦了，我看了官方的wp发现好简单，这里贴一下官方的wp

一个残缺的rsa私钥，一个加密后的key，一个exe结尾的压缩包从key中提取c


  from Crypto.Util.number import * 
  c = open('flag.enc', 'rb').read() 
  c = bytes_to_long(c) 
  print(c)

首先将拥有的两部分分别base64解密然后转16进制数观察
t = '''...''' t = t.strip()
import base64 from Crypto.Util.number import *
tb = base64.b64decode(t)
print(tb.hex())
前半部分： 308204a502010002820101009db757827029f6d0a0ac70c502ab948c28f31804dc8901c619fb9edd 3a87ab39c2cb2306ad6ccf39bc1a9b645b02c5a3c945c1daa523bf9a68ee5b68e41e48f32b38865a 9007c171964fb0a210a5b53b4c5d19ceed7685050deab3213ce767d7bff1469ea568a50584aa89ef a906c52a5cc67d05fe0200e1b0adc9af5ef8150fe28c565eb18977ddc74591ac0f483ee41a02366d 0a5efd5627f6e2e6887081263fc8288af96d5f253ce9edfde2aafa7ab3b1cb123da382e89ea829fd 30347620d35774c111a8ebd53054e460f7c3c9d20def8f8bb43dc5a4cf0d0844b84478a4a3ab734d 6437144792cebda0d8e0f6f628b700d96f55e1815e11fb2b83197205020301000102820101009a4f 9a857b2cf3da687a8fd392ab422a689e80afb0ff340719c1014cbf49a2945f2cd5d660b487849bb1 04bd09f70a5d183ef24ef528a6fd731153caaaf79eb49d632ec1490eed8c2f5f45192c64958fb145 9e4cc236262c
后半部分： 46e1d53f202d341b0bd0dfa681072c58926f9e04931e0f3a9093a51e536e2c11a10281801621b755 1fa5b9b70fbd739502f92df9cf2b17aa9c99795bd8e83484dc03276ae85fdf1aa8e572a7901618fa 00d6c888e3f44f229d4295f13028c8eb6ae9f5649b3f8ca7a669093685d856910fba4229ef5096c0 d4baa64dc46629ad5e998fd4e88f1c7ba37af5eeef7abd4357823fba049a8499594ff251c85e6a8f 5b9567f9
由ASN.1的标签原理可将两部分整理为：
308204a5020100 02820101:009db757827029f6d0a0ac70c502ab948c28f31804dc8901c619fb9edd3a87ab39c2cb230 6ad6ccf39bc1a9b645b02c5a3c945c1daa523bf9a68ee5b68e41e48f32b38865a9007c171964fb0a21 0a5b53b4c5d19ceed7685050deab3213ce767d7bff1469ea568a50584aa89efa906c52a5cc67d05fe0 200e1b0adc9af5ef8150fe28c565eb18977ddc74591ac0f483ee41a02366d0a5efd5627f6e2e688708 1263fc8288af96d5f253ce9edfde2aafa7ab3b1cb123da382e89ea829fd30347620d35774c111a8ebd 53054e460f7c3c9d20def8f8bb43dc5a4cf0d0844b84478a4a3ab734d6437144792cebda0d8e0f6f62 8b700d96f55e1815e11fb2b83197205 0203:010001 02820101:009a4f9a857b2cf3da687a8fd392ab422a689e80afb0ff340719c1014cbf49a2945f2cd5d6 60b487849bb104bd09f70a5d183ef24ef528a6fd731153caaaf79eb49d632ec1490eed8c2f5f45192c 64958fb1459e4cc236262c... ... ...46e1d53f202d341b0bd0dfa681072c58926f9e04931e0f3a9093a51e536e2c11a1 028180:1621b7551fa5b9b70fbd739502f92df9cf2b17aa9c99795bd8e83484dc03276ae85fdf1aa8e5 72a7901618fa00d6c888e3f44f229d4295f13028c8eb6ae9f5649b3f8ca7a669093685d856910fba42 29ef5096c0d4baa64dc46629ad5e998fd4e88f1c7ba37af5eeef7abd4357823fba049a8499594ff251c 85e6a8f5b9567f9
可知前半段完整的数据为n和c，后半段完整的数据为qInv
直接解密

from Crypto.Util.number import * import gmpy2 
n = 0x9db757827029f6d0a0ac70c502ab948c28f31804dc8901c619fb9edd3a87ab39c2cb2306ad6ccf 39bc1a9b645b02c5a3c945c1daa523bf9a68ee5b68e41e48f32b38865a9007c171964fb0a210a5b5 3b4c5d19ceed7685050deab3213ce767d7bff1469ea568a50584aa89efa906c52a5cc67d05fe0200 e1b0adc9af5ef8150fe28c565eb18977ddc74591ac0f483ee41a02366d0a5efd5627f6e2e6887081 263fc8288af96d5f253ce9edfde2aafa7ab3b1cb123da382e89ea829fd30347620d35774c111a8eb d53054e460f7c3c9d20def8f8bb43dc5a4cf0d0844b84478a4a3ab734d6437144792cebda0d8e0f6 f628b700d96f55e1815e11fb2b83197205 
e = 0x10001 
q = 0x1621b7551fa5b9b70fbd739502f92df9cf2b17aa9c99795bd8e83484dc03276ae85fdf1aa8e572 a7901618fa00d6c888e3f44f229d4295f13028c8eb6ae9f5649b3f8ca7a669093685d856910fba42 29ef5096c0d4baa64dc46629ad5e998fd4e88f1c7ba37af5eeef7abd4357823fba049a8499594ff2 51c85e6a8f5b9567f9 
p = n//q 
d = gmpy2.invert(e,(p-1)*(q-1)) 
print(long_to_bytes(pow(c,d,n)))

nocCRT:

额，这个虽然是CRT，但是模数却不互质，所以会想到扩展CRT

扩展CRT的原理：
gcd(ni,nj) != 1, 单独讨论，设gcd(n1,n2) = g

x = c1 mod n1 , x = c2 mod n2

则，x = k1n1 + c1, x = k2 n2 + c2, 合并移项得到：

n1k1 - k2n2 = (c2 - c1),然后翡镯定理可知：

(n1/g) k1 = (c2 - c1) /g mod (n2 / g)

k1 = (c2 - c1) /g * (n1 / g)^{-1} mod (n2 / g)

k1 = (c2 - c1) /g * (n1 / g)^{-1} + k3 * n2 / g

带入可得：x = ( (c2 - c1) /g * (n1 / g)^{-1} + k3 * n2 / g) n1 + c1

然后就有： x = ( (c2 - c1) /g * (n1 / g)^{-1} + k3 * n2 / g) n1 + c1(mod n1n2/g)


from Crypto.Util.number import *

import gmpy2

  

n = 5        # 同余方程个数

a = [1259284928311091851012441581576, 1501691203352712190922548476321, 1660842626322200346728249202857, 657314037433265072289232145909, 2056630082529583499248887436721]

m = [1921232050179818686537976490035, 2050175089402111328155892746480, 1960810970476421389691930930824, 1797713136323968089432024221276, 2326915607951286191807212748022]  # 模数


def exgcd(a, b):

    if 0 == b:

        return 1, 0, a

    x, y, q = exgcd(b, a % b)

    x, y = y, (x - a // b * y)

    return x, y, q

def CRT():

    if n == 1 :

        if m[0] > a[0]:

            return a[0];

        else:

            return -1;

    for i in range(n):

        if m[i] <= a[i] :

            return -1;

  

        x, y, d = exgcd(m[0], m[i])

        if (a[i] - a[0]) % d != 0:

            return -1;    

        t = m[i] // d;

        x = (a[i] - a[0]) // d * x % t

        a[0] = x * m[0] + a[0];

        m[0] = m[0] * m[i] // d;

        a[0] = (a[0] % m[0] + m[0]) % m[0]

    return a[0];

  

ans = CRT()

print(ans)

from Crypto.Util.number import long_to_bytes

print(long_to_bytes(ans))

#SYC{wha+s_wr0n9!_CRT_bu+_n0+_<0mpr1me!}

nc:

nc连接，没什么，就是暴力破解，把四个未知字符爆破出来就好了，打开容器，然后

打开命令管理器，输入nc nc1.ctfplus.cn 19218，nc连接，

会出现这一行，然后把后面的字符串部分给输入下面根据题目编写的爆破脚本里面

会爆破出四个未知的字符

再把其输入

然后1回车，2回车 …………………………直到得到flag

SYC{MAYB3_Y0U_KN0W_A1AN-B3<K3R?}

import hashlib
import itertools
import string

# 目标哈希
target_hash = "74da6891e993963ec9c34c6f481159548566a68df7db60ca10e0176cdca86840"
# proof后面的字符串部分
proof_suffix = "8DYJCztWXcPPoNab"  # 这是从服务器获得的 proof[4:]

# 字符集（包括大小写字母和数字）
charset = string.ascii_letters + string.digits
# 生成所有可能的 4 字符组合
for combination in itertools.product(charset, repeat=4):

    candidate = ''.join(combination)  # 组合成 4 字符的字符串

    candidate_input = candidate + proof_suffix  # 拼接 XXXX 和 proof[4:]

    # 计算哈希

    hash_result = hashlib.sha256(candidate_input.encode()).hexdigest()

    # 检查是否与目标哈希匹配

    if hash_result == target_hash:

        print(f"Found matching XXXX: {candidate}")

        break  # 找到匹配的 XXXX 后退出

LLL:

根据题目的提示，

可以知道是三L的知识点，浅浅的了解一下格密码的知识点就好了，这题不需要了解那么的深，

格知识：

参数

模： $p$

私钥： $(f, g)$

公钥： $h = f^{- 1} g % p$

临时密钥 $r$

加密：

解密：

因为 b = (a*m + c) % p，且 a, b ,p 都已经知道，所以直接构造格子，一开始无脑构造二维，

发现不行，观察一下数据，发现位数太大了，

超范围了，所以考虑构造三维格子

构造好直接在sagemath中跑即可


from Crypto.Util.number import *

from gmpy2 import *

  

a = 169790849804323540946197204708402762862586197604183102589270741859708550301920348112941305999764092197996929298474590062625556806793613268527763774013772685954699561684244945434843656515307801882995934869499880288594142919381501796488815033294127591623260894764750214588993456840404443515671353802614450411717

  

b = 87985708831523238980948938165414984318379459926002798504435964538203443877988599888615810231215118828138306895572062833107988965151522391460216837691927960249874511818878134399363147008042562222910234739940697553852540265617603482995091203105040187460485673579382171260197291783748886765929376179151804062085

  

p = 131724494512065658801039766546788821718063963144467818735768040631367069153816254855229655449559099188694403260044990366292026916085340250077198735215774149087025577263769846650728593180101073940507285459917860726551385227481715873503612683249433020201729190862430476054822102865441136763977415824331858801617

  

M = Matrix(ZZ,[[p,0,0],[-a,1,0],[b,0,2**400]])

c,m,k = M.LLL()[0]

flag = long_to_bytes(abs(m))  

print(flag)

  

#SYC{1e989433efffd767589e989ad0f091075c06}

ecc:

还是看题目就知道知识点，椭圆曲线加密，去网上搜一下对应的知识点浅浅学一下就可以了

加密过程：

1、选一条椭圆曲线Ep(a,b),并取椭圆曲线上一点作为基点P（这里的a、b是方程中x一次项系数和常数项）

2、选定一个大数k作为私钥，并生成公钥Q=kP,

3、加密:选择随机数r，将消息M生成密文C,密文是一个点对，即C=(rP, M+rQ)。

4、解密:M+rQ-k(rP)=M+r(kP)-k(rp)=M

性质：

A(x1,y1),B(x2,y2)以及C(x3,y3)均在同一椭圆曲线y^2=x^3+ax+b上,且在有限域p上，其中C是AB直线与曲线的交点的关于x轴的对称点，即 A+B=C ）有如下关系：

1、坐标的计算:

x3=k^2-x1-x2(modp)

y3=k(x1-x3)-y(modp)

2、斜率计算

若A=B，则k=(3x2+a)/2y1(A=B即要计算A点切线，需要求导)

若A≠B，则k=(y2-y1)/(x2-x1)

大致了解一下就可以看题了

K = k * G

c1 = m + rK

c2 = r * G

G为基点，k是私钥，K是公钥，我们已经知道c1,c2,具体解密过程如下脚本，最后用c1 - kc2就可以得到m了,直接在sagemath上跑


from Crypto.Util.number import *

p = 93202687891394085633786409619308940289806301885603002539703165565954917915237

a = 93822086754590882682502837744000915992590989006575416134628106376590825652793

b = 80546187587527518012258369984400999843218609481640396827119274116524742672463

k = 58946963503925758614502522844777257459612909354227999110879446485128547020161

cipher_left = 34210996654599605871773958201517275601830496965429751344560373676881990711573

cipher_right = 62166121351090454316858010748966403510891793374784456622783974987056684617905

E = EllipticCurve(GF(p),[a,b])#先E = EllipticCurve(GF(p),[a,b])建立一个椭圆曲线，要先建立后面才能用

c1 = E([40485287784577105052142632380297282223290388901294496494726004092953216846111 , 81688798450940847410572480357702533480504451191937977779652402489509511335169])

c2 = E( [51588540344302003527882762117190244240363885481651104291377049503085003152858 , 77333747801859674540077067783932976850711668089918703995609977466893496793359])

#这里要加上E说明是在E下上的点

m = c1 - k * c2

ml=cipher_left//m[0]

mr=cipher_right//m[1]

M=long_to_bytes(int(ml))+long_to_bytes(int(mr))

print(M)

#SYC{Ecc$is!sO@HaRd}

ezRSA:

这个是coppersmith的题，e 只有3 ，m = (m >> bits) << bits，可以知道是m的低位泄露，

了解一下sagemath：

PolynomialRing ：构造多项式环

Zmod(n) ：模运算

implementation='NTL' ：执行 NTL

small_roots(self, X=None , beta=1.0 , epsilon=None)：计算多项式的小整数根

其中的参数：X —— 根的绝对边界， beta( $β$ ) —— compute a root mod p where p is a factor of N and $p \geq N^{β}$

(程序默认值是 1.0 ，此时 p=N)，epsilon （默认： $β / 8$ )

monic()：用于将多项式的首项系数归一化为1。它接受一个多项式作为参数，然后返回一个新的多项式，其中首项系数已经被归一化为1。这个过程可以简化多项式的表达式，使其更易于计算和分析

Dense univariate polynomials over (\ZZ/n\ZZ), implemented using NTL - Polynomials贴一个sagemath的用法

回归本题：

创建一个模n的多项式环，构造一个多项式f = 2024x - 2023 - h

然后在给定的范围内去寻找小根（范围需要直接去调参数，但是我还不怎么会调参数，这个参数一般都是0.4，但只是一般），然后再构建一个新的多项式，f = (mhigh + y)^3 - c，尝试恢复m，直接在sagemath上面跑就好，恢复出m然后long_to_bytes,即可领取一份flag！！

from Crypto.Util.number import *
import gmpy2

n = 98776098002891477120992675696155328927086322526307976337988006606436135336004472363084175941067711391936982491358233723506086793155908108571814951698009309071244571404116817767749308434991695075517682979438837852005396491907180020541510210086588426719828012276157990720969176680296088209573781988504138607511

c = 9379399412697943604731810117788765980709097637865795846842608472521416662350816995261599566999896411508374352899659705171307916591351157861393506101348972544843696221631571188094524310759046142743046919075577350821523746192424192386688583922197969461446371843309934880019670502610876840610213491163201385965

h = 111518648179416351438603824560360041496706848494616308866057817087295675324528913254309319829895222661760009533326673551072163865

e = 3

bits = 150

PR.<x> = PolynomialRing(Zmod(n))

f = 2024*x - 2023 - h

f = f.monic()

roots = f.small_roots(X = 2^500,beta = 0.4)

if roots:

    mhigh = roots[0]

PR.<y> = PolynomialRing(Zmod(n))

f = (mhigh + y)^3 - c

f = f.monic()

roots = f.small_roots(X = 2^151,beta=0.4)

if roots:

    m = mhigh + roots[0]

    print(m)

m = 55098146333703730947926790790691720107068601034889480665048328600442527334253416918933209078294725467152711309939550589383549

print(long_to_bytes(m))

#SYC{crypto_is_very_interesting_why_dont_you_join_us}

easy_LLL:

额，这道题，看起来和week3的LLL好像，但是多了：

result = [encrypt(m) for i in range(5)]

c = [x[0] for x in result]

a = [x[1] for x in result]

虽然看起来直接上了难度，但是也可以直接构造三维格子，只选一组数据带入去出，过程和LLL的一模一样，非预期给出了，而且我试了每组数据，发现都可以出，格密码的话我是在NSS工坊上面瞎学的，做的很勉强。

from Crypto.Util.number import *

p = 114770017142688382362918268558878024848633097928402093647914503696492833723966801545716194546592346338592062332306371502256979159033965343002132956304625610134374822329402499359634555710129039614275145668904822690744696925414716152630310915301980153974374009140517084226870950134327432658087834138202887501571

c = 75827517416784647262997004080634347924631190865715212882627791181841845414253117114184423517850773219376565782814219713490136873921446382123059696483594598328510450811390866671002685611755205236016843942407419858592870716928648777362367108239158432436307113173823883182666320180058177554647020175991566479974

a = 154147211832384364492785997490497428696214843927503185938896425556028644075902949520267734189423717477702286854849502563505554965833703544305651488482204719931055591825164774932532116940955079750398001376723036214113076925445019856194390932639722726924707396244454184674407094860919513514591518499956074524561

M = Matrix(ZZ,[[p,0,0],[-a,1,0],[c,0,2**400]])

c,m,k = M.LLL()[0]

flag = long_to_bytes(abs(m))  

print(flag)

#SYC{125b-5c7b19c7-90e2-8d87c8a8}

下面是非预期的方法：
（虽然我现在还是看不太懂）

-b = -c+a*m+kp

$(k 1, k 2, k 3, k 4, k 5, m, - 1) [\begin{matrix} p & 0 & 0 & 0 & 0 & 0 & 0 \0 & p & 0 & 0 & 0 & 0 & 0 \0 & 0 & p & 0 & 0 & 0 & 0 0 & 0 & 0 & p & 0 & 0 & 0 \0 & 0 & 0 & 0 & p & 0 & 0 \a 0 & a 1 & a 2 & a 3 & a 4 & r & 0 \c 0 & c 1 & c 2 & c 3 & c 4 & 0 & 2^{255} \end{matrix}] = (- b 1, - b 2, - b 3, - b 4, - b 5, m, - 2^{255})$

from Crypto.Util.number import * 
p = 11477001714268838236291826855887802484863309792840209364791450369649283372396680 15457161945465923463385920623323063715022569791590339653430021329563046256101343 74822329402499359634555710129039614275145668904822690744696925414716152630310915 301980153974374009140517084226870950134327432658087834138202887501571 
c = [2526915767412008250032358545184292856040462596793266290851792270487182851339723 38586150059681240174286398539605504685428942704518716124966316451750158262034932 65945456529848647562285359912541672751550625137876486033809099678631009005979648 033707322772087110235116987698692692467320479776960630479772236446980, 75827517416784647262997004080634347924631190865715212882627791181841845414253117 11418442351785077321937656578281421971349013687392144638212305969648359459832851 04508113908666710026856117552052360168439424074198585928707169286487773623671082 39158432436307113173823883182666320180058177554647020175991566479974, 40004397317197465344043603398406750064538475824927459799822216246602968059960442 39209286181541462187650487112017410839740281883027081539802479046385802021188067 65619059473461992793303215453474217538078389555984131852004514411356216424771791 5766667365412215754183668349398802684299015216478025166881475794536, 16711257143606850336586355581909703391105580636095435863487225535083010317005439 43537510580064102411213812181024220712744301103620954496763312398363601515308984 38152873706465650717840020981830214898820464926094417083615507867528577735652528 21037805549119284258373739189052221307754872723967188683410620808193, 10651222799904898854353754234563652892559410712812503063500266598057470900655884 04461890173576236818286779351250121446899637988659717829147046167982394519713705 11961281779438306334353650663495164449411037055054859128957955413918744183200858 441122917851347996800797164614883188302584586112732819164555910532500] 
a = [1778761639208387205854746405113912490514188278533723873426352453414957924688261 99544624082182728094652999191797576747605771062756630817438777653951772485569478 51632490395611330919079562225834682464339000483539727288925669608735623951588145 9115499360779486974615331766141255410923960657795391638070660994726539, 15414721183238436449278599749049742869621484392750318593889642555602864407590294 95202677341894237174777022868548495025635055549658337035443056514884822047199310 55591825164774932532116940955079750398001376723036214113076925445019856194390932 639722726924707396244454184674407094860919513514591518499956074524561, 16223691031241644830331607928462613145244435229011047762013584288512500349306817 23307661742259970490940808366856178369114756385082839185763045025828488470974672 51286819613975600023439985149604495163647781268904127545271241114039490048103188 362740808427663167350820948490766499995036870926879430699822216419877, 15632433064946585686520565264291911655148061006083045632336151476178340661316282 65550663672158227471451092235303816897806250357950044589192623624203752255607904 67893332585836287433463308447660726674632677063603419250881619682710122472587150 879771212601074942044613408069114640355658551759306352327418458216623, 94727349364308455432706991721504607810501329870619614073375570944298709074650444 44213935631885480908192562500951610297851817034352572662714912365533225352941829 24407470734636151065015301339307500102900512267659061942103729043234608842386651 94406125116885468971886527174150462509520345910607640580833401931201] 
Ge = Matrix(ZZ,[ 
	[p,0,0,0,0,0,0], 
	[0,p,0,0,0,0,0], 
	[0,0,p,0,0,0,0], 
	[0,0,0,p,0,0,0], 
	[0,0,0,0,p,0,0], 
	[a[0],a[1],a[2],a[3],a[4],1,0], 
	[c[0],c[1],c[2],c[3],c[4],0,2^255] 
]) 
for i in Ge.LLL(): 
	if abs(i[-1]) == 2^255: 
	m = abs(i[-2]) 
	flag = long_to_bytes(int(m)) 
	print(flag) 
	#b'SYC{125b-5c7b19c7-90e2-8d87c8a8}

highlow:

这道题，真的绷不住了，开始看错题了，一直一直出不来，当时人都崩溃了，后面才发现len(flag) == 44，他的长度是352，太悲伤了，

这道题就是将flag的高位泄露，p异或752位，得到pxor，flag<<400,后400全都是0，异或不会变，所以pxor，相当于pxor的后四百位就是p，前272位也是没有变话的，这道题相当于已经知道了p的高低位，相当于高低位泄露，copper，直接打！


from gmpy2 import *
from Crypto.Util.number import *

pxor = 124229245244085791439650934438639686782423445183921252684721764061493908790073948877623812930339081158169421854801552819088679937157357924845248082716160727839419054107753000815066526032809275137495740454967765165248115412626716101315676902716808647904092798908601183908297141420793614426863816161203796966951

n = 14091206320622523674847720139761543154822190879035380245424481267482550932229611965964424965958386255076593911062804299275581742665134207390532802109700225140999812698020838683697375891035625255222001884477214361835101442288725383073334392995186053867261497679234362794914108033574681292656522807928680812726462195077833184018122369579002715900477290345396065912536529290811962117814900448319776590712946259540382461632468634827959957286905806432005632864663985014872365672653476822833921870071851313424903481282350342304819149894610089804321405589433980650340610521659031234826823369114800150883988613877877881069579

c = 11017336122691034053241992293963114590816319844384287448629663672049205892828600396465505710922907685545939978376321927394655458727494361852952898280905220963163625482295222856129164172619564344634365520328815972232825639292605311741655988427166811406091329613627961070231457035303298793651546412496975662225857123805867756651901374507447803198638466304862480202099076813471571495380132563252630789218173007275890600746758285415274434393381125742526014986039652677605642226576741424053749512280825231217420239089105794080707322357602941046822659335487420672699022969372037662958497832065752272061853723653365171768556

e = 65537

p_low = pxor & ((1<<400)-1)

p_high = pxor>>752

pbits = 1024

kbits = pbits - p_high.nbits()-p_low.nbits()

PR.<x> = PolynomialRing(Zmod(n))

f = p_high * 2 ^ 752 + x * 2 ^400 + p_low

p0 = f.monic().small_roots(X = 2 ^ 352,beta = 0.4)[0]

p = p_high * 2 ^ 752 + p0 * 2 ^400 + p_low

#这里注意一下，一定要把位数给加上啊，就是p_high * 2 ^ 752后面这个，2^752不要忘记了，我当时就给忘记了，导致一直出不来，而且最后的时候我直接print（p_high + p0 + p_low）了，当时真的出不来，明明就差一点，一直看不出来问题，悲伤，后来晚上从图书馆回来才发现了这个低级的错误。

  

print(p)#124229245244085791439650934438639686782423445183921252684721764061493908790073948870293397345832864335042475785728973083585314359884258149847711858397514908905613053545356103378803167400248049812953390169260920922532533579636992286759858305144877980057956578200502522929436738617263575249394601586523948318247

#这里求出p之后直接rsa出flag

q = n // p

phi = (q-1)*(p-1)

d = pow(e,-1,phi)

m = pow(c,d,n)

print(long_to_bytes(m))

#SYC{2f521b13bc9d6e932e9f5cbe511112df9e3a9c6}

LinkedListModular：

此题为re+crypto，前面的部分是re的部分，是队友(vstral)做的，下面放一下vstral的过程：

本题⽤到了⼤量的 GMP 库函数和链表结构

check 函数中，涉及到了多个加密和数学运算

在 write_enc_cmp_to_file 函数中，将传⼊的数据（以 a2 作为起始地址， a3 为⻓度的字节数据写⼊到⽂件中，并以特定格式保存）。主要作⽤是将⼀个字节流按⼗六进制格式写⼊⽂件

a1 ：这是⼀个整数，表⽰⽂件名的编号。
a2 ：这是⼀个指针，指向要写⼊⽂件的数据的起始地址。
a3 ：表⽰要写⼊的数据的字节数。
⾸先分配了空间，然后通过 sprintf 将 a1 格式化⽣成⽂件名，然后⽤ fprintf 将每个字节以⼗六进制格式写⼊到⽂件中

逆向过程：

先将加密后写⼊⽂件的数据提取，然后将 enc 通过密钥解密并且分段输出 p ， q ， e ， c


enc0 = 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 

enc1 = 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 

enc2 = 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 

enc3 = [0x39,0x71,0x5e,0x17,0x15,0x69,0x59,0x16,0x7a,0x0a,0x59,0x04,0x20,0x41,0x4b,0x40,0x17,0x5a,0x7c,0x7e,0x0d,0x0c,0x43,0x3b,0x0c,0x40,0x2e,0x0a,0x53,0x5c,0x7a,0x45,0x4f,0x44,0x47,0x59,0x7c,0x73,0x5c,0x0b,0x12,0x6a,0x5c,0x11,0x7b,0x5e,0x5c,0x5d,0x7b,0x17,0x1d,0x40,0x11,0x0d,0x78,0x2e,0x5b,0x0d,0x40,0x6b,0x59,0x10,0x7e,0x5d,0x59,0x06,0x71,0x4a,0x40,0x40,0x11,0x0b,0x2c,0x7e,0x5e,0x0b,0x4e,0x68,0x57,0x11,0x7f,0x0b,0x5c,0x06,0x22,0x10,0x4e,0x44,0x45,0x5b,0x78,0x7e,0x5f,0x56,0x12,0x6d,0x5c,0x42,0x7a,0x58,0x0d,0x50,0x72,0x47,0x4c,0x49,0x44,0x5d,0x7e,0x72,0x08,0x0d,0x43,0x3f,0x0e,0x4d,0x7a,0x5e,0x5f,0x04,0x7a,0x46,0x1d,0x40,0x4d,0x57,0x7c,0x2a,0x0b,0x0b,0x11,0x38,0x5b,0x17,0x79,0x5d,0x5e,0x5d,0x73,0x41,0x49,0x47,0x40,0x5e,0x7f,0x2a,0x0f,0x5e,0x4f,0x61,0x57,0x45,0x7d,0x5c,0x0e,0x54,0x71,0x17,0x4b,0x45,0x47,0x59,0x7a,0x7e,0x5e,0x09,0x11,0x38,0x5b,0x44,0x74,0x5b,0x5b,0x56,0x72,0x44,0x48,0x12,0x41,0x09,0x2b,0x79,0x5a,0x5c,0x4f,0x69,0x5c,0x40,0x7d,0x5f,0x5b,0x56,0x71,0x45,0x1c,0x42,0x45,0x09,0x7d,0x7b,0x5f,0x5a,0x13,0x6d,0x0d,0x42,0x7f,0x50,0x58,0x01,0x7a,0x42,0x1f,0x16,0x12,0x5a,0x79,0x2d,0x56,0x0b,0x42,0x3b,0x0a,0x46,0x28,0x08,0x5b,0x5d,0x22,0x10,0x1c,0x16,0x15,0x0c,0x7b,0x78,0x56,0x56,0x4f,0x38,0x5a,0x40,0x28,0x0a,0x5d,0x55,0x22,0x40,0x4d,0x44,0x43,0x5f,0x2a,0x73,0x08,0x0a,0x40,0x6f,0x0a,0x17,0x6c,0x18,0x51,0x55,0x3b,0x13,0x1d,0x11,0x16,0x0c,0x7f,0x7a,0x5b,0x5e,0x4e,0x3c,0x56,0x43,0x78,0x5b,0x5e,0x54,0x74,0x17,0x1b,0x45,0x17,0x5d,0x28,0x7f,0x0f,0x5b,0x46,0x6f,0x09,0x46,0x78,0x0a,0x5d,0x57,0x22,0x40,0x1a,0x41,0x11,0x0e,0x7f,0x2e,0x0d,0x5b,0x14,0x68,0x0b,0x45,0x28,0x5d,0x5d,0x5d,0x75,0x13,0x40,0x45,0x17,0x5c,0x78,0x2f,0x0d,0x0e,0x15,0x3b,0x0e,0x47,0x2f,0x58,0x58,0x03,0x22,0x4a,0x4b,0x41,0x43,0x0a,0x78,0x7f,0x08,0x0c,0x42,0x6d,0x5e,0x4d,0x79,0x5e,0x09,0x5d,0x77,0x44,0x1d,0x48,0x4c,0x5f,0x7b,0x7c,0x0f,0x5e,0x11,0x6e,0x0c,0x13,0x29,0x0c,0x52,0x04,0x21,0x13,0x49,0x47,0x41,0x58,0x2a,0x78,0x0f,0x56,0x45,0x38,0x0d,0x40,0x7b,0x50,0x5b,0x5d,0x75,0x47,0x41,0x46,0x17,0x5f,0x2c,0x7e,0x5c,0x0d,0x4f,0x68,0x57,0x41,0x28,0x0c,0x5b,0x06,0x75,0x4b,0x18,0x15,0x42,0x58,0x7d,0x29,0x5e,0x5c,0x43,0x6a,0x5b,0x47,0x2e,0x59,0x0f,0x07,0x77,0x42,0x40,0x45,0x41,0x09,0x2b,0x2d,0x08,0x0b,0x14,0x61,0x5b,0x10,0x29,0x51,0x5c,0x50,0x21,0x4b,0x1b,0x48,0x40,0x09,0x7e,0x2f,0x5c,0x0d,0x11,0x6a,0x5d,0x4d,0x7f,0x5f,0x0d,0x55,0x70,0x11,0x4a,0x40,0x40,0x0d,0x7a,0x7f,0x08,0x0a,0x16,0x6b,0x0a,0x14,0x7c,0x5a,0x5c,0x56,0x72,0x13,0x1f,0x49,0x41,0x0b,0x70,0x7a,0x5b,0x59,0x15,0x60,0x5e,0x14,0x7d,0x5f,0x0d,0x03,0x7a,0x41,0x4a,0x48,0x12,0x0a,0x71,0x2a,0x57,0x09,0x14,0x3c,0x5d,0x41,0x79,0x5b,0x5e,0x07,0x75,0x47,0x4c,0x40,0x4d,0x4f,0x2c,0x71,0x5e,0x17,0x13,0x6b,0x0c,0x47,0x6c,0x0a,0x51,0x55,0x3b,0x43,0x48,0x48,0x42,0x5e,0x7e,0x7e,0x5c,0x0e,0x4f,0x3a,0x0e,0x40,0x74,0x0b,0x5c,0x52,0x74,0x41,0x49,0x40,0x11,0x09,0x2b,0x28,0x56,0x0a,0x13,0x69,0x5a,0x10,0x2d,0x5c,0x5f,0x01,0x73,0x11,0x4a,0x42,0x42,0x0c,0x2b,0x2d,0x0c,0x5a,0x40,0x6a,0x0e,0x45,0x2e,0x0d,0x59,0x54,0x25,0x17,0x4c,0x11,0x46,0x5f,0x78,0x72,0x5f,0x0d,0x47,0x6a,0x09,0x45,0x7b,0x0b,0x0e,0x52,0x22,0x17,0x18,0x49,0x15,0x0c,0x78,0x72,0x5c,0x0e,0x13,0x3b,0x0a,0x41,0x74,0x5e,0x59,0x53,0x22,0x42,0x18,0x40,0x47,0x0c,0x2a,0x28,0x56,0x5c,0x4f,0x6a,0x58,0x40,0x28,0x5e,0x08,0x51,0x25,0x11,0x1d,0x40,0x16,0x57,0x71,0x7a,0x5a,0x5a,0x40,0x61,0x09,0x14,0x28,0x5d,0x5c,0x00,0x22,0x42,0x4d,0x40,0x43,0x5d,0x71,0x7a,0x0c,0x0d,0x15,0x6b,0x5f,0x40,0x75,0x5f,0x5b,0x06,0x75,0x46,0x4c,0x43,0x12,0x09,0x2f,0x73,0x5f,0x5a,0x4f,0x6d,0x0c,0x10,0x7f,0x5a,0x5f,0x04,0x71,0x41,0x40,0x46,0x44,0x0c,0x7f,0x78,0x5c,0x5e,0x14,0x6e,0x5d,0x45,0x28,0x08,0x5c,0x07,0x72,0x4b,0x49,0x42,0x4c,0x5e,0x2a,0x7d,0x0a,0x5b,0x41,0x3b,0x5a,0x40,0x29,0x5a,0x09,0x04,0x21,0x42,0x4d,0x49,0x47,0x0b,0x71,0x2d,0x08,0x59,0x14,0x3f,0x5c,0x4d,0x79,0x5c,0x0a,0x57,0x22,0x44,0x49,0x43,0x4d,0x0c,0x7b,0x7e,0x57,0x09,0x4e,0x6c,0x5a,0x10,0x7e,0x0a,0x09,0x01,0x25,0x10,0x40,0x14,0x17,0x0e,0x2b,0x28,0x56,0x5e,0x44,0x38,0x0b,0x4d,0x7e,0x0a,0x5e,0x56,0x70,0x41,0x1d,0x45,0x47,0x5a,0x70,0x7e,0x5e,0x0a,0x45,0x61,0x0c,0x13,0x2e,0x5b,0x0a,0x04,0x76,0x47,0x4f,0x13,0x47,0x5d,0x7e,0x7b,0x5e,0x09,0x41,0x6d,0x0c,0x47,0x2e,0x5f,0x58,0x55,0x76,0x4a,0x1a,0x16,0x47,0x5c,0x2f,0x2e,0x58,0x0c,0x43,0x69,0x59,0x42,0x7b,0x08,0x52,0x56,0x27,0x4a,0x1a,0x48,0x44,0x0e,0x79,0x2d,0x57,0x0b,0x15,0x6f,0x5f,0x11,0x2d,0x58,0x08,0x00,0x7a,0x42,0x48,0x48,0x46,0x0a,0x2d,0x7b,0x0f,0x0e,0x41,0x3c,0x0b,0x46,0x7f,0x08,0x59,0x04,0x7a,0x40,0x41,0x48,0x10,0x0c,0x70,0x2d,0x08,0x5e,0x4e,0x3c,0x0b,0x4c,0x2e,0x0f,0x0d,0x07,0x76,0x4b,0x40,0x49,0x45,0x0d,0x2d,0x2f,0x08,0x5d,0x4f,0x6b,0x0d,0x46,0x74,0x08,0x5e,0x5c,0x27,0x17,0x1d,0x42,0x43,0x09,0x7e,0x7a,0x0d,0x5b,0x12,0x6c,0x0a,0x45,0x7d,0x50,0x5b,0x00,0x76,0x40,0x4d,0x44,0x10,0x56,0x78,0x7d,0x5b,0x5a,0x41,0x3a,0x5f,0x46,0x7f,0x0b,0x09,0x57,0x22,0x43,0x41,0x44,0x17,0x57,0x7e,0x28,0x08,0x5e,0x4e,0x6b,0x5c,0x13,0x2e,0x51,0x5c,0x5c,0x73,0x40,0x1f,0x11,0x4d,0x0b,0x2b,0x29,0x5b,0x0e,0x13,0x6c,0x09,0x4c,0x7e,0x5e,0x59,0x51,0x7a,0x10,0x1c,0x14,0x12,0x0e,0x7d,0x29,0x5a,0x58,0x12,0x6c,0x0e,0x46,0x7b,0x50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 

def crypt(data):

    key = "IKnowYouLikeCrypto"

    v10 = len(data)

    result = ""

    for i in range(v10):

        data[i] ^= ord(key[i % 18])

        data[i] = chr(data[i])

        result += data[i]

    return result

def turn(data):

    location = []

    for i in range(len(data)):

        if data[i] == " ":

            location.append(i)

    p = data[:location[0]]

    q = data[location[0]+1:location[1]]

    e = data[location[1]+1:location[2]]

    c = data[location[2]+1:]

    p = int(p[2:],16)

    q = int(q[2:],16)

    e = int(e[2:],16)

    c = int(c[2:],16)

    return p, q, e, c

def output(data):

    data = turn(crypt(data))

    print("p = "+ str(data[0]))

    print("q = "+ str(data[1]))

    print("e = "+ str(data[2]))

    print("c = "+ str(data[3]))

print("\nenc0的结果为：")    
output(enc0)
print("\nenc1的结果为：")    
output(enc1)
print("\nenc2的结果为：")
output(enc2)
print("\nenc3的结果为：")
output(enc3)

密码部分：

这里得到了四组p,q,e,c,但是题目提醒了注意e和phi，就知道了e和phi不互质，所以很简单的


enc0�Ľ��Ϊ��

p = 122270861290599578639045245855962543705730032630755185606879757206179711733753365147966963024978794275592619421652783145619018195586521233885682978373729799794310979858883407180862687141564775908815997081768279744281601978220649648570076155855057872071934647568517429940869409265220173220394958396829102922747

q = 145895839511071843328632553786312889050300087771516738662432330420448872465700664308309226114037360915656319628912705547767935389266619849150150441056011167554962792718447234998480891798333165313729924211768140920745823520187297704741938920310919784476251310206400453334805818184285230822153394899548512014363 

e = 54050
c = 14030135514635598302794617424716291835816328683662639736248038724984547291821817259473592149765257404252757655788615493983340073369799283614267799122874068386483545369841536424720586065630488365125416323465659628220325230113595380678935117166076914407010605138044768627282116054287141762067469166512857628490621045347308961162010242614814641531660373325719065999031562248458476856972067741768151921225579744016300062897152743302145692689352793324795135401136498046047188121063247708395640777371548996062674320708390110447311929288206121760762410872526924579641318919472258601562794025337256187068455307563612901185023

enc1�Ľ��Ϊ��

p = 86010241306828695711315128924599735140271626506041901817784997674189297941853729364282050885284675170170821186690061333205093054758809155320992207244551279236901389783439594217221100781360422444116377542166763838467239709394816251633118839088171402103700954007772937216244712571543562096482264586237749875819

q = 135870661274724260761777656982848828868032325851948061776635650311711582931870434507059024120746651393755948511526211104639055783183197136990340032725326244689949152528703158984146195397743387805547644375182877781616481448740610009562687655041910892459867395655316615742128949396867439946390178637509880411633

e = 4270

c = 11598574270272630669714907360583133378706899622553239805344812166774743123206830402943691200503759695399591347094422594967770567057117248269726758467264312401133127715780819831501882369655128752151118270781981866484643528591517003660647218224315002479901416782305051851693962233968104493864345621514513889489988026466245490355297184507186588531101596026437796555886217059247337756496652510943309401703243002842025454251768433665133067967206291033448828322668417084410333992791255746157109331097709760150924142238375746555792880008101876932484326448226098739524995637966296227393399377617119263161025919164346792544561

enc2�Ľ��Ϊ��
p = 113996636277299533570182302103805270404328287588039359480117725799731468903729837871758089945224543747241332872053035409977686988828746257770577119441654973432267840909581546222729272312791011601321369417719086022464065509525953388164636951042845396223299823034634171347201104613118094684282180610968181763283

q = 179012430125443495243486068136135361221504157943761632954371910668630940409423204711355123693305413255114948345339274042082038162457949049106824669564814439501181874462870236637605932234845161478752403949944328541754810406699391028626685666316761083909928278039334844695628682088512654628951908784094717201679

e = 49233
c = 1322321374946524818652023720096939051768018158840573773662739229912717016596391549713276689400149946305546387707907323058608844811344367201330349195046716581714778505096098070215006674487247507334946756165794643707736833654141976843194322370706461949417335615521534451266831579666968022184721632924700931501823301459381927293501659690181802933722359094313659106322298471487330133269235036148809591432757124529202915730121032198174486512328435816936036772002230322906874285577303764890953815215779772380996339638756319626274531454590050226922831941419968900991324918796869830028048552466209680423495017341743205088786

enc3�Ľ��Ϊ��

p = 123888812740641928836168042326359816394975720485951831888838275274076617471071546431769283439173465908911537906450091397803657135440556050932230723993937394896061320470381209514964704945423660616750442602267012194575389456964026217997304524759607303104433720633388080425678353978941287094242434193955055630059
  
q = 121955917457499976375150103626761474166113111625393616609019476776085394161196238468185025413954055886598242407538128364904706744043310319400281880426622607966177170554459739351838972332013408544407052828843515726885278465751980156963483293699874070321302525920625454568002421588162549519570398808647199511817

e = 53954
c = 2212173972744047970990379804298933258334252251912942437533545314979358687449975375840376152127572741500667242076551697163529870996537676041038948227338266260926612489035397580103113172682245696555495676218185074197392305638733600231895438792687753022367399336104584002419428238425292877036252656179628594398438480118648680608266982741701265182418582127103364633630337738597016983834833305453465805338763525698753576429977848403451927698484187503450184378049543359003342835467612662169578981988940376898818772304771375667079867314152601395767452321011385597804726251559017443302250435057446590994585807118833643004990


from Crypto.Util.number import *
import gmpy2

p = 123888812740641928836168042326359816394975720485951831888838275274076617471071546431769283439173465908911537906450091397803657135440556050932230723993937394896061320470381209514964704945423660616750442602267012194575389456964026217997304524759607303104433720633388080425678353978941287094242434193955055630059

q = 121955917457499976375150103626761474166113111625393616609019476776085394161196238468185025413954055886598242407538128364904706744043310319400281880426622607966177170554459739351838972332013408544407052828843515726885278465751980156963483293699874070321302525920625454568002421588162549519570398808647199511817
e = 53954
c = 2212173972744047970990379804298933258334252251912942437533545314979358687449975375840376152127572741500667242076551697163529870996537676041038948227338266260926612489035397580103113172682245696555495676218185074197392305638733600231895438792687753022367399336104584002419428238425292877036252656179628594398438480118648680608266982741701265182418582127103364633630337738597016983834833305453465805338763525698753576429977848403451927698484187503450184378049543359003342835467612662169578981988940376898818772304771375667079867314152601395767452321011385597804726251559017443302250435057446590994585807118833643004990

n = q*p
e1 = e // 2
phi = (q - 1)
a = gmpy2.gcd(e,phi)
print(a)
d1 = gmpy2.invert(e1,phi）
m1 = pow(c,d1,q)
s = gmpy2.iroot(m1,2)[0]
print(s)  

#这里只放了一个脚本，其他的三组和这个一模一样，但是有一组是互素的我记得，那就更简单了，没必要都列出来，然后就到了恶心的环节了

这个得出四组值之后，没有看到题目的提示，无论如何我都出不来啊当时，后面试错了无数次之后才看到题目的提示，然后老是想把拼一块去转hex，但发现就是不能够256，后面突然灵光乍现，觉得会不会是把四个m分别转hex然后加一块，wow~ ~~ 刚好是256，当时一早上都没有听课，一直在想这个，太悲伤了，然后打包转md5值,还有没注意flag的题头居然是flag而不是SYC，太悲伤了，当时差点破防，哈哈哈


import hashlib

enc0 =47278196062167541585455853280941033307618230193432284377294672427930775790940

enc1 = 6649739036131957280788032150850684705382994434372821202484300458163206934369

enc3 = 9168988683066850946264592811590831830063906088604167160562565578372116763563

enc4 =48464490238122893740346191745565682578473536991653209626280563589974254605724

enc0 = hex(enc0)[2:]
enc1 = hex(enc1)[2:]
enc3 = hex(enc3)[2:] 
enc4 = hex(enc4)[2:]

print(enc0, enc1, enc3, enc4)

print("sum = %d", len(enc0) + len(enc1) + len(enc3) + len(enc4))

result = enc0 + enc1 + enc3 + enc4

result_md5 = hashlib.md5(result.encode()).hexdigest()
  
print("result = " + result_md5)

#flag{d3f06717efc6c0daf454ffeac9764687}

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