NSSCTF—Crypyo 2.0版本 ԅ(≖‿≖ԅ) （待续……）
[鹤城杯 2021]easy_crypto
题目：
公正公正公正诚信文明公正民主公正法治法治诚信民主自由敬业公正友善公正平等平等法治民主平等平等和谐敬业自由诚信平等和谐平等公正法治法治平等平等爱国和谐公正平等敬业公正敬业自由敬业平等自由法治和谐平等文明自由诚信自由平等富强公正敬业平等民主公正诚信和谐公正文明公正爱国自由诚信自由平等文明公正诚信富强自由法治法治平等平等自由平等富强法治诚信和谐
解题：
社会主义价值观解码，+.+
flag{IlUqU9O5guX6YiITsRNPiQmbhNRjGuTP}
[强网拟态 2021]拟态签到题
题目：
ZmxhZ3tHYXFZN0t0RXRyVklYMVE1b1A1aUVCUkNZWEVBeThyVH0=题解：
后面有等号，大概率base加密，按照密文的组成数字和字母范围大概是base64
大概啊都是大概，我基础不太行，(#^.^#)
flag{GaqY7KtEtrVIX1Q5oP5iEBRCYXEAy8rT}
[SWPUCTF 2021 新生赛]crypto8
题目：
73E-30U1&>V-H965S95]I<U]P;W=E<GT`题解：
有< > = 等符号和数字还有字母组成，是Uuncode编码，嘎嘎嘎，还好我先去看了遍编码 *^-^*
NSSCTF{cheese_is_power}
[SWPUCTF 2021 新生赛]crypto7
题目：
69f7906323b4f7d1e4e972acf4abfbfc,得到的结果用NSSCTF{}包裹。题解：
MD5加密,别问理由，不知道嘞！这不是有标签提示么。
NSSCTF{md5yyds}
[SWPUCTF 2021 新生赛]crypto6
题目：
var="************************************"
flag='NSSCTF{' + base64.b16encode(base64.b32encode(base64.b64encode(var.encode()))) + '}'
print(flag)
小明不小心泄露了源码，输出结果为：4A5A4C564B36434E4B5241544B5432454E4E32465552324E47424758534D44594C4657564336534D4B5241584F574C4B4B463245365643424F35485649534C584A5A56454B4D4B5049354E47593D3D3D，你能还原出var的正确结果吗？题目：
审计代码，需要进行base64，base32，base16解密
需要解密的密文是：4A5A4C564B36434E4B5241544B5432454E4E32465552324E47424758534D44594C4657564336534D4B5241584F574C4B4B463245365643424F35485649534C584A5A56454B4D4B5049354E47593D3D3D
采用魔法棒一把梭，哎嘿
NSSCTF{5e110989-dc43-1bd3-00b4-9009206158fe}
[SWPUCTF 2021 新生赛]ez_caesar
题目：
import base64
def caesar(plaintext):
str_list = list(plaintext)
i = 0
while i < len(plaintext):
if not str_list[i].isalpha():
str_list[i] = str_list[i]
else:
a = "A" if str_list[i].isupper() else "a"
str_list[i] = chr((ord(str_list[i]) - ord(a) + 5) % 26 + ord(a) or 5)
i = i + 1
return ''.join(str_list)
flag = "*************************"
str = caesar(flag)
print(str)
#str="U1hYSFlLe2R0em1mYWpwc3RiaGZqeGZ3fQ=="题解：
开头引用base64的库，密文两个等号
试试魔法棒
SXXHYK{dtzmfajpstbhfjxfw}
[SWPUCTF 2021 新生赛]pigpig
题目：
题解：
可爱的猪猪，下面的一看就是猪圈密码
照着表对，最近才开始用工具来着，以前都习惯去对表格（挠头.jpg）
NSSCTF{whenthepigwanttoeat}
[鹤城杯 2021]A_CRYPTO
题目：
4O595954494Q32515046324757595N534R52415653334357474R4N575955544R4O5N4Q46434S4O59474253464Q5N444R4Q51334557524O5N4S424944473542554O595N44534O324R49565746515532464O49345649564O464R4R494543504N35
题解：
rot13，base16，base32，base64，base85
到这边一切顺利，但是在Base85出现了问题
是的，它Base85不行了，然后我跑去找我师傅了，我师傅剥个瓜子（边啃边剥.jpg）
在师傅指导下，我解出来了。◍˃ᵕ˂◍
flag{W0w_y0u_c4n_rea11y_enc0d1ng!}
[SWPUCTF 2021 新生赛]ez_rsa
题目：
p = 1325465431
q = 152317153
e = 65537
计算出d,将d用MD5加密后包裹NSSCTF{}提交解题：
啊哈哈哈哈，我爱直白的题目。˙Ⱉ˙ฅ
import gmpy2
import hashlib
p = 1325465431
q = 152317153
e = 65537
phi = (p-1) * (q-1)
d = gmpy2.invert(e, phi)
print(d)
#43476042047970113
k = hex(d).encode('utf-8')
print(k)
#b'0x9a753ada825b41'
flag = "flag{" + hashlib.md5(k).hexdigest() + "}"
print(flag)
#flag{3a249418d1fa200066dbebedf8b164d9}
[BJDCTF 2020]base??
题目：
dict:{0: 'J', 1: 'K', 2: 'L', 3: 'M', 4: 'N', 5: 'O', 6: 'x', 7: 'y', 8: 'U', 9: 'V', 10: 'z', 11: 'A', 12: 'B', 13: 'C', 14: 'D', 15: 'E', 16: 'F', 17: 'G', 18: 'H', 19: '7', 20: '8', 21: '9', 22: 'P', 23: 'Q', 24: 'I', 25: 'a', 26: 'b', 27: 'c', 28: 'd', 29: 'e', 30: 'f', 31: 'g', 32: 'h', 33: 'i', 34: 'j', 35: 'k', 36: 'l', 37: 'm', 38: 'W', 39: 'X', 40: 'Y', 41: 'Z', 42: '0', 43: '1', 44: '2', 45: '3', 46: '4', 47: '5', 48: '6', 49: 'R', 50: 'S', 51: 'T', 52: 'n', 53: 'o', 54: 'p', 55: 'q', 56: 'r', 57: 's', 58: 't', 59: 'u', 60: 'v', 61: 'w', 62: '+', 63: '/', 64: '='}
chipertext:
FlZNfnF6Qol6e9w17WwQQoGYBQCgIkGTa9w3IQKw题解：
算是base吧。反正肯定不是我想的那个。给了dict，应该是把base的表换了，直接找代码么。
import base64
a = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/='
dict={0: 'J', 1: 'K', 2: 'L', 3: 'M', 4: 'N', 5: 'O', 6: 'x', 7: 'y', 8: 'U', 9: 'V', 10: 'z', 11: 'A', 12: 'B', 13: 'C', 14: 'D', 15: 'E', 16: 'F', 17: 'G', 18: 'H', 19: '7', 20: '8', 21: '9', 22: 'P', 23: 'Q', 24: 'I', 25: 'a', 26: 'b', 27: 'c', 28: 'd', 29: 'e', 30: 'f', 31: 'g', 32: 'h', 33: 'i', 34: 'j', 35: 'k', 36: 'l', 37: 'm', 38: 'W', 39: 'X', 40: 'Y', 41: 'Z', 42: '0', 43: '1', 44: '2', 45: '3', 46: '4', 47: '5', 48: '6', 49: 'R', 50: 'S', 51: 'T', 52: 'n', 53: 'o', 54: 'p', 55: 'q', 56: 'r', 57: 's', 58: 't', 59: 'u', 60: 'v', 61: 'w', 62: '+', 63: '/', 64: '='}
d=''
for i in range(65):
d+=dict[i]
b = 'FlZNfnF6Qol6e9w17WwQQoGYBQCgIkGTa9w3IQKw'
l=[]
for i in range(len(b)):
l.append(d.index(b[i]))
m=''
for ll in l:
m+=a[ll]
f=base64.b64decode(m)
print(f)
#b'BJD{D0_Y0u_kNoW_Th1s_b4se_map}'
[LitCTF 2023]梦想是红色的 (初级)
题目：
自由友善公正公正敬业法治自由自由和谐平等自由自由公正法治诚信民主诚信自由自由诚信民主爱国友善平等诚信富强友善爱国自由诚信民主敬业爱国诚信民主友善爱国平等爱国爱国敬业敬业友善爱国公正敬业爱国敬业和谐文明诚信文明友善爱国自由诚信民主爱国爱国诚信和谐友善爱国自由友善平等爱国友善平等友善自由诚信自由平等爱国爱国敬业敬业友善爱国敬业敬业友善自由友善平等诚信自由法治诚信和谐
题解：
嗯，标准的社会主义核心价值观
LitCTF{为之则易,不为则难}
[SWPUCTF 2021 新生赛]traditional
题目：
西方的二进制数学的发明者莱布尼茨，从中国的八卦图当中受到启发，演绎并推论出了数学矩阵，
最后创造的二进制数学。二进制数学的诞生为计算机的发明奠定了理论基础。而计算机现在改变
了我们整个世界，改变了我们生活，而他的源头却是来自于八卦图。现在，给你一组由八卦图方位
组成的密文，你能破解出其中的含义吗？
震坤艮 震艮震 坤巽坤 坤巽震 震巽兑 震艮震 震离艮 震离艮
格式：NSSCTF{}解题：
不是我说，这个八卦阵有毒，它要倒着看。我是一个个对出来的，呜呜呜。（其实，有代码来着）
整为1半为0得出
001000100 001100001 000110000 000110001 001110011 001100001 001101100 001101100
9位二进制是不能转变的，你会发现每组的第一位是0，可以去掉，变成了8位二进制，进行ACSII码转换（二进制转十六进制）。
NSSCTF{Da01sall}
[LitCTF 2023]Hex？Hex！(初级)
题目：
4c69744354467b746169313131636f6f6c6c616161217d题解：
题目提示很明显咧！੯‧̀͡u\
HEX！十六进制解码！（这我不用出解题过程吧੧ᐛ੭，我不会告诉你是我懒得截图的！）
LitCTF{tai111coollaaa!}
[SWPUCTF 2021 新生赛]crypto2
题目：
from gmpy2 import *
from Crypto.Util.number import *
flag = '***************'
p = getPrime(512)
q = getPrime(512)
m1 = bytes_to_long(bytes(flag.encode()))
n = p*q
e1 = getPrime(32)
e2 = getPrime(32)
print()
flag1 = pow(m1,e1,n)
flag2 = pow(m1,e2,n)
print('flag1= '+str(flag1))
print('flag2= '+str(flag2))
print('e1= ' +str(e1))
print('e2= '+str(e2))
print('n= '+str(n))
#flag1= 100156221476910922393504870369139942732039899485715044553913743347065883159136513788649486841774544271396690778274591792200052614669235485675534653358596366535073802301361391007325520975043321423979924560272762579823233787671688669418622502663507796640233829689484044539829008058686075845762979657345727814280
#flag2= 86203582128388484129915298832227259690596162850520078142152482846864345432564143608324463705492416009896246993950991615005717737886323630334871790740288140033046061512799892371429864110237909925611745163785768204802056985016447086450491884472899152778839120484475953828199840871689380584162839244393022471075
#e1= 3247473589
#e2= 3698409173
#n= 103606706829811720151309965777670519601112877713318435398103278099344725459597221064867089950867125892545997503531556048610968847926307322033117328614701432100084574953706259773711412853364463950703468142791390129671097834871371125741564434710151190962389213898270025272913761067078391308880995594218009110313题解：
两个flag，两个e，一个n
共模攻击喽？
from Crypto.Util.number import *
from gmpy2 import gmpy2
flag1= 100156221476910922393504870369139942732039899485715044553913743347065883159136513788649486841774544271396690778274591792200052614669235485675534653358596366535073802301361391007325520975043321423979924560272762579823233787671688669418622502663507796640233829689484044539829008058686075845762979657345727814280
flag2= 86203582128388484129915298832227259690596162850520078142152482846864345432564143608324463705492416009896246993950991615005717737886323630334871790740288140033046061512799892371429864110237909925611745163785768204802056985016447086450491884472899152778839120484475953828199840871689380584162839244393022471075
e1= 3247473589
e2= 3698409173
n= 103606706829811720151309965777670519601112877713318435398103278099344725459597221064867089950867125892545997503531556048610968847926307322033117328614701432100084574953706259773711412853364463950703468142791390129671097834871371125741564434710151190962389213898270025272913761067078391308880995594218009110313
r,x,y = gmpy2.gcdext(e1, e2)
m = pow(flag1,x,n) * pow(flag2,y,n) % n
if gmpy2.gcd(e1,e2) ==1:
print(long_to_bytes(m))
else :
m = gmpy2.iroot(m,r)[0]
print(long_to_bytes(m))
#b'NSSCTF{xxxxx******xxxxx}'你没看错，那个像是报错的flag真的是flag
[羊城杯 2021]Bigrsa
题目：
from Crypto.Util.number import *
from flag import *
n1 = 103835296409081751860770535514746586815395898427260334325680313648369132661057840680823295512236948953370895568419721331170834557812541468309298819497267746892814583806423027167382825479157951365823085639078738847647634406841331307035593810712914545347201619004253602692127370265833092082543067153606828049061
n2 = 115383198584677147487556014336448310721853841168758012445634182814180314480501828927160071015197089456042472185850893847370481817325868824076245290735749717384769661698895000176441497242371873981353689607711146852891551491168528799814311992471449640014501858763495472267168224015665906627382490565507927272073
e = 65537
m = bytes_to_long(flag)
c = pow(m, e, n1)
c = pow(c, e, n2)
print("c = %d" % c)
# output
# c = 60406168302768860804211220055708551816238816061772464557956985699400782163597251861675967909246187833328847989530950308053492202064477410641014045601986036822451416365957817685047102703301347664879870026582087365822433436251615243854347490600004857861059245403674349457345319269266645006969222744554974358264题解：
哦——
两个n哪！是个共享素数！n1与n2有个公因数。
import gmpy2
from Crypto.Util.number import *
n1 = 103835296409081751860770535514746586815395898427260334325680313648369132661057840680823295512236948953370895568419721331170834557812541468309298819497267746892814583806423027167382825479157951365823085639078738847647634406841331307035593810712914545347201619004253602692127370265833092082543067153606828049061
n2 = 115383198584677147487556014336448310721853841168758012445634182814180314480501828927160071015197089456042472185850893847370481817325868824076245290735749717384769661698895000176441497242371873981353689607711146852891551491168528799814311992471449640014501858763495472267168224015665906627382490565507927272073
e = 65537
c = 60406168302768860804211220055708551816238816061772464557956985699400782163597251861675967909246187833328847989530950308053492202064477410641014045601986036822451416365957817685047102703301347664879870026582087365822433436251615243854347490600004857861059245403674349457345319269266645006969222744554974358264
q=gmpy2.gcd(n1,n2)
q1=n1//q
q2=n2//q
d1=gmpy2.invert(e,(q1-1)*(q-1))
d2=gmpy2.invert(e,(q2-1)*(q-1))
m1=pow(c,d2,n2)
m2=pow(m1,d1,n1)
flag=long_to_bytes(m2)
print(flag)
#b'SangFor{qSccmm1WrgvIg2Uq_cZhmqNfEGTz2GV8}'
[SWPU 2020]happy
题目：
('c=', '0x7a7e031f14f6b6c3292d11a41161d2491ce8bcdc67ef1baa9eL')
('e=', '0x872a335')
#q + q*p^3 =1285367317452089980789441829580397855321901891350429414413655782431779727560841427444135440068248152908241981758331600586
#qp + q *p^2 = 1109691832903289208389283296592510864729403914873734836011311325874120780079555500202475594解密：
出现了q和p的关系式，用方程解，可以试试文心一言生成，自己人肉搜索也行。˙ϖ˙
import gmpy2
from Crypto.Util.number import *
n1 = 103835296409081751860770535514746586815395898427260334325680313648369132661057840680823295512236948953370895568419721331170834557812541468309298819497267746892814583806423027167382825479157951365823085639078738847647634406841331307035593810712914545347201619004253602692127370265833092082543067153606828049061
n2 = 115383198584677147487556014336448310721853841168758012445634182814180314480501828927160071015197089456042472185850893847370481817325868824076245290735749717384769661698895000176441497242371873981353689607711146852891551491168528799814311992471449640014501858763495472267168224015665906627382490565507927272073
e = 65537
c = 60406168302768860804211220055708551816238816061772464557956985699400782163597251861675967909246187833328847989530950308053492202064477410641014045601986036822451416365957817685047102703301347664879870026582087365822433436251615243854347490600004857861059245403674349457345319269266645006969222744554974358264
q=gmpy2.gcd(n1,n2)
q1=n1//q
q2=n2//q
d1=gmpy2.invert(e,(q1-1)*(q-1))
d2=gmpy2.invert(e,(q2-1)*(q-1))
m1=pow(c,d2,n2)
m2=pow(m1,d1,n1)
flag=long_to_bytes(m2)
print(flag)
#b'SangFor{qSccmm1WrgvIg2Uq_cZhmqNfEGTz2GV8}'
[AFCTF 2018]BASE
题目：
题目就不给了，它太多惹—_—
题解：
标题给了，是base呀！Σ(๑º㉨º๑ ) d(ŐдŐ๑) Σ(ﾟдﾟlll) 但是它真的好多好多，这比俄罗斯套娃还过分！！！
温馨提示：它太太太多次解密了，我就截图了开头和结尾。魔法棒不一定从头到尾都出现捏，自己稍微试试。
adctf{U_5h0u1d_Us3_T00l5}#魔法棒给我关了，我对图手打的flag，应该大概是对滴吧？¦•ˇ₃ˇ•｡)
[LitCTF 2023]你是我的关键词(Keyworld) (初级)
题目：
IFRURC{X0S_YP3_JX_HBXV0PA}题解：
题目已经给了加密方法啦！关键字加密，嗯，加密的密钥是‘你’：you
‘你’是关键词Keyworld，你真的，我哭死 ・ࡇ・ (看到flag悄咪咪说句：咦~什么肉麻的话)
LITCTF{YOU_AR3_MY_KEYWORD}
[安洵杯 2019]JustBase
题目：
VGhlIGdlb@xvZ#kgb@YgdGhlIEVhcnRoJ#Mgc#VyZmFjZSBpcyBkb@!pbmF)ZWQgYnkgdGhlIHBhcnRpY#VsYXIgcHJvcGVydGllcyBvZiB#YXRlci$gUHJlc@VudCBvbiBFYXJ)aCBpbiBzb@xpZCwgbGlxdWlkLCBhbmQgZ@FzZW(!cyBzdGF)ZXMsIHdhdGVyIGlzIGV$Y@VwdGlvbmFsbHkgcmVhY#RpdmUuIEl)IGRpc#NvbHZlcywgdHJhbnNwb#J)cywgYW%kIHByZWNpcGl)YXRlcyBtYW%%IGNoZW!pY@FsIGNvbXBvdW%kcyBhbmQgaXMgY@(uc#RhbnRseSBtb@RpZnlpbmcgdGhlIGZhY@Ugb@YgdGhlIEVhcnRoLiBFdmFwb#JhdGVkIGZyb@)gdGhlIG(jZWFucywgd@F)ZXIgdmFwb#IgZm(ybXMgY@xvdWRzLCBzb@!lIG(mIHdoaWNoIGFyZSB)cmFuc#BvcnRlZCBieSB#aW%kIG(@ZXIgdGhlIGNvbnRpbmVudHMuIENvbmRlbnNhdGlvbiBmcm(tIHRoZSBjbG(!ZHMgcHJvdmlkZXMgdGhlIGVzc@VudGlhbCBhZ@VudCBvZiBjb@%)aW%lbnRhbCBlcm(zaW(uOiByYWluLlRoZSByYXRlIGF)IHdoaWNoIGEgbW(sZWN!bGUgb@Ygd@F)ZXIgcGFzc@VzIHRob#VnaCB)aGUgY#ljbGUgaXMgbm()IHJhbmRvbQpBbmQgdGhlIGZsYWcgaXM^IENURnsyMi!RV)VSVFlVSU*tUExLSkhHRkRTLUFaWENWQk%NfQ==
题解：
扔进了Cyber，它报不出来，看了下，发现这段base多了')!@#$%^&*('，并且没有数字，那就进行码表替换（我薅走了Jeager的wp 中的运行脚本 ଲଇଉକ）
import base64
m = 'VGhlIGdlb@xvZ#kgb@YgdGhlIEVhcnRoJ#Mgc#VyZmFjZSBpcyBkb@!pbmF)ZWQgYnkgdGhlIHBhcnRpY#VsYXIgcHJvcGVydGllcyBvZiB#YXRlci$gUHJlc@VudCBvbiBFYXJ)aCBpbiBzb@xpZCwgbGlxdWlkLCBhbmQgZ@FzZW(!cyBzdGF)ZXMsIHdhdGVyIGlzIGV$Y@VwdGlvbmFsbHkgcmVhY#RpdmUuIEl)IGRpc#NvbHZlcywgdHJhbnNwb#J)cywgYW%kIHByZWNpcGl)YXRlcyBtYW%%IGNoZW!pY@FsIGNvbXBvdW%kcyBhbmQgaXMgY@(uc#RhbnRseSBtb@RpZnlpbmcgdGhlIGZhY@Ugb@YgdGhlIEVhcnRoLiBFdmFwb#JhdGVkIGZyb@)gdGhlIG(jZWFucywgd@F)ZXIgdmFwb#IgZm(ybXMgY@xvdWRzLCBzb@!lIG(mIHdoaWNoIGFyZSB)cmFuc#BvcnRlZCBieSB#aW%kIG(@ZXIgdGhlIGNvbnRpbmVudHMuIENvbmRlbnNhdGlvbiBmcm(tIHRoZSBjbG(!ZHMgcHJvdmlkZXMgdGhlIGVzc@VudGlhbCBhZ@VudCBvZiBjb@%)aW%lbnRhbCBlcm(zaW(uOiByYWluLlRoZSByYXRlIGF)IHdoaWNoIGEgbW(sZWN!bGUgb@Ygd@F)ZXIgcGFzc@VzIHRob#VnaCB)aGUgY#ljbGUgaXMgbm()IHJhbmRvbQpBbmQgdGhlIGZsYWcgaXM^IENURnsyMi!RV)VSVFlVSU*tUExLSkhHRkRTLUFaWENWQk%NfQ=='
c = ')!@#$%^&*('
for i in m:
for j in range(10):
if i == c[j]:
m = m.replace(i,str(j))
print(base64.b64decode(m))输出：b"The geology of the Earth's surface is dominated by the particular properties of water. Present on Earth in solid, liquid, and gaseous states, water is exceptionally reactive. It dissolves, transports, and precipitates many chemical compounds and is constantly modifying the face of the Earth. Evaporated from the oceans, water vapor forms clouds, some of which are transported by wind over the continents. Condensation from the clouds provides the essential agent of continental erosion: rain.The rate at which a molecule of water passes though the cycle is not random\nAnd the flag is: CTF{22-QWERTYUIO-PLKJHGFDS-AZXCVBNM}"
[LitCTF 2023]家人们！谁懂啊，RSA签到都不会 (初级)
题目：
from Crypto.Util.number import *
from secret import flag
m = bytes_to_long(flag)
p = getPrime(512)
q = getPrime(512)
e = 65537
n = p*q
c = pow(m,e,n)
print(f'p = {p}')
print(f'q = {q}')
print(f'c = {c}')
'''
p = 12567387145159119014524309071236701639759988903138784984758783651292440613056150667165602473478042486784826835732833001151645545259394365039352263846276073
q = 12716692565364681652614824033831497167911028027478195947187437474380470205859949692107216740030921664273595734808349540612759651241456765149114895216695451
c = 108691165922055382844520116328228845767222921196922506468663428855093343772017986225285637996980678749662049989519029385165514816621011058462841314243727826941569954125384522233795629521155389745713798246071907492365062512521474965012924607857440577856404307124237116387085337087671914959900909379028727767057
'''题目：
标准的大素数解码，这不能不会！不然，不然，不然我就再想想怎么做。
import gmpy2
from Crypto.Util.number import *
e = 65537
p = 12567387145159119014524309071236701639759988903138784984758783651292440613056150667165602473478042486784826835732833001151645545259394365039352263846276073
q = 12716692565364681652614824033831497167911028027478195947187437474380470205859949692107216740030921664273595734808349540612759651241456765149114895216695451
c = 108691165922055382844520116328228845767222921196922506468663428855093343772017986225285637996980678749662049989519029385165514816621011058462841314243727826941569954125384522233795629521155389745713798246071907492365062512521474965012924607857440577856404307124237116387085337087671914959900909379028727767057
n = p*q
phi = (p-1) * (q-1)
d = gmpy2.invert(e,phi)
m = pow(c,d,n)
print(long_to_bytes(m))
#b'LitCTF{it_is_easy_to_solve_question_when_you_know_p_and_q}'
[鹤城杯 2021]Crazy_Rsa_Tech
题目：
Why I need to send messages to so many people…
from Crypto.Util.number import *
from Crypto.Util.Padding import *
FLAG = bytes_to_long(pad(b"flag{??????}",64))
def init_key():
p, q = getPrime(512), getPrime(512)
n = p*q
e = 9
while(GCD((p-1)*(q-1),e)!=1):
p, q = getPrime(512), getPrime(512)
n = p*q
d = inverse(e,(p-1)*(q-1))
return n,e,d
n_list=list()
c_list=list()
for i in range(9):
N,e,d=init_key()
n_list.append(N)
c=pow(FLAG,e,N)
c_list.append(pow(FLAG,e,N))
assert(pow(c,d,N)==FLAG)
print("n_list:",n_list)
print("c_list:",c_list)
n_list: [71189786319102608575263218254922479901008514616376166401353025325668690465852130559783959409002115897148828732231478529655075366072137059589917001875303598680931962384468363842379833044123189276199264340224973914079447846845897807085694711541719515881377391200011269924562049643835131619086349617062034608799, 92503831027754984321994282254005318198418454777812045042619263533423066848097985191386666241913483806726751133691867010696758828674382946375162423033994046273252417389169779506788545647848951018539441971140081528915876529645525880324658212147388232683347292192795975558548712504744297104487514691170935149949, 100993952830138414466948640139083231443558390127247779484027818354177479632421980458019929149817002579508423291678953554090956334137167905685261724759487245658147039684536216616744746196651390112540237050493468689520465897258378216693418610879245129435268327315158194612110422630337395790254881602124839071919, 59138293747457431012165762343997972673625934330232909935732464725128776212729547237438509546925172847581735769773563840639187946741161318153031173864953372796950422229629824699580131369991913883136821374596762214064774480548532035315344368010507644630655604478651898097886873485265848973185431559958627423847, 66827868958054485359731420968595906328820823695638132426084478524423658597714990545142120448668257273436546456116147999073797943388584861050133103137697812149742551913704341990467090049650721713913812069904136198912314243175309387952328961054617877059134151915723594900209641163321839502908705301293546584147, 120940513339890268554625391482989102665030083707530690312336379356969219966820079510946652021721814016286307318930536030308296265425674637215009052078834615196224917417698019787514831973471113022781129000531459800329018133248426080717653298100515701379374786486337920294380753805825328119757649844054966712377, 72186594495190221129349814154999705524005203343018940547856004977368023856950836974465616291478257156860734574686154136925776069045232149725101769594505766718123155028300703627531567850035682448632166309129911061492630709698934310123778699316856399909549674138453085885820110724923723830686564968967391721281, 69105037583161467265649176715175579387938714721653281201847973223975467813529036844308693237404592381480367515044829190066606146105800243199497182114398931410844901178842049915914390117503986044951461783780327749665912369177733246873697481544777183820939967036346862056795919812693669387731294595126647751951, 76194219445824867986050004226602973283400885106636660263597964027139613163638212828932901192009131346530898961165310615466747046710743013409318156266326090650584190382130795884514074647833949281109675170830565650006906028402714868781834693473191228256626654011772428115359653448111208831188721505467497494581]
c_list: [62580922178008480377006528793506649089253164524883696044759651305970802215270721223149734532870729533611357047595181907404222690394917605617029675103788705320032707977225447998111744887898039756375876685711148857676502670812333076878964148863713993853526715855758799502735753454247721711366497722251078739585, 46186240819076690248235492196228128599822002268014359444368898414937734806009161030424589993541799877081745454934484263188270879142125136786221625234555265815513136730416539407710862948861531339065039071959576035606192732936477944770308784472646015244527805057990939765708793705044236665364664490419874206900, 85756449024868529058704599481168414715291172247059370174556127800630896693021701121075838517372920466708826412897794900729896389468152213884232173410022054605870785910461728567377769960823103334874807744107855490558726013068890632637193410610478514663078901021307258078678427928255699031215654693270240640198, 14388767329946097216670270960679686032536707277732968784379505904021622612991917314721678940833050736745004078559116326396233622519356703639737886289595860359630019239654690312132039876082685046329079266785042428947147658321799501605837784127004536996628492065409017175037161261039765340032473048737319069656, 1143736792108232890306863524988028098730927600066491485326214420279375304665896453544100447027809433141790331191324806205845009336228331138326163746853197990596700523328423791764843694671580875538251166864957646807184041817863314204516355683663859246677105132100377322669627893863885482167305919925159944839, 2978800921927631161807562509445310353414810029862911925227583943849942080514132963605492727604495513988707849133045851539412276254555228149742924149242124724864770049898278052042163392380895275970574317984638058768854065506927848951716677514095183559625442889028813635385408810698294574175092159389388091981, 16200944263352278316040095503540249310705602580329203494665614035841657418101517016718103326928336623132935178377208651067093136976383774189554806135146237406248538919915426183225265103769259990252162411307338473817114996409705345401251435268136647166395894099897737607312110866874944619080871831772376466376, 31551601425575677138046998360378916515711528548963089502535903329268089950335615563205720969393649713416910860593823506545030969355111753902391336139384464585775439245735448030993755229554555004154084649002801255396359097917380427525820249562148313977941413268787799534165652742114031759562268691233834820996, 25288164985739570635307839193110091356864302148147148153228604718807817833935053919412276187989509493755136905193728864674684139319708358686431424793278248263545370628718355096523088238513079652226028236137381367215156975121794485995030822902933639803569133458328681148758392333073624280222354763268512333515]题解：
英文提示……它不重要，我不会！（叉腰.jpg 理直气壮.jpg）
有众多的n与c，e=9，那么低指数加密e=9，然后对这群众多的n与c进行爆破，典型的低加密指数攻击，呀哈哈哈哈（反派笑声）
import gmpy2
from functools import reduce
from Crypto.Util.number import long_to_bytes
# 中国剩余定理
def CRT(cipher, n):
N = reduce(lambda x, y: x * y, (i for i in n))
result = 0
data = zip(cipher, n)
for ci, ni in data:
Ni = N // ni
di = gmpy2.invert(Ni, ni)
result += ci * Ni * di
return result % N, N
e = 9
n = [71189786319102608575263218254922479901008514616376166401353025325668690465852130559783959409002115897148828732231478529655075366072137059589917001875303598680931962384468363842379833044123189276199264340224973914079447846845897807085694711541719515881377391200011269924562049643835131619086349617062034608799, 92503831027754984321994282254005318198418454777812045042619263533423066848097985191386666241913483806726751133691867010696758828674382946375162423033994046273252417389169779506788545647848951018539441971140081528915876529645525880324658212147388232683347292192795975558548712504744297104487514691170935149949, 100993952830138414466948640139083231443558390127247779484027818354177479632421980458019929149817002579508423291678953554090956334137167905685261724759487245658147039684536216616744746196651390112540237050493468689520465897258378216693418610879245129435268327315158194612110422630337395790254881602124839071919, 59138293747457431012165762343997972673625934330232909935732464725128776212729547237438509546925172847581735769773563840639187946741161318153031173864953372796950422229629824699580131369991913883136821374596762214064774480548532035315344368010507644630655604478651898097886873485265848973185431559958627423847, 66827868958054485359731420968595906328820823695638132426084478524423658597714990545142120448668257273436546456116147999073797943388584861050133103137697812149742551913704341990467090049650721713913812069904136198912314243175309387952328961054617877059134151915723594900209641163321839502908705301293546584147, 120940513339890268554625391482989102665030083707530690312336379356969219966820079510946652021721814016286307318930536030308296265425674637215009052078834615196224917417698019787514831973471113022781129000531459800329018133248426080717653298100515701379374786486337920294380753805825328119757649844054966712377, 72186594495190221129349814154999705524005203343018940547856004977368023856950836974465616291478257156860734574686154136925776069045232149725101769594505766718123155028300703627531567850035682448632166309129911061492630709698934310123778699316856399909549674138453085885820110724923723830686564968967391721281, 69105037583161467265649176715175579387938714721653281201847973223975467813529036844308693237404592381480367515044829190066606146105800243199497182114398931410844901178842049915914390117503986044951461783780327749665912369177733246873697481544777183820939967036346862056795919812693669387731294595126647751951, 76194219445824867986050004226602973283400885106636660263597964027139613163638212828932901192009131346530898961165310615466747046710743013409318156266326090650584190382130795884514074647833949281109675170830565650006906028402714868781834693473191228256626654011772428115359653448111208831188721505467497494581]
c = [62580922178008480377006528793506649089253164524883696044759651305970802215270721223149734532870729533611357047595181907404222690394917605617029675103788705320032707977225447998111744887898039756375876685711148857676502670812333076878964148863713993853526715855758799502735753454247721711366497722251078739585, 46186240819076690248235492196228128599822002268014359444368898414937734806009161030424589993541799877081745454934484263188270879142125136786221625234555265815513136730416539407710862948861531339065039071959576035606192732936477944770308784472646015244527805057990939765708793705044236665364664490419874206900, 85756449024868529058704599481168414715291172247059370174556127800630896693021701121075838517372920466708826412897794900729896389468152213884232173410022054605870785910461728567377769960823103334874807744107855490558726013068890632637193410610478514663078901021307258078678427928255699031215654693270240640198, 14388767329946097216670270960679686032536707277732968784379505904021622612991917314721678940833050736745004078559116326396233622519356703639737886289595860359630019239654690312132039876082685046329079266785042428947147658321799501605837784127004536996628492065409017175037161261039765340032473048737319069656, 1143736792108232890306863524988028098730927600066491485326214420279375304665896453544100447027809433141790331191324806205845009336228331138326163746853197990596700523328423791764843694671580875538251166864957646807184041817863314204516355683663859246677105132100377322669627893863885482167305919925159944839, 2978800921927631161807562509445310353414810029862911925227583943849942080514132963605492727604495513988707849133045851539412276254555228149742924149242124724864770049898278052042163392380895275970574317984638058768854065506927848951716677514095183559625442889028813635385408810698294574175092159389388091981, 16200944263352278316040095503540249310705602580329203494665614035841657418101517016718103326928336623132935178377208651067093136976383774189554806135146237406248538919915426183225265103769259990252162411307338473817114996409705345401251435268136647166395894099897737607312110866874944619080871831772376466376, 31551601425575677138046998360378916515711528548963089502535903329268089950335615563205720969393649713416910860593823506545030969355111753902391336139384464585775439245735448030993755229554555004154084649002801255396359097917380427525820249562148313977941413268787799534165652742114031759562268691233834820996, 25288164985739570635307839193110091356864302148147148153228604718807817833935053919412276187989509493755136905193728864674684139319708358686431424793278248263545370628718355096523088238513079652226028236137381367215156975121794485995030822902933639803569133458328681148758392333073624280222354763268512333515]
x, N = CRT(c, n)
m = gmpy2.iroot(gmpy2.mpz(x), e)[0]
print(m)
print(long_to_bytes(m))
#b'flag{H0w_Fun_13_HAstads_broadca5t_AtTack!}\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16\x16'
[AFCTF 2018]你能看出这是什么加密么
题目：
p=0x928fb6aa9d813b6c3270131818a7c54edb18e3806942b88670106c1821e0326364194a8c49392849432b37632f0abe3f3c52e909b939c91c50e41a7b8cd00c67d6743b4f
q=0xec301417ccdffa679a8dcc4027dd0d75baf9d441625ed8930472165717f4732884c33f25d4ee6a6c9ae6c44aedad039b0b72cf42cab7f80d32b74061
e=0x10001
c=0x70c9133e1647e95c3cb99bd998a9028b5bf492929725a9e8e6d2e277fa0f37205580b196e5f121a2e83bc80a8204c99f5036a07c8cf6f96c420369b4161d2654a7eccbdaf583204b645e137b3bd15c5ce865298416fd5831cba0d947113ed5be5426b708b89451934d11f9aed9085b48b729449e461ff0863552149b965e22b6 题解：
大素数解码
import gmpy2
from Crypto.Util.number import *
p=0x928fb6aa9d813b6c3270131818a7c54edb18e3806942b88670106c1821e0326364194a8c49392849432b37632f0abe3f3c52e909b939c91c50e41a7b8cd00c67d6743b4f
q=0xec301417ccdffa679a8dcc4027dd0d75baf9d441625ed8930472165717f4732884c33f25d4ee6a6c9ae6c44aedad039b0b72cf42cab7f80d32b74061
e=0x10001
c=0x70c9133e1647e95c3cb99bd998a9028b5bf492929725a9e8e6d2e277fa0f37205580b196e5f121a2e83bc80a8204c99f5036a07c8cf6f96c420369b4161d2654a7eccbdaf583204b645e137b3bd15c5ce865298416fd5831cba0d947113ed5be5426b708b89451934d11f9aed9085b48b729449e461ff0863552149b965e22b6
n = p*q
phi = (p-1) * (q-1)
d = gmpy2.invert(e,phi)
m = pow(c,d,n)
print(long_to_bytes(m))
#afctf{R54_|5_$0_$imp13}
[SWPUCTF 2021 新生赛]crypto9
题目：
AKKPLX{qv5x0021-7n8w-wr05-x25w-7882ntu5q984}
脚本给你了，去解吧
letter_list = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' # 字母表
# 根据输入的key生成key列表
def Get_KeyList(key):
key_list = []
for ch in key:
key_list.append(ord(ch.upper()) - 65)
return key_list
# 加密函数
def Encrypt(plaintext, key_list):
ciphertext = ""
i = 0
for ch in plaintext: # 遍历明文
if 0 == i % len(key_list):
i = 0
if ch.isalpha(): # 明文是否为字母,如果是,则判断大小写,分别进行加密
if ch.isupper():
ciphertext += letter_list[(ord(ch) - 65 + key_list[i]) % 26]
i += 1
else:
ciphertext += letter_list[(ord(ch) - 97 + key_list[i]) % 26].lower()
i += 1
else: # 如果密文不为字母,直接添加到密文字符串里
ciphertext += ch
return ciphertext
# 解密函数
def Decrypt(ciphertext, key):
plaintext = ""
i = 0
for ch in ciphertext: # 遍历密文
if 0 == i % len(key_list):
i = 0
if ch.isalpha(): # 密文为否为字母,如果是,则判断大小写,分别进行解密
if ch.isupper():
plaintext += letter_list[(ord(ch) - 65 - key_list[i]) % 26]
i += 1
else:
plaintext += letter_list[(ord(ch) - 97 - key_list[i]) % 26].lower()
i += 1
else: # 如果密文不为字母,直接添加到明文字符串里
plaintext += ch
return plaintext
if __name__ == '__main__':
print("加密请按D,解密请按E:")
user_input = input();
while (user_input != 'D' and user_input != 'E'): # 输入合法性判断
print("输入有误!请重新输入:")
user_input = input()
print("请输入密钥:")
key = input()
while (False == key.isalpha()): # 输入合法性判断
print("输入有误!密钥为字母,请重新输入:")
key = input()
key_list = Get_KeyList(key)
if user_input == 'D':
# 加密
print("请输入明文:")
plaintext = input()
ciphertext = Encrypt(plaintext, key_list)
print("密文为:\n%s" % ciphertext)
else:
# 解密
print("请输入密文:")
ciphertext = input()
plaintext = Decrypt(ciphertext, key_list)
print("明文为:\n%s" % plaintext)题目提示：
密钥长度为3
题解：
哦！好长的代码！不过功能很明确，试出密钥解密。
密钥是：NSS（好颠啊~）
加密请按D,解密请按E:
E
请输入密钥:
NSS
请输入密文:
AKKPLX{qv5x0021-7n8w-wr05-x25w-7882ntu5q984}
明文为:
NSSCTF{dd5f0021-7a8e-ee05-f25e-7882abc5d984}
[NSSCTF 2022 Spring Recruit]classic
题目：
UZZJAM{UIXETGR7TMWD42SKTCWEP4AG_mhrlmshnayfihzl}题解：
一眼命中，凯撒。
import base64
def caesar(ciphertext, shift=ord('U')-ord('N')):
str_list = list(ciphertext)
i = 0
while i < len(ciphertext):
if str_list[i].isalpha():
a = "A" if str_list[i].isupper() else "a"
str_list[i] = chr((ord(str_list[i]) - ord(a) - shift) % 26 + ord(a))
i += 1
return ''.join(str_list)
str='UZZJAM{UIXETGR7TMWD42SKTCWEP4AG_mhrlmshnayfihzl}'
kstr = caesar(str)
print(kstr)
#NSSCTF{NBQXMZK7MFPW42LDMVPXI4TZ_fakeflagtrybase}
[SWPUCTF 2021 新生赛]crypto4
题目：
from gmpy2 import *
from Crypto.Util.number import *
flag = '**********'
p = getPrime(512)
q = next_prime(p)
m1 = bytes_to_long(bytes(flag.encode()))
e = 0x10001
n = p*q
flag1 = pow(m1,e,n)
print('flag= '+str(flag1))
print('n= '+str(n))
flag= 10227915341268619536932290456122384969242151167487654201363877568935534996454863939953106193665663567559506242151019201314446286458150141991211233219320700112533775367958964780047682920839507351492644735811096995884754664899221842470772096509258104067131614630939533042322095150722344048082688772981180270243
n= 52147017298260357180329101776864095134806848020663558064141648200366079331962132411967917697877875277103045755972006084078559453777291403087575061382674872573336431876500128247133861957730154418461680506403680189755399752882558438393107151815794295272358955300914752523377417192504702798450787430403387076153题解：
大素数解码，你懂得。别被那个flag吓到啦——
import gmpy2
from Crypto.Util.number import *
flag = 10227915341268619536932290456122384969242151167487654201363877568935534996454863939953106193665663567559506242151019201314446286458150141991211233219320700112533775367958964780047682920839507351492644735811096995884754664899221842470772096509258104067131614630939533042322095150722344048082688772981180270243
n = 52147017298260357180329101776864095134806848020663558064141648200366079331962132411967917697877875277103045755972006084078559453777291403087575061382674872573336431876500128247133861957730154418461680506403680189755399752882558438393107151815794295272358955300914752523377417192504702798450787430403387076153
q = 7221289171488727827673517139597844534869368289455419695964957239047692699919030405800116133805855968123601433247022090070114331842771417566928809956044421
p = 7221289171488727827673517139597844534869368289455419695964957239047692699919030405800116133805855968123601433247022090070114331842771417566928809956045093
e = 0x10001
phi = (p-1) * (q-1)
d = gmpy2.invert(e,phi)
m = pow(flag,d,n)
print(long_to_bytes(m))
#b'NSSCTF{no_why}'
[LitCTF 2023]Is this only base?
题目：
SWZxWl=F=DQef0hlEiSUIVh9ESCcMFS9NF2NXFzM
今年是本世纪的第23年呢题解：
Is this only base？
base？不止哦，一般base会在最后有等号，看来是移位了，栅栏加密，key是23呢。
栅栏解码：SWZxWlFDe0liUV9ScF9FNFMzX2NSMCEhISEhfQ==
base64解密：IfqZQC{IbQ_Rp_E4S3_cR0!!!!!}
[SWPUCTF 2021 新生赛]crypto5
题目：
flag= 25166751653530941364839663846806543387720865339263370907985655775152187319464715737116599171477207047430065345882626259880756839094179627032623895330242655333
n= 134109481482703713214838023035418052567000870587160796935708584694132507394211363652420160931185332280406437290210512090663977634730864032370977407179731940068634536079284528020739988665713200815021342700369922518406968356455736393738946128013973643235228327971170711979683931964854563904980669850660628561419题解：
只给了c和n，小明文爆破（emo中）
from gmpy2 import iroot
from Crypto.Util.number import long_to_bytes
c = 25166751653530941364839663846806543387720865339263370907985655775152187319464715737116599171477207047430065345882626259880756839094179627032623895330242655333
n = 134109481482703713214838023035418052567000870587160796935708584694132507394211363652420160931185332280406437290210512090663977634730864032370977407179731940068634536079284528020739988665713200815021342700369922518406968356455736393738946128013973643235228327971170711979683931964854563904980669850660628561419
e = 0
while 1:
e = e + 1
if(iroot(c,e)[1]==True):
m = iroot(c,e)[0]
flag = long_to_bytes(m)
if(flag[:6]==b'NSSCTF'):
print(long_to_bytes(m))
break
print(flag)
#b'NSSCTF{because_i_like}'
[AFCTF 2018]Vigenère
题目：
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Wr, zqmrrlqjy, phr Xnxrrygfnwtvbna os zjw ojigkm Atnzgk ib Azkaqcn, op Yyjeegu Koamtwmo, Afynubykf, sjlenrrvg gu vzy Oucxnue Wafyy kf gnn eoerf xin tuk amcgovmxa os udz iazgfneoay, mw, ia zjw Hwmr, gwl bl Gwlbkrvzh wf gng yikd Ckxxlr uh lbasr Ixtoaogk, mklrswty caddcoh ntm leprcjy, Phnz cpefk wfcpeq Ixtoaogk une, ntm wf Eoizn kutnc bo ok Hjya aaj Rvdrvgfxang Ycitry, vzup tukh irr Gdkihvrj ozoz gnd Uhlrmrinpk vg nde Oxrbifn Ejisn, ntm bhnz cdf loyocqcnr eghjepzrwn okvoyan gnnu aaj vzy Otnzn wf Txgsn Xrvzjqn, vy cfx kutnc bo ok vgnwlye mqsfunnyz; aaj cpag gu Xlae ntm Qnqkrwhzeaz Bbagku, lbay ugem fhrn Hisee zx teie Ysl, yoaiucdr Vgswa, cbtczapz Cdfeaaina, efzctfesu Ixumrxew, ujd gu mw ayr qlbar Nica aaj Vzcjgf cqqcu Opvyleajnvt Fzclyo mne xn rvmjl xk. — Aaj owr gng kolpbxc wf gnkk Xacygaitvup, ocph n lrzm eknaujcr uw bhr vtgnacgoxv os Jkncje Cxxdiqkpuy, se zaccayra hfadtk cw enij gndee udz Lvbgk, iqr Suabuaku, shz ohx bicekf Zijoe.#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main()
{
freopen("flag.txt","r",stdin);
freopen("flag_encode.txt","w",stdout);
char key[] = /*SADLY SAYING! Key is eaten by Monster!*/;
int len = strlen(key);
char ch;
int index = 0;
while((ch = getchar()) != EOF){
if(ch>='a'&&ch<='z'){
putchar((ch-'a'+key[index%len]-'a')%26+'a');
++index;
}else if(ch>='A'&&ch<='Z'){
putchar((ch-'A'+key[index%len]-'a')%26+'A');
++index;
}else{
putchar(ch);
}
}
return 0;
}题解：
不用看了，维吉尼亚爆破就行。
爆破结果：
When in the Course of human events it becomes necessary for one people to dissolve the political bands which have connected them with another and to assume among the powers of the earth, the separate and equal station to which the Laws of Nature and of Nature's God entitle them, a decent respect to the opinions of mankind requires that they should declare the causes which impel them to the separation.
We hold these truths to be self-evident, that all men are created equal, that they are endowed by their Creator with certain unalienable Rights, that among these are Life, Liberty and the pursuit of Happiness. — That to secure these rights, Governments are instituted among Men, deriving their just powers from the consent of the governed, — That whenever any Form of Government becomes destructive of these ends, it is the Right of the People to alter or to abolish it, and to institute new Government, laying its foundation on such principles and organizing its powers in such form, as to them shall seem most likely to effect their Safety and Happiness. Prudence, indeed, will dictate that Governments long established should not be changed for light and transient causes; and accordingly all experience hath shewn that mankind are more disposed to suffer, while evils are sufferable than to right themselves by abolishing the forms to which they are accustomed. But when a long train of abuses and usurpations, pursuing invariably the same Object evinces a design to reduce them under absolute Despotism, it is their right, it is their duty, to throw off such Government, and to provide new Guards for their future security. — Such has been the patient sufferance of these Colonies; and such is now the necessity which constrains them to alter their former Systems of Government. The history of the present King of Great Britain is a history of repeated injuries and usurpations, all having in direct object the establishment of an absolute Tyranny over these States. To prove this, let Facts be submitted to a candid world.
He has refused his Assent to Laws, the most wholesome and necessary for the public good.
He has forbidden his Governors to pass Laws of immediate and pressing importance, unless suspended in their operation till his Assent should be obtained; and when so suspended, he has utterly neglected to attend to them.
He has refused to pass other Laws for the accommodation of large districts of people, unless those people would relinquish the right of Representation in the Legislature, a right inestimable to them and formidable to tyrants only.
He has called together legislative bodies at places unusual, uncomfortable, and distant from the depository of their Public Records, for the sole purpose of fatiguing them into compliance with his measures.
He has dissolved Representative Houses repeatedly, for opposing with manly firmness his invasions on the rights of the people.
He has refused for a long time, after such dissolutions, to cause others to be elected, whereby the Legislative Powers, incapable of Annihilation, have returned to the People at large for their exercise; the State remaining in the mean time exposed to all the dangers of invasion from without, and convulsions within.
He has endeavoured to prevent the population of these States; for that purpose obstructing the Laws for Naturalization of Foreigners; refusing to pass others to encourage their migrations hither, and raising the conditions of new Appropriations of Lands.
He has obstructed the Administration of Justice by refusing his Assent to Laws for establishing Judiciary Powers.
He has made Judges dependent on his Will alone for the tenure of their offices, and the amount and payment of their salaries.
flag is afctf{Whooooooo_U_Gotcha!}
He has erected a multitude of New Offices, and sent hither swarms of Officers to harass our people and eat out their substance.
He has kept among us, in times of peace, Standing Armies without the Consent of our legislatures.
He has affected to render the Military independent of and superior to the Civil Power.
He has combined with others to subject us to a jurisdiction foreign to our constitution, and unacknowledged by our laws; giving his Assent to their Acts of pretended Legislation:
For quartering large bodies of armed troops among us:
For protecting them, by a mock Trial from punishment for any Murders which they should commit on the Inhabitants of these States:
For cutting off our Trade with all parts of the world:
For imposing Taxes on us without our Consent:
For depriving us in many cases, of the benefit of Trial by Jury:
For transporting us beyond Seas to be tried for pretended offences:
For abolishing the free System of English Laws in a neighbouring Province, establishing therein an Arbitrary government, and enlarging its Boundaries so as to render it at once an example and fit instrument for introducing the same absolute rule into these Colonies
For taking away our Charters, abolishing our most valuable Laws and altering fundamentally the Forms of our Governments:
For suspending our own Legislatures, and declaring themselves invested with power to legislate for us in all cases whatsoever.
He has abdicated Government here, by declaring us out of his Protection and waging War against us.
He has plundered our seas, ravaged our coasts, burnt our towns, and destroyed the lives of our people.
He is at this time transporting large Armies of foreign Mercenaries to compleat the works of death, desolation, and tyranny, already begun with circumstances of Cruelty & Perfidy scarcely paralleled in the most barbarous ages, and totally unworthy the Head of a civilized nation.
He has constrained our fellow Citizens taken Captive on the high Seas to bear Arms against their Country, to become the executioners of their friends and Brethren, or to fall themselves by their Hands.
He has excited domestic insurrections amongst us, and has endeavoured to bring on the inhabitants of our frontiers, the merciless Indian Savages whose known rule of warfare, is an undistinguished destruction of all ages, sexes and conditions.
In every stage of these Oppressions We have Petitioned for Redress in the most humble terms: Our repeated Petitions have been answered only by repeated injury. A Prince, whose character is thus marked by every act which may define a Tyrant, is unfit to be the ruler of a free people.
Nor have We been wanting in attentions to our British brethren. We have warned them from time to time of attempts by their legislature to extend an unwarrantable jurisdiction over us. We have reminded them of the circumstances of our emigration and settlement here. We have appealed to their native justice and magnanimity, and we have conjured them by the ties of our common kindred to disavow these usurpations, which would inevitably interrupt our connections and correspondence. They too have been deaf to the voice of justice and of consanguinity. We must, therefore, acquiesce in the necessity, which denounces our Separation, and hold them, as we hold the rest of mankind, Enemies in War, in Peace Friends.
We, therefore, the Representatives of the united States of America, in General Congress, Assembled, appealing to the Supreme Judge of the world for the rectitude of our intentions, do, in the Name, and by Authority of the good People of these Colonies, solemnly publish and declare, That these united Colonies are, and of Right ought to be Free and Independent States, that they are Absolved from all Allegiance to the British Crown, and that all political connection between them and the State of Great Britain, is and ought to be totally dissolved; and that as Free and Independent States, they have full Power to levy War, conclude Peace, contract Alliances, establish Commerce, and to do all other Acts and Things which Independent States may of right do. — And for the support of this Declaration, with a firm reliance on the protection of Divine Providence, we mutually pledge to each other our Lives, our Fortunes, and our sacred Honor. 中间那句话：flag is afctf{Whooooooo_U_Gotcha!}
[BJDCTF 2020]rsa
题目：
from Crypto.Util.number import getPrime,bytes_to_long
flag=open("flag","rb").read()
p=getPrime(1024)
q=getPrime(1024)
assert(e<100000)
n=p*q
m=bytes_to_long(flag)
c=pow(m,e,n)
print c,n
print pow(294,e,n)
p=getPrime(1024)
n=p*q
m=bytes_to_long("BJD"*32)
c=pow(m,e,n)
print c,n
'''
output:
12641635617803746150332232646354596292707861480200207537199141183624438303757120570096741248020236666965755798009656547738616399025300123043766255518596149348930444599820675230046423373053051631932557230849083426859490183732303751744004874183062594856870318614289991675980063548316499486908923209627563871554875612702079100567018698992935818206109087568166097392314105717555482926141030505639571708876213167112187962584484065321545727594135175369233925922507794999607323536976824183162923385005669930403448853465141405846835919842908469787547341752365471892495204307644586161393228776042015534147913888338316244169120 13508774104460209743306714034546704137247627344981133461801953479736017021401725818808462898375994767375627749494839671944543822403059978073813122441407612530658168942987820256786583006947001711749230193542370570950705530167921702835627122401475251039000775017381633900222474727396823708695063136246115652622259769634591309421761269548260984426148824641285010730983215377509255011298737827621611158032976420011662547854515610597955628898073569684158225678333474543920326532893446849808112837476684390030976472053905069855522297850688026960701186543428139843783907624317274796926248829543413464754127208843070331063037
381631268825806469518166370387352035475775677163615730759454343913563615970881967332407709901235637718936184198930226303761876517101208677107311006065728014220477966000620964056616058676999878976943319063836649085085377577273214792371548775204594097887078898598463892440141577974544939268247818937936607013100808169758675042264568547764031628431414727922168580998494695800403043312406643527637667466318473669542326169218665366423043579003388486634167642663495896607282155808331902351188500197960905672207046579647052764579411814305689137519860880916467272056778641442758940135016400808740387144508156358067955215018
979153370552535153498477459720877329811204688208387543826122582132404214848454954722487086658061408795223805022202997613522014736983452121073860054851302343517756732701026667062765906277626879215457936330799698812755973057557620930172778859116538571207100424990838508255127616637334499680058645411786925302368790414768248611809358160197554369255458675450109457987698749584630551177577492043403656419968285163536823819817573531356497236154342689914525321673807925458651854768512396355389740863270148775362744448115581639629326362342160548500035000156097215446881251055505465713854173913142040976382500435185442521721 12806210903061368369054309575159360374022344774547459345216907128193957592938071815865954073287532545947370671838372144806539753829484356064919357285623305209600680570975224639214396805124350862772159272362778768036844634760917612708721787320159318432456050806227784435091161119982613987303255995543165395426658059462110056431392517548717447898084915167661172362984251201688639469652283452307712821398857016487590794996544468826705600332208535201443322267298747117528882985955375246424812616478327182399461709978893464093245135530135430007842223389360212803439850867615121148050034887767584693608776323252233254261047
'''题解：
只需要前半段的，pow（294，e，n）可以求出e为多少，在进行正常解密
from Crypto.Util.number import *
import gmpy2
c1 = 12641635617803746150332232646354596292707861480200207537199141183624438303757120570096741248020236666965755798009656547738616399025300123043766255518596149348930444599820675230046423373053051631932557230849083426859490183732303751744004874183062594856870318614289991675980063548316499486908923209627563871554875612702079100567018698992935818206109087568166097392314105717555482926141030505639571708876213167112187962584484065321545727594135175369233925922507794999607323536976824183162923385005669930403448853465141405846835919842908469787547341752365471892495204307644586161393228776042015534147913888338316244169120
n1 = 13508774104460209743306714034546704137247627344981133461801953479736017021401725818808462898375994767375627749494839671944543822403059978073813122441407612530658168942987820256786583006947001711749230193542370570950705530167921702835627122401475251039000775017381633900222474727396823708695063136246115652622259769634591309421761269548260984426148824641285010730983215377509255011298737827621611158032976420011662547854515610597955628898073569684158225678333474543920326532893446849808112837476684390030976472053905069855522297850688026960701186543428139843783907624317274796926248829543413464754127208843070331063037
f = 381631268825806469518166370387352035475775677163615730759454343913563615970881967332407709901235637718936184198930226303761876517101208677107311006065728014220477966000620964056616058676999878976943319063836649085085377577273214792371548775204594097887078898598463892440141577974544939268247818937936607013100808169758675042264568547764031628431414727922168580998494695800403043312406643527637667466318473669542326169218665366423043579003388486634167642663495896607282155808331902351188500197960905672207046579647052764579411814305689137519860880916467272056778641442758940135016400808740387144508156358067955215018
c2 = 979153370552535153498477459720877329811204688208387543826122582132404214848454954722487086658061408795223805022202997613522014736983452121073860054851302343517756732701026667062765906277626879215457936330799698812755973057557620930172778859116538571207100424990838508255127616637334499680058645411786925302368790414768248611809358160197554369255458675450109457987698749584630551177577492043403656419968285163536823819817573531356497236154342689914525321673807925458651854768512396355389740863270148775362744448115581639629326362342160548500035000156097215446881251055505465713854173913142040976382500435185442521721
n2 = 12806210903061368369054309575159360374022344774547459345216907128193957592938071815865954073287532545947370671838372144806539753829484356064919357285623305209600680570975224639214396805124350862772159272362778768036844634760917612708721787320159318432456050806227784435091161119982613987303255995543165395426658059462110056431392517548717447898084915167661172362984251201688639469652283452307712821398857016487590794996544468826705600332208535201443322267298747117528882985955375246424812616478327182399461709978893464093245135530135430007842223389360212803439850867615121148050034887767584693608776323252233254261047
q = gmpy2.gcd(n1,n2)
p = n1 // q
phi = (q-1)*(p-1)
for e in range(1,100000):
if gmpy2.gcd(e,phi)==1:
k = pow(294,e,n1)
if(k==f):
d = gmpy2.invert(e,phi)
m = pow(c1,d,n1)
print(long_to_bytes(m))
#b'BJD{p_is_common_divisor}'