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NSS Crypto-03

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本文字数： 76k 阅读时长 ≈ 1:09

这是base??

题目

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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

分析

刚开始以为是python字典 然后把这个进行替换 让ai去解密 但是不行

然后提取出 字典所有值，放到CyberChef中 base64替换 解密即可 BJD{D0_Y0u_kNoW_Th1s_b4se_map}

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base64_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: '='
}

all_values_str = ''.join(base64_dict.values())

print(all_values_str)

# JKLMNOxyUVzABCDEFGH789PQIabcdefghijklmWXYZ0123456RSTnopqrstuvw+/=

[NSSCTF 2022 Spring Recruit]rrrsssaaa

题目

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from Crypto.Util.number import *
import gmpy2
from functools import reduce
from secret import flag

p = getPrime(1024)
i = 0
while True:
    r = p * 5 + i
    if isPrime(r):
        i = 0
        break
    else:
        i += 1
while True:
    q = p * 10 + i
    if isPrime(q):
        break
    else:
        i += 1

n = p * q * r
e = 65537
c = pow(bytes_to_long(flag.encode()), e, n)
print('c=' + str(c))
print('p3=' + str(pow(p, 3, n)))
print('q3=' + str(pow(q, 3, n)))
print('r3=' + str(pow(r, 3, n)))
# n = 44571911854174527304485400947383944661319242813524818888269963870884859557542264803774212076803157466539443358890313286282067621989609252352994203884813364011659788234277369629312571477760818634118449563652776213438461157699447304292906151410018017960605868035069246651843561595572415595568705784173761441087845248621463389786351743200696279604003824362262237505386409700329605140703782099240992158439201646344692107831931849079888757310523663310273856448713786678014221779214444879454790399990056124051739535141631564534546955444505648933134838799753362350266884682987713823886338789502396879543498267617432600351655901149380496067582237899323865338094444822339890783781705936546257971766978222763417870606459677496796373799679580683317833001077683871698246143179166277232084089913202832193540581401453311842960318036078745448783370048914350299341586452159634173821890439194014264891549345881324015485910286021846721593668473
# c = 11212699652154912414419576042130573737460880175860430868241856564678915039929479534373946033032215673944727767507831028500814261134142245577246925294110977629353584372842303558820509861245550773062016272543030477733653059813274587939179134498599049035104941393508776333632172797303569396612594631646093552388772109708942113683783815011735472088985078464550997064595366458370527490791625688389950370254858619018250060982532954113416688720602160768503752410505420577683484807166966007396618297253478916176712265476128018816694458551219452105277131141962052020824990732525958682439071443399050470856132519918853636638476540689226313542250551212688215822543717035669764276377536087788514506366740244284790716170847347643593400673746020474777085815046098314460862593936684624708574116108322520985637474375038848494466480630236867228454838428542365166285156741433845949358227546683144341695680712263215773807461091898003011630162481
# p3 = 891438237083490546089708018947678893226384856270496377765399277417697191150845296075484241536063149330788867177806265725641352439792185047059884077696267280233195764685547392586251429555216372682368991273055524268769223153988946085858123028200360359212117360701384933036871231911448311911374115683475228820531478240539549424647154342506853356292956506486091063660095505979187297020928573605860329881982122478494944846700224611808246427660214535971723459345029873385956677292979041143593821672034573140001092625650099257402018634684516092489263998517027205660003413512870074652126328536906790020794659204007921147300771594986038917179253827432120018857213350120695302091483756021206199805521083496979628811676116525321724267588515105188480380865374667274442027086789352802613365511142499668793725505110436809024171752137883546327359935102833441492430652019931999144063825010678766130335038975376834579129516127516820037383067
# q3 = 44571911854174527304485400947383944661319242813524818888269963870884859557542264803774212076803157466539443358890313286282067621989609252352994203884813364011659788234277369629312571477760818634118449563652776213438461157699447304292906151410018017960605868035069246651843561595572415595568705784173761440671033435053531971051698504592848580356684103015611323747688216493729331061402058160819388999663041629882482138465124920580049057123360829897432472221079140360215664537272316836767039948368780837985855835419681893347839311156887660438769948501100287062738217966360434291369179859862550767272985972263442512061098317471708987686120577904202391381040801620069987103931326500146536990700234262413595295698193570184681785854277656410199477649697026112650581343325348837547631237627207304757407395388155701341044939408589591213693329516396531103489233367665983149963665364824119870832353269655933102900004362236232825539480774
# r3 = 22285955927087263652242700473691972330659621406762409444134981935442429778771132401887106038401578733269721679445156643141033810994804626176497101942406682005829894117138684814656285738880409317059224781826388106719230578849723652146453075705009008980302934017534623325921780797786207797784352892086880720749202442492937918619992591614713131681306874944356693778359565004415437554407990089293135634916859631279984463829118336826115430997439527110961309956466956650522900331263720500751112297418506140413317489683875995326726992533904683800042127871963320754241310699432792081707870167598822650064976439270556418985242630368723264289700246406905189810458354474959276748887369363592834205660349184660073395182450526542246354364903399132116153732074081050985584216815493617906868615192465631416955706457835185743023758573279838341229835613609332206338401219168119635681832981552328638132500079074010106995297184587143613134093145

分析

p = getPrime(1024)

r = p * 5 + i 判断r是否为素数，如果是素数 i清零 否则j++ 直到r为素数

q = p * 10 + i q也同理r

n = p * q * r e = 65537 c = pow(bytes_to_long(flag.encode()), e, n)

求m

p3 q3 r3 均为 X ^3 mod n 优先考虑低加密指数攻击

C 三 m^e mod n C = k*n + m^e , k∈Z

p3 = pow(p, 3, n) q3 = pow(q, 3, n) r3 = pow(r, 3, n)

c = pow(m,e,n)

然后低加密指数求出对应的 p q r

n = q * p * r

WP

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import gmpy2
import libnum
n = 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
c = 0x10652cdfaa6b63f6d7bd1109da08181e500e5643f5b240a9024bfa84d5f2cac9310562978347bb232d63e7289283871efab83d84ff5a7b64a94a79d34cfbd4ef121723ba1f663e514f83f6f01492b4e13e1bb4296d96ea5a353d3bf2edd2f449c03c4a3e995237985a596908adc741f32365
e = 3
k = 0
while 1:
    res = gmpy2.iroot(c+k*n,e)
    if(res[1] == True):
        print(libnum.n2s(int(res[0])))
        break
    k = k+1

用这个脚本 带入 q3 p3 r3 可以求出 p q ，r没求出来。

但是知道 p q n 也就可以求出r来

r 可根据原题中p去求出来

求出后就正常求phi d m 即可得到flag

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import gmpy2
from Crypto.Util.number import *

n = 44571911854174527304485400947383944661319242813524818888269963870884859557542264803774212076803157466539443358890313286282067621989609252352994203884813364011659788234277369629312571477760818634118449563652776213438461157699447304292906151410018017960605868035069246651843561595572415595568705784173761441087845248621463389786351743200696279604003824362262237505386409700329605140703782099240992158439201646344692107831931849079888757310523663310273856448713786678014221779214444879454790399990056124051739535141631564534546955444505648933134838799753362350266884682987713823886338789502396879543498267617432600351655901149380496067582237899323865338094444822339890783781705936546257971766978222763417870606459677496796373799679580683317833001077683871698246143179166277232084089913202832193540581401453311842960318036078745448783370048914350299341586452159634173821890439194014264891549345881324015485910286021846721593668473
c = 11212699652154912414419576042130573737460880175860430868241856564678915039929479534373946033032215673944727767507831028500814261134142245577246925294110977629353584372842303558820509861245550773062016272543030477733653059813274587939179134498599049035104941393508776333632172797303569396612594631646093552388772109708942113683783815011735472088985078464550997064595366458370527490791625688389950370254858619018250060982532954113416688720602160768503752410505420577683484807166966007396618297253478916176712265476128018816694458551219452105277131141962052020824990732525958682439071443399050470856132519918853636638476540689226313542250551212688215822543717035669764276377536087788514506366740244284790716170847347643593400673746020474777085815046098314460862593936684624708574116108322520985637474375038848494466480630236867228454838428542365166285156741433845949358227546683144341695680712263215773807461091898003011630162481
p3 = 891438237083490546089708018947678893226384856270496377765399277417697191150845296075484241536063149330788867177806265725641352439792185047059884077696267280233195764685547392586251429555216372682368991273055524268769223153988946085858123028200360359212117360701384933036871231911448311911374115683475228820531478240539549424647154342506853356292956506486091063660095505979187297020928573605860329881982122478494944846700224611808246427660214535971723459345029873385956677292979041143593821672034573140001092625650099257402018634684516092489263998517027205660003413512870074652126328536906790020794659204007921147300771594986038917179253827432120018857213350120695302091483756021206199805521083496979628811676116525321724267588515105188480380865374667274442027086789352802613365511142499668793725505110436809024171752137883546327359935102833441492430652019931999144063825010678766130335038975376834579129516127516820037383067
q3 = 44571911854174527304485400947383944661319242813524818888269963870884859557542264803774212076803157466539443358890313286282067621989609252352994203884813364011659788234277369629312571477760818634118449563652776213438461157699447304292906151410018017960605868035069246651843561595572415595568705784173761440671033435053531971051698504592848580356684103015611323747688216493729331061402058160819388999663041629882482138465124920580049057123360829897432472221079140360215664537272316836767039948368780837985855835419681893347839311156887660438769948501100287062738217966360434291369179859862550767272985972263442512061098317471708987686120577904202391381040801620069987103931326500146536990700234262413595295698193570184681785854277656410199477649697026112650581343325348837547631237627207304757407395388155701341044939408589591213693329516396531103489233367665983149963665364824119870832353269655933102900004362236232825539480774
r3 = 22285955927087263652242700473691972330659621406762409444134981935442429778771132401887106038401578733269721679445156643141033810994804626176497101942406682005829894117138684814656285738880409317059224781826388106719230578849723652146453075705009008980302934017534623325921780797786207797784352892086880720749202442492937918619992591614713131681306874944356693778359565004415437554407990089293135634916859631279984463829118336826115430997439527110961309956466956650522900331263720500751112297418506140413317489683875995326726992533904683800042127871963320754241310699432792081707870167598822650064976439270556418985242630368723264289700246406905189810458354474959276748887369363592834205660349184660073395182450526542246354364903399132116153732074081050985584216815493617906868615192465631416955706457835185743023758573279838341229835613609332206338401219168119635681832981552328638132500079074010106995297184587143613134093145

e = 3
k = 0

# 低加密指数攻击
# while 1:
#     res = gmpy2.iroot(p3+k*n,e)
#     if(res[1] == True):
#         print('p=',res[0])
#         # print(libnum.n2s(int(res[0])))
#         break
#     k = k+1
p= 96241803526087516516438618680574139229212699224895199026126947479609515703069904259770933066463243844738712136916991719874179296797623802919752542053959297743706931240798992583332970879091497936378700193716012227086531507335444090574605921869576355704757589370608232721639204280020820678250787751406162350723


# while 1:
#     res = gmpy2.iroot(q3+k*n,e)
#     if(res[1] == True):
#         print('q=',res[0])
#         # print(libnum.n2s(int(res[0])))
#         break
#     k = k+1
q = 962418035260875165164386186805741392292126992248951990261269474796095157030699042597709330664632438447387121369169917198741792967976238029197525420539592977437069312407989925833329708790914979363787001937160122270865315073354440905746059218695763557047575893706082327216392042800208206782507877514061623507521

# while 1:
#     res = gmpy2.iroot(r3+k*n,e)
#     if(res[1] == True):
#         print('r=',res[0])
#         # print(libnum.n2s(int(res[0])))
#         break
#     k = k+1
# r求不出来

# 按原题给出的p去求出r
i = 0
while True:
    r = p * 5 + i
    if isPrime(r):
        i = 0
        break
    else:
        i += 1

e2 = 65537
phi_n = (q-1)*(p-1)*(r-1)

d = gmpy2.invert(e2,(p-1)*(q-1)*(r-1))
m = pow(c,d,n)
print(long_to_bytes(m))

# b'NSSCTF{cc10786a-cc59-a07d-5c9f-df1c55b18cd4}'

[LitCTF 2023]P_Leak

题目

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from Crypto.Util.number import *
e=65537
m=bytes_to_long(b'xxxx')
p=getPrime(512)
q=getPrime(512)
n=p*q
phi=(p-1)*(q-1)
d=inverse(e,phi)
dp=d%(p-1)
c=pow(m,e,n)
print("dp=",dp)
print("n=",n)
print("c=",c)
#dp= 5892502924236878675675338970704766304539618343869489297045857272605067962848952532606770917225218534430490745895652561015493032055636004130931491316020329
#n= 50612159190225619689404794427464916374543237300894011803225784470008992781409447214236779975896311093686413491163221778479739252804271270231391599602217675895446538524670610623369953168412236472302812808639218392319634397138871387898452935081756580084070333246950840091192420542761507705395568904875746222477
#c= 39257649468514605476432946851710016346016992413796229928386230062780829495844059368939749930876895443279723032641876662714088329296631207594999580050131450251288839714711436117326769029649419789323982613380617840218087161435260837263996287628129307328857086987521821533565738409794866606381789730458247531619

分析

已知dp,n,c,e， dp泄露类

已知 e n c dp 倒推：求m -> 求d -> 求phi -> p q d = gmpy2.invert(e,phi_n) phi_n = (p-1)(q-1)

由 dp*e 三 e*d mod (p-1) -> e*d = k(p-1)+dp*e， k∈Z 由 e*d 三 1 mod (p-1)(q-1) -> e*d = Ω(p-1)(q-1)+1 ，Ω∈Z 根据两个e*等式 -》 k(p-1)+dp*e = Ω(p-1)(q-1)+1 ， k∈Z，Ω∈Z 进而得到 dp * e = (p-1)* (Ω(q-1)-k ) + 1 也就说 dp*e = β(p-1) + 1 ，β∈Z ∵ dp < (p-1) ∴ e>β 即 1<β<e 所以存在一个β 满足 dp*e = β(p-1) + 1 所以遍历β 找到后可求出 p-1 进而求出 p q

可以直接套模板

WP

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from Crypto.Util.number import long_to_bytes
import gmpy2

e = 65537
dp= 5892502924236878675675338970704766304539618343869489297045857272605067962848952532606770917225218534430490745895652561015493032055636004130931491316020329
n= 50612159190225619689404794427464916374543237300894011803225784470008992781409447214236779975896311093686413491163221778479739252804271270231391599602217675895446538524670610623369953168412236472302812808639218392319634397138871387898452935081756580084070333246950840091192420542761507705395568904875746222477
c= 39257649468514605476432946851710016346016992413796229928386230062780829495844059368939749930876895443279723032641876662714088329296631207594999580050131450251288839714711436117326769029649419789323982613380617840218087161435260837263996287628129307328857086987521821533565738409794866606381789730458247531619


for i in range(1, e):  # 在范围(1,e)之间进行遍历
    if (dp * e - 1) % i == 0:
        if n % (((dp * e - 1) // i) + 1) == 0:  # 存在p，使得n能被p整除
            p = ((dp * e - 1) // i) + 1
            q = n // (((dp * e - 1) // i) + 1)
            phi = (q - 1) * (p - 1)  # 欧拉定理
            d = gmpy2.invert(e, phi)  # 求模逆
            m = pow(c, d, n)  # 快速求幂取模运算

print(long_to_bytes(m))
# b'LitCTF{Prim3_1s_Le@k!!!!!}'

[CISCN 2023 初赛]Sign_in_passwd

题目

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j2rXjx8yjd=YRZWyTIuwRdbyQdbqR3R9iZmsScutj2iqj3/tidj1jd=D
GHI3KLMNJOPQRSTUb%3DcdefghijklmnopWXYZ%2F12%2B406789VaqrstuvwxyzABCDEF5

第二行带%3D 先进行url解码 解码后65位 推测是base64置换表 直接上 CyberChef 对第一行解密

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GHI3KLMNJOPQRSTUb=cdefghijklmnopWXYZ/12+406789VaqrstuvwxyzABCDEF5
截图

得到flag flag{8e4b2888-6148-4003-b725-3ff0d93a6ee4}

[GDOUCTF 2023]Absolute_Baby_Encrytpion

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Alice用脚本向Bob发送了以下消息。 这是用JS写的。 Bob对JS一无所知。 你能帮忙看看留言吗？

+}!q")hiim)#}-nvm)i-$#mvn#0mnbm)im#n+}!qnm8)i-$#mvnoc#0nz<$9inm!>-n1:1-nm8)i-$~c58n!}qhij#0[noic##m8nc8n?!8c}w!n]>&

js脚本

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let messagetoEncrypt = prompt("Enter a string: ").toLowerCase();
let charArray = messagetoEncrypt.split("");
let encryptedString = "";
let hasInvalidCharacter = false;

for (let i = 0; i < charArray.length; i++) {
    switch (charArray[i]) {
        case 'a':
            encryptedString = encryptedString.concat('!')
            break;
        case 'b':
            encryptedString = encryptedString.concat('1')
            break;
        case 'c':
            encryptedString = encryptedString.concat(')')
            break;
        case 'd':
            encryptedString = encryptedString.concat('v')
            break;
        case 'e':
            encryptedString = encryptedString.concat('m')
            break;
        case 'f':
            encryptedString = encryptedString.concat('+')
            break;
        case 'g':
            encryptedString = encryptedString.concat('q')
            break;
        case 'h':
            encryptedString = encryptedString.concat('0')
            break;
        case 'i':
            encryptedString = encryptedString.concat('c')
            break;
        case 'j':
            encryptedString = encryptedString.concat(']')
            break;
        case 'k':
            encryptedString = encryptedString.concat('(')
            break;
        case 'l':
            encryptedString = encryptedString.concat('}')
            break;
        case 'm':
            encryptedString = encryptedString.concat('[')
            break;
        case 'n':
            encryptedString = encryptedString.concat('8')
            break;
        case 'o':
            encryptedString = encryptedString.concat('5')
            break;
        case 'p':
            encryptedString = encryptedString.concat('$')
            break;
        case 'q':
            encryptedString = encryptedString.concat('*')
            break;
        case 'r':
            encryptedString = encryptedString.concat('i')
            break;
        case 's':
            encryptedString = encryptedString.concat('>')
            break;
        case 't':
            encryptedString = encryptedString.concat('#')
            break;
        case 'u':
            encryptedString = encryptedString.concat('<')
            break;
        case 'v':
            encryptedString = encryptedString.concat('?')
            break;
        case 'w':
            encryptedString = encryptedString.concat('o')
            break;
        case 'x':
            encryptedString = encryptedString.concat('^')
            break;
        case 'y':
            encryptedString = encryptedString.concat('-')
            break;
        case 'z':
            encryptedString = encryptedString.concat('_')
            break;
        case '0':
            encryptedString = encryptedString.concat('h')
            break;
        case '1':
            encryptedString = encryptedString.concat('w')
            break;
        case '2':
            encryptedString = encryptedString.concat('e')
            break;
        case '3':
            encryptedString = encryptedString.concat('9')
            break;
        case '4':
            encryptedString = encryptedString.concat('g')
            break;
        case '5':
            encryptedString = encryptedString.concat('z')
            break;
        case '6':
            encryptedString = encryptedString.concat('d')
            break;
        case '7':
            encryptedString = encryptedString.concat('~')
            break;
        case '8':
            encryptedString = encryptedString.concat('=')
            break;
        case '9':
            encryptedString = encryptedString.concat('x')
            break;
        case '!':
            encryptedString = encryptedString.concat('j')
            break;
        case '@':
            encryptedString = encryptedString.concat(':')
            break;
        case '#':
            encryptedString = encryptedString.concat('4')
            break;
        case '$':
            encryptedString = encryptedString.concat('b')
            break;
        case '%':
            encryptedString = encryptedString.concat('`')
            break;
        case '^':
            encryptedString = encryptedString.concat('l')
            break;
        case '&':
            encryptedString = encryptedString.concat('3')
            break;
        case '*':
            encryptedString = encryptedString.concat('t')
            break;
        case '(':
            encryptedString = encryptedString.concat('6')
            break;
        case ')':
            encryptedString = encryptedString.concat('s')
            break;
        case '_':
            encryptedString = encryptedString.concat('n')
            break;
        case '+':
            encryptedString = encryptedString.concat(';')
            break;

        case '-':
            encryptedString = encryptedString.concat('\'')
            break;
        case '=':
            encryptedString = encryptedString.concat('r')
            break;
        case '`':
            encryptedString = encryptedString.concat('k')
            break;
        case '~':
            encryptedString = encryptedString.concat('p')
            break;
        case '{':
            encryptedString = encryptedString.concat('\"')
            break;
        case '}':
            encryptedString = encryptedString.concat('&')
            break;
        case '[':
            encryptedString = encryptedString.concat('/')
            break;
        case ']':
            encryptedString = encryptedString.concat('\\')
            break;
        case '|':
            encryptedString = encryptedString.concat('2')
            break;
        case ':':
            encryptedString = encryptedString.concat('.')
            break;
        case ';':
            encryptedString = encryptedString.concat('%')
            break;
        case '\"':
            encryptedString = encryptedString.concat('|')
            break;
        case '\'':
            encryptedString = encryptedString.concat(',')
            break;
        case '<':
            encryptedString = encryptedString.concat('@')
            break;
        case '>':
            encryptedString = encryptedString.concat('{')
            break;
        case ',':
            encryptedString = encryptedString.concat('u')
            break;
        case '.':
            encryptedString = encryptedString.concat('7')
            break;
        case '?':
            encryptedString = encryptedString.concat('y')
            break;
        case '/':
            encryptedString = encryptedString.concat('a')
            break;

        default:
            hasInvalidCharacter = true;
    }
}

if (hasInvalidCharacter) {
    encryptedString = "Invalid String!";
} else {
    console.log(`Your encoded string is ${encryptedString}`);
}

根据js脚本 是一个字母符号置换表 py转成字典

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list = {'!': 'a', '1': 'b', ')': 'c', 'v': 'd', 'm': 'e', ' ': 'f',
        'q': 'g', '0': 'h', 'c': 'i', ']': 'j', '(': 'k', '}': 'l', '[': 'm', '8': 'n',
        '5': 'o', '$': 'p', '*': 'q', 'i': 'r', '>': 's', '#': 't', '<': 'u', '?': 'v', 'o': 'w', '^': 'x', '-': 'y',
        '_': 'z', 'h': '0', 'w': '1', 'e': '2', '9': '3', 'g': '4', 'z': '5', 'd': '6', '~': '7', '=': '8', 'x': '9', 'j': '!', ':': '@', '4': '#', 'b': '$', '`': '%', 'l': '^',
        '3': '&', 't': '*', '6': '(', 's': ')', 'n': '_', ';': ' ', '\\': '-', '&': '}', '/': '[', '\\\\': '{', '2': 'l', '.': ':', '%': ';', '|': '\\',
        ',': '\\', '@': '<', '{': '>', 'u': ',', '7': '.', 'y': '?', 'a': '//', '"': '{'}

cipher =' }!q")hiim)#}-nvm)i-$#mvn#0mnbm)im#n }!qnm8)i-$#mvnoc#0nz<$9inm!>-n1:1-nm8)i-$~c58n!}qhij#0[noic##m8nc8n?!8c}w!n]>&'

m = ''

for i in cipher:
    m  += list[i]

print(m)

# flag{c0rrectly_decrypted_the_$ecret_flag_encrypted_with_5up3r_easy_b@by_encryp7ion_alg0r!thm_written_in_vanil1a_js}

[NISACTF 2022]normal 【套娃提 Ook!】

题目

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..... ..... ..... ...!? !!.?. ..... ..... ..... ..?.? !.?.. ..... .....
..... ..... !.?.. ..... ....! ?!!.? ..... ..... ?.?!. ?!.?. ..... .....
..... .!?!! .?!!! !!!!! !!!!! !!!?. ?!.?! !!!!! !!!!! .!... ..... .....
!.!!! .?... ..... ..... !?!!. ?.... ..... ...?. ?!.?. ..... !.?.. .....
....! ?!!.? ..... ..... ?.?!. ?!.?. ..... ..... ..... .!?!! .?!!! !!!!!
!!!!! !!!?. ?!.?! !!!!! !!!!! .!... ..... .!... ....! .?... ..... .....
!?!!. ?.... ..... ...?. ?!.?. .!.?. ..... ..... !?!!. ?.... ..... .?.?!
.?!.? ..... ..... ..... ..!?! !.?!! !!!!! !!!!! !!!!? .?!.? !!!!! !!!!!
!.!.. ..... ..... .!.?. ..... ..... ..!?! !.?.. ..... ..... ?.?!. ?....
..... ..... ....! .!!!! !!!!! !!!!! !.?.. ..... ....! ?!!.? ..... .....
?.?!. ?!.?. ..... ..... ..... .!?!! .?!!! !!!!! !!!!! !!!?. ?!.?! !!!!!
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?!.?. ..... ..... ..... ....! .?... ..... ...!? !!.?. ..... ....? .?!.?
!.?.. ..... ..... ..... !?!!. ?!!!! !!!!! !!!!! !!?.? !.?!! !!!!! !!!!.
!.... ..... ..!.? ..... ..... ...!? !!.?. ..... ..... .?.?! .?... .....
..... ...!. !!!!! !!!!! !.?.. ..... ....! ?!!.? ..... ..... ?.?!. ?!.?.
..... ..... ..... .!?!! .?!!! !!!!! !!!!! !!!?. ?!.?! !!!!! !!!!! .!...
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.?... ..... ..... .!.!! !!!!! !!!!. ?.... ..... ..!?! !.?.. ..... ...?.
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.?!!! !!!!! !!!!! !!!?. ?!.?! !!!!! !!!!! .!... ..... ..... !.!!! !!!!.
?.... ..... ....! ?!!.? ..... ..... ..?.? !.?.. ..... ...!. ?.... .....
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..... ..... ....! .!.?. ..... ..... ..!?! !.?.. ..... ..... ?.?!. ?..!.
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?.... ..... ....! ?!!.? ..... ..... ..?.? !.?.. ..... .!.?. ..... .....
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..?.? !.?!. ?.... ..... ..... ...!? !!.?! !!!!! !!!!! !!!!! ?.?!. ?!!!!
!!!!! !!.!. ..... ..... !.... ...!. ?.... ..... ....! ?!!.? ..... .....
..?.? !.?!. ?.... ..... ..!?! !.?.. ..... ...?. ?!.?! .?... ..... .....
....! ?!!.? !!!!! !!!!! !!!!! !?.?! .?!!! !!!!! !!!.! ..... ..... ...!.
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!!.?. ..... ....? .?!.? !.?.. ..... ..... ..... !?!!. ?!!!! !!!!! !!!!!
!!?.? !.?!! !!!!! !!!!. !.... ...!. ?.... ..... ..... .!?!! .?... .....
..... .?.?! .?!.? .

分析

只要三种元素 !?. 比较符合Ook加密

套娃题 没啥意思 不想着了

[LitCTF 2023]yafu (中级)

题目

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from Crypto.Util.number import *
from secret import flag

m = bytes_to_long(flag)
n  = 1
for i in range(15):
    n *=getPrime(32)
e = 65537
c = pow(m,e,n)
print(f'n = {n}')
print(f'c = {c}')
'''
n = 15241208217768849887180010139590210767831431018204645415681695749294131435566140166245881287131522331092026252879324931622292179726764214435307
c = 12608550100856399369399391849907846147170257754920996952259023159548789970041433744454761458030776176806265496305629236559551086998780836655717
'''

分析

yafu分解n

截图

WP

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'''
from Crypto.Util.number import *
from secret import flag

m = bytes_to_long(flag)
n  = 1
for i in range(15):
    n *=getPrime(32)
e = 65537
c = pow(m,e,n)
print(f'n = {n}')
print(f'c = {c}')
# n = 15241208217768849887180010139590210767831431018204645415681695749294131435566140166245881287131522331092026252879324931622292179726764214435307
# c = 12608550100856399369399391849907846147170257754920996952259023159548789970041433744454761458030776176806265496305629236559551086998780836655717
'''

import gmpy2
from Crypto.Util.number import long_to_bytes

e = 65537
c = 12608550100856399369399391849907846147170257754920996952259023159548789970041433744454761458030776176806265496305629236559551086998780836655717
n = 15241208217768849887180010139590210767831431018204645415681695749294131435566140166245881287131522331092026252879324931622292179726764214435307

P1 = 2201440207
P2 = 3354884521
P3 = 2719600579
P4 = 4171911923
P5 = 2906576131
P6 = 2767137487
P7 = 4021078331
P8 = 2585574697
P9 = 4044505687
P10 = 2151018733
P11 = 2315495107
P12 = 2758708999
P13 = 3355651511
P14 = 2923522073
P15 = 3989697563

phi = (P1-1)*(P2-1)*(P3-1)*(P4-1)*(P5-1)*(P6-1)*(P7-1)*(P8-1)*(P9-1)*(P10-1)*(P11-1)*(P12-1)*(P13-1)*(P14-1)*(P15-1)
d  = gmpy2.invert(e,phi)
m = pow(c,d,n)
print(long_to_bytes(m))

# b'LitCTF{Mu1tiple_3m4ll_prim5_fac7ors_@re_uns4f5}'

[LitCTF 2023]md5的破解

题目

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from Crypto.Util.number import *
from hashlib import md5
from secret import flag

#flag全是由小写字母及数字组成
m=md5(flag).hexdigest()
print(flag[:13]+flag[15:18]+flag[19:34]+flag[35:38])
print(m)
# b'LitCTF{md5can??3de?rypt213thoughcr?sh}?
# 496603d6953a15846cd7cc476f146771

# b'
LitCTF{md5can
3de
rypt213thoughc
rsh}'

分析

缺少第13、14、18、34位 爆破大小写字母数字即可

WP

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from hashlib import md5
from string import ascii_lowercase,digits

# LitCTF{md5can??3de?rypt213thoughcr?sh}

for i in ascii_lowercase+digits:
    for j in ascii_lowercase + digits:
        for k in ascii_lowercase + digits:
            for l in ascii_lowercase + digits:
                f = 'LitCTF{md5can'+i+j+'3de'+k+'rypt213thoughcr'+l+'sh}'
                h = md5(f.encode()).hexdigest()
                # print(h)
                if h == '496603d6953a15846cd7cc476f146771':
                    print(f)
                    exit()
                    
# LitCTF{md5can123dexrypt213thoughcrpsh}

[HDCTF 2023]爬过小山去看云

题目

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密文：ymyvzjtxswwktetpyvpfmvcdgywktetpyvpfuedfnzdjsiujvpwktetpyvnzdjpfkjssvacdgywktetpyvnzdjqtincduedfpfkjssne
在山的那头，有3个人，4只鸟，19只羊，11朵云

没见过 不会整 直接看下wp记录下

【希尔+云影】

希尔密码 hill cipher

替换密码，使用矩阵和矩阵乘法来混合明文

http://www.metools.info/code/hillcipher243.html 在线解密

得到 yourpiniseightfourtwozeroeightfourtwoonezeroeighteightfourzerotwofourzeroeightfourzeroonezeroonetwofourx

对应数字 eightfourtwozeroeightfourtwoonezeroeighteightfourzerotwofourzeroeightfourzeroonezeroonetwofour

842084210884024084010124 只包含01248云影密码

NOTFLAG

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# 云影密码加解密脚本
# https://www.jianshu.com/p/b5aa5cf60f83
def de_code(c):
    dic = [chr(i) for i in range(ord("A"), ord("Z") + 1)]
    flag = []
    c2 = [i for i in c.split("0")]
    for i in c2:
        c3 = 0
        for j in i:
            c3 += int(j)
        flag.append(dic[c3 - 1])
    return flag

def encode(plaintext):
    dic = [chr(i) for i in range(ord("A"), ord("Z") + 1)]
    m = [i for i in plaintext]
    tmp = [];flag = []
    for i in range(len(m)):
        for j in range(len(dic)):
            if m[i] == dic[j]:
                tmp.append(j + 1)
    for i in tmp:
        res = ""
        if i >= 8:
            res += int(i/8)*"8"
        if i%8 >=4:
            res += int(i%8/4)*"4"
        if i%4 >=2:
            res += int(i%4/2)*"2"
        if i%2 >= 1:
            res += int(i%2/1)*"1"
        flag.append(res + "0")
    print ("".join(flag)[:-1])

c = input("输入要解密的数字串:")
print (de_code(c))
m_code = input("请输入要加密的数字串:")
encode(m_code)

[LitCTF 2023]e的学问【e phi不互素】

题目描述:当e不是65537了，该怎么办呢？

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from Crypto.Util.number import *
m=bytes_to_long(b'xxxxxx')
p=getPrime(256)
q=getPrime(256)
e=74
n=p*q
c=pow(m,e,n)
print("p=",p)
print("q=",q)
print("c=",c)
#p= 86053582917386343422567174764040471033234388106968488834872953625339458483149
#q= 72031998384560188060716696553519973198388628004850270102102972862328770104493
#c= 3939634105073614197573473825268995321781553470182462454724181094897309933627076266632153551522332244941496491385911139566998817961371516587764621395810123

WP

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from Crypto.Util.number import long_to_bytes
import gmpy2
# e phi【p-1 q-1】 均互素   去公约数
e=74
p= 86053582917386343422567174764040471033234388106968488834872953625339458483149
q= 72031998384560188060716696553519973198388628004850270102102972862328770104493
c= 3939634105073614197573473825268995321781553470182462454724181094897309933627076266632153551522332244941496491385911139566998817961371516587764621395810123
n= p*q
phi_n = (q-1)*(p-1)
t = gmpy2.gcd(e,phi_n)
#
e1 = e // t
d= gmpy2.invert(e1,phi_n)
m= pow(c,d,n)
# 求的m后 对 m^t开t次方即可
msg = gmpy2.iroot(m,t)
if msg[1]:
    print(long_to_bytes(msg[0]))

# b'LitCTF{e_1s_n0t_@_Prime}'

风二西工具也可直出m 然后longtobyte即可得到flag

### [湖湘杯 2021]signin - 忘做了后面做

题目

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from Crypto.Util.number import *
from secret import flag
import random

m1 = bytes_to_long(flag[:len(flag) // 2])
m2 = bytes_to_long(flag[len(flag) // 2:])

def gen(pbits, qbits):
    p1, q1 = getPrime(pbits), getPrime(qbits)
    n1 = p1**4*q1
    q2 = getPrime(qbits)
    bound = p1 // (8*q1*q2) + 1
    p2 = random.randrange(p1, p1 + bound)
    while not isPrime(p2):
        p2 = random.randrange(p1, p1 + bound)
    n2 = p2**4*q2
    return (n1, n2), (p1, q1), (p2, q2)

e = 0x10001
pbits = int(360)
qbits = int(128)
pk, sk1, sk2 = gen(pbits, qbits)
c1 = pow(m1, e, pk[0])
c2 = pow(m2, e, pk[1])
print(f'pk = {pk}')
print(f'c1, c2 = {c1, c2}')

"""
pk = (1150398070565459492080597718626032792435556703413923483458704675295997646493249759818468321328556510074044954676615760446708253531839417036997811506222349194302791943489195718713797322878586379546657275419261647635859989280700191441312691274285176619391539387875252135478424580680264554294179123254566796890998243909286508189826458854346825493157697201495100628216832191035903848391447704849808577310612723700318670466035077202673373956324725108350230357879374234418393233, 1242678737076048096780023147702514112272319497423818488193557934695583793070332178723043194823444815153743889740338870676093799728875725651036060313223096288606947708155579060628807516053981975820338028456770109640111153719903207363617099371353910243497871090334898522942934052035102902892149792570965804205461900841595290667647854346905445201396273291648968142608158533514391348407631818144116768794595226974831093526512117505486679153727123796834305088741279455621586989)
c1, c2 = (361624030197288323178211941746074961985876772079713896964822566468795093475887773853629454653096485450671233584616088768705417987527877166166213574572987732852155320225332020636386698169212072312758052524652761304795529199864805108000796457423822443871436659548626629448170698048984709740274043050729249408577243328282313593461300703078854044587993248807613713896590402657788194264718603549894361488507629356532718775278399264279359256975688280723740017979438505001819438, 33322989148902718763644384246610630825314206644879155585369541624158380990667828419255828083639294898100922608833810585530801931417726134558845725168047585271855248605561256531342703212030641555260907310067120102069499927711242804407691706542428236208695153618955781372741765233319988193384708525251620506966304554054884590718068210659709406626033891748214407992041364462525367373648910810036622684929049996166651416565651803952838857960054689875755131784246099270581394)
"""

[NSSCTF 2022 Spring Recruit]Vigenere

不知道key 利用在线网站爆破 https://www.guballa.de/vigenere-solver

image-20240221093344960

[AFCTF 2018]可怜的RSA

题目给了一个 flag.enc 和 public.key

openssl rsa -pubin -text -modulus -in warmup -in public.key > 123.txt

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Public-Key: (2070 bit)
Modulus:
    25:b1:8b:f5:f3:89:09:7d:17:23:78:66:bb:51:cf:
    f8:de:92:24:53:74:9e:bc:40:3b:09:95:c9:7c:0e:
    38:6d:46:c1:61:ca:df:f7:7c:69:86:0d:ae:47:91:
    c2:14:cf:84:87:aa:aa:9f:26:e9:20:a9:77:83:49:
    06:03:8a:ef:b5:c3:08:27:df:cf:3f:c9:e9:76:95:
    44:f9:4e:07:cd:fe:08:72:03:9a:3a:62:62:11:66:
    78:b2:61:fb:2d:6b:9d:32:53:9e:92:a1:53:b3:67:
    56:29:ba:b3:94:2e:7d:35:e3:0f:7e:ef:5a:bf:1c:
    50:d7:97:d0:cc:88:e1:bd:cc:fd:1a:12:ea:6f:7e:
    f7:5c:37:27:db:df:2e:78:0f:34:28:ae:8f:7a:4f:
    b7:a8:9f:18:4a:36:50:32:b1:53:f8:42:5e:84:57:
    50:eb:2b:7a:bc:02:dc:15:ce:02:07:50:7a:a9:50:
    86:3b:b8:48:0a:78:02:8d:d6:29:79:94:4d:6c:63:
    3f:af:a1:03:e4:db:28:ce:87:f5:a0:c6:ed:4a:2f:
    26:64:42:7f:56:5c:77:81:ab:61:91:45:6d:97:1c:
    7f:fa:39:52:72:37:4c:ec:01:55:e5:f9:11:89:db:
    74:2e:4c:28:b0:3a:0f:a1:1c:ff:b0:31:73:d2:a4:
    cc:e6:ae:53
Exponent: 65537 (0x10001)
Modulus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
-----BEGIN PUBLIC KEY-----
MIIBJDANBgkqhkiG9w0BAQEFAAOCAREAMIIBDAKCAQMlsYv184kJfRcjeGa7Uc/4
3pIkU3SevEA7CZXJfA44bUbBYcrf93xphg2uR5HCFM+Eh6qqnybpIKl3g0kGA4rv
tcMIJ9/PP8npdpVE+U4Hzf4IcgOaOmJiEWZ4smH7LWudMlOekqFTs2dWKbqzlC59
NeMPfu9avxxQ15fQzIjhvcz9GhLqb373XDcn298ueA80KK6Pek+3qJ8YSjZQMrFT
+EJehFdQ6yt6vALcFc4CB1B6qVCGO7hICngCjdYpeZRNbGM/r6ED5Nsozof1oMbt
Si8mZEJ/Vlx3gathkUVtlxx/+jlScjdM7AFV5fkRidt0LkwosDoPoRz/sDFz0qTM
5q5TAgMBAAE=
-----END PUBLIC KEY-----

n转10进制

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n = 79832181757332818552764610761349592984614744432279135328398999801627880283610900361281249973175805069916210179560506497075132524902086881120372213626641879468491936860976686933630869673826972619938321951599146744807653301076026577949579618331502776303983485566046485431039541708467141408260220098592761245010678592347501894176269580510459729633673468068467144199744563731826362102608811033400887813754780282628099443490170016087838606998017490456601315802448567772411623826281747245660954245413781519794295336197555688543537992197142258053220453757666537840276416475602759374950715283890232230741542737319569819793988431443
e = 65537
p = 3133337 
q = 25478326064937419292200172136399497719081842914528228316455906211693118321971399936004729134841162974144246271486439695786036588117424611881955950996219646807378822278285638261582099108339438949573034101215141156156408742843820048066830863814362379885720395082318462850002901605689761876319151147352730090957556940842144299887394678743607766937828094478336401159449035878306853716216548374273462386508307367713112073004011383418967894930554067582453248981022011922883374442736848045920676341361871231787163441467533076890081721882179369168787287724769642665399992556052144845878600126283968890273067575342061776244939

flag.enc中是经过base64加密的 所以需要先解密一下才能得到c的值

之前没遇到过这种，解密完cipher就不知道咋做了 看了下提示和别人的wp 需要加密时使用了padding

image-20240221093440173 
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import base64
import gmpy2
from Crypto.Cipher import PKCS1_OAEP
from Crypto.PublicKey import RSA

n = 79832181757332818552764610761349592984614744432279135328398999801627880283610900361281249973175805069916210179560506497075132524902086881120372213626641879468491936860976686933630869673826972619938321951599146744807653301076026577949579618331502776303983485566046485431039541708467141408260220098592761245010678592347501894176269580510459729633673468068467144199744563731826362102608811033400887813754780282628099443490170016087838606998017490456601315802448567772411623826281747245660954245413781519794295336197555688543537992197142258053220453757666537840276416475602759374950715283890232230741542737319569819793988431443
e = 65537
p = 3133337
q = 25478326064937419292200172136399497719081842914528228316455906211693118321971399936004729134841162974144246271486439695786036588117424611881955950996219646807378822278285638261582099108339438949573034101215141156156408742843820048066830863814362379885720395082318462850002901605689761876319151147352730090957556940842144299887394678743607766937828094478336401159449035878306853716216548374273462386508307367713112073004011383418967894930554067582453248981022011922883374442736848045920676341361871231787163441467533076890081721882179369168787287724769642665399992556052144845878600126283968890273067575342061776244939
phi_n = (q-1)*(p-1)
d = gmpy2.invert(e,phi_n)
# print(d)
cipher ='GVd1d3viIXFfcHapEYuo5fAvIiUS83adrtMW/MgPwxVBSl46joFCQ1plcnlDGfL19K/3PvChV6n5QGohzfVyz2Z5GdTlaknxvHDUGf5HCukokyPwK/1EYU7NzrhGE7J5jPdi0Aj7xi/Odxy0hGMgpaBLd/nL3N8O6i9pc4Gg3O8soOlciBG/6/xdfN3SzSStMYIN8nfZZMSq3xDDvz4YB7TcTBh4ik4wYhuC77gmT+HWOv5gLTNQ3EkZs5N3EAopy11zHNYU80yv1jtFGcluNPyXYttU5qU33jcp0Wuznac+t+AZHeSQy5vk8DyWorSGMiS+J4KNqSVlDs12EqXEqqJ0uA=='

cipher = base64.b64decode(cipher)
# print(cipher)

key_info = RSA.construct((n, e, int(d), p, q))
key = RSA.importKey(key_info.exportKey())
key = PKCS1_OAEP.new(key)
flag = key.decrypt(cipher)
print(flag)
# b'afctf{R54_|5_$0_B0rin9}'

[AFCTF 2018]BASE

本来py base64库一点点解码的 发现是疯狂套娃有点麻烦

看了下wp 直接循环

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import base64
file = open("flag_encode.txt",'r')
text = file.read()
while(1):
    try:
        text = base64.b32decode(text ).decode()
    except:
        try:
            text = base64.b64decode(text ).decode()
        except:
            try:
                text = base64.b16decode(text ).decode()
            except:
                print("text！",text)
                break
#afctf{U_5h0u1d_Us3_T00l5}

[HGAME 2023 week1]RSA

题目

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from Crypto.Util.number import *

flag = open('flag.txt', 'rb').read()

p = getPrime(512)
q = getPrime(512)
n=p*q
e = 65537
m = bytes_to_long(flag)
c = pow(m, e, n)
print(f"c={c}")
print(f"n={n}")

"""
c=110674792674017748243232351185896019660434718342001686906527789876264976328686134101972125493938434992787002915562500475480693297360867681000092725583284616353543422388489208114545007138606543678040798651836027433383282177081034151589935024292017207209056829250152219183518400364871109559825679273502274955582
n=135127138348299757374196447062640858416920350098320099993115949719051354213545596643216739555453946196078110834726375475981791223069451364024181952818056802089567064926510294124594174478123216516600368334763849206942942824711531334239106807454086389211139153023662266125937481669520771879355089997671125020789
"""

WP

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import gmpy2
from Crypto.Util.number import long_to_bytes

c=110674792674017748243232351185896019660434718342001686906527789876264976328686134101972125493938434992787002915562500475480693297360867681000092725583284616353543422388489208114545007138606543678040798651836027433383282177081034151589935024292017207209056829250152219183518400364871109559825679273502274955582
n=135127138348299757374196447062640858416920350098320099993115949719051354213545596643216739555453946196078110834726375475981791223069451364024181952818056802089567064926510294124594174478123216516600368334763849206942942824711531334239106807454086389211139153023662266125937481669520771879355089997671125020789

e = 65537
p = 12022912661420941592569751731802639375088427463430162252113082619617837010913002515450223656942836378041122163833359097910935638423464006252814266959128953
q = 11239134987804993586763559028187245057652550219515201768644770733869088185320740938450178816138394844329723311433549899499795775655921261664087997097294813
phi_n = (q-1)*(p-1)
d = gmpy2.invert(e,phi_n)
m = pow(c,d,n)
print(long_to_bytes(m))
# b'hgame{factordb.com_is_strong!}'

[LitCTF 2023]Euler

题目

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from Crypto.Util.number import *
from secret import flag

m = bytes_to_long(flag)
p = getPrime(512)
q = getPrime(512)
n = p*q
c = pow(m,n-p-q+3,n)
print(f'n = {n}')
print(f'c = {c}')
"""
n = 115140122725890943990475192890188343698762004010330526468754961357872096040956340092062274481843042907652320664917728267982409212988849109825729150839069369465433531269728824368749655421846730162477193420534803525810831025762500375845466064264837531992986534097821734242082950392892529951104643690838773406549
c = 406480424882876909664869928877322864482740577681292497936198951316587691545267772748204383995815523935005725558478033908575228532559165174398668885819826720515607326399097899572022020453298441
"""

WP

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import gmpy2
from Crypto.Util.number import long_to_bytes

# phi = (p-1)(q-1) = pq-p-q+1 = n-p-q+1 
# c = pow(m,phi+2,n) = m^(phi+2) mod n = m^phi mod n  *  m^2 mod n
# 欧拉定理  m^phi 三 1 mod n  ->  c = m^2 mod n

n = 115140122725890943990475192890188343698762004010330526468754961357872096040956340092062274481843042907652320664917728267982409212988849109825729150839069369465433531269728824368749655421846730162477193420534803525810831025762500375845466064264837531992986534097821734242082950392892529951104643690838773406549
c = 406480424882876909664869928877322864482740577681292497936198951316587691545267772748204383995815523935005725558478033908575228532559165174398668885819826720515607326399097899572022020453298441

m = gmpy2.iroot(c,2)[0]
# print(m)
print(long_to_bytes(m))
# b'LitCTF{a1a8887793acfc199182a649e905daab}'

[HDCTF 2023]Math_Rsa

题目

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from Crypto.Util.number import *
from shin import flag


m=bytes_to_long(flag)
r=getPrime(1024)
assert r%4==3
p=getPrime(1024)
assert pow(p,(r-1)//2,r)==1
q=getPrime(1024)
e=65537
n=p*q
a=pow(p,2,r)
c=pow(m,e,n)
print(f"n = {n}")
print(f"r = {r}")
print(f"a = {a}")
print(f"c = {c}")
'''
n = 14859096721972571275113983218934367817755893152876205380485481243331724183921836088288081702352994668073737901001999266644597320501510110156000004121260529706467596723314403262665291609405901413014268847623323618322794733633701355018297180967414569196496398340411723555826597629318524966741762029358820546567319749619243298957600716201084388836601266780686983787343862081546627427588380349419143512429889606408316907950943872684371787773262968532322073585449855893701828146080616188277162144464353498105939650706920663343245426376506714689749161228876988380824497513873436735960950355105802057279581583149036118078489
r = 145491538843334216714386412684012043545621410855800637571278502175614814648745218194962227539529331856802087217944496965842507972546292280972112841086902373612910345469921148426463042254195665018427080500677258981687116985855921771781242636077989465778056018747012467840003841693555272437071000936268768887299
a = 55964525692779548127584763434439890529728374088765597880759713360575037841170692647451851107865577004136603179246290669488558901413896713187831298964947047118465139235438896930729550228171700578741565927677764309135314910544565108363708736408337172674125506890098872891915897539306377840936658277631020650625
c = 12162333845365222333317364738458290101496436746496440837075952494841057738832092422679700884737328562151621948812616422038905426346860411550178061478808128855882459082137077477841624706988356642870940724988156263550796637806555269282505420720558849717265491643392140727605508756229066139493821648882251876933345101043468528015921111395602873356915520599085461538265894970248065772191748271175288506787110428723281590819815819036931155215189564342305674107662339977581410206210870725691314524812137801739246685784657364132180368529788767503223017329025740936590291109954677092128550252945936759891497673970553062223608
'''

分析

a=pow(p,2,r)

a = p^2 mod r ，a r 已知 可以进一步求出p 求出p后 n//p 得到q 然后在求phi d m

WP

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import gmpy2
from Crypto.Util.number import long_to_bytes

n = 14859096721972571275113983218934367817755893152876205380485481243331724183921836088288081702352994668073737901001999266644597320501510110156000004121260529706467596723314403262665291609405901413014268847623323618322794733633701355018297180967414569196496398340411723555826597629318524966741762029358820546567319749619243298957600716201084388836601266780686983787343862081546627427588380349419143512429889606408316907950943872684371787773262968532322073585449855893701828146080616188277162144464353498105939650706920663343245426376506714689749161228876988380824497513873436735960950355105802057279581583149036118078489
r = 145491538843334216714386412684012043545621410855800637571278502175614814648745218194962227539529331856802087217944496965842507972546292280972112841086902373612910345469921148426463042254195665018427080500677258981687116985855921771781242636077989465778056018747012467840003841693555272437071000936268768887299
a = 55964525692779548127584763434439890529728374088765597880759713360575037841170692647451851107865577004136603179246290669488558901413896713187831298964947047118465139235438896930729550228171700578741565927677764309135314910544565108363708736408337172674125506890098872891915897539306377840936658277631020650625
c = 12162333845365222333317364738458290101496436746496440837075952494841057738832092422679700884737328562151621948812616422038905426346860411550178061478808128855882459082137077477841624706988356642870940724988156263550796637806555269282505420720558849717265491643392140727605508756229066139493821648882251876933345101043468528015921111395602873356915520599085461538265894970248065772191748271175288506787110428723281590819815819036931155215189564342305674107662339977581410206210870725691314524812137801739246685784657364132180368529788767503223017329025740936590291109954677092128550252945936759891497673970553062223608

e=65537

# p^((r-1)//2) mod r = 1
# a=pow(p,2,r)
# a = p^2 mod r   a r 已知 可以进一步求出p  求出p后 n//p 得到q  然后在求phi d m
# r%4==3

p = 135098300162574110032318082604507116145598393187097375349178563291884099917465443655846455456198422625358836544141120445250413758672683505731015242196083913722084539762488109001442453793004455466844129788221721833309756439196036660458760461237225684006072689852654273913614912604470081753828559417535710077291
p2 = 10393238680760106682068330079504927400023017668703262222099938883730714731279774539115772083330909231443250673803376520592094213873608775241097598890818459890825805707433039425020588461191209551582950712455537148377360546659885111322482174840763781771983328894358193926388929089085190683242441518733058810008


# sage
# a = p^2 mod r 求p
# r = 145491538843334216714386412684012043545621410855800637571278502175614814648745218194962227539529331856802087217944496965842507972546292280972112841086902373612910345469921148426463042254195665018427080500677258981687116985855921771781242636077989465778056018747012467840003841693555272437071000936268768887299
# a = 55964525692779548127584763434439890529728374088765597880759713360575037841170692647451851107865577004136603179246290669488558901413896713187831298964947047118465139235438896930729550228171700578741565927677764309135314910544565108363708736408337172674125506890098872891915897539306377840936658277631020650625
# R.<x>=PolynomialRing(Zmod(r))
# f=(x^2)-a
# ans=f.roots()
# print(ans)
# p = ans[0][0]
# print(p)
# [(135098300162574110032318082604507116145598393187097375349178563291884099917465443655846455456198422625358836544141120445250413758672683505731015242196083913722084539762488109001442453793004455466844129788221721833309756439196036660458760461237225684006072689852654273913614912604470081753828559417535710077291, 1), (10393238680760106682068330079504927400023017668703262222099938883730714731279774539115772083330909231443250673803376520592094213873608775241097598890818459890825805707433039425020588461191209551582950712455537148377360546659885111322482174840763781771983328894358193926388929089085190683242441518733058810008, 1)]

# pow(p,(r-1)//2,r)==1
assert pow(p,(r-1)//2,r)==1
# assert pow(p2,(r-1)//2,r)==1
# sage 求出 p 有两个解 但是题目给出条件  assert pow(p2,(r-1)//2,r)==1  p2不满足 所以只有一个p

q = n // p
phi = (q-1)*(p-1)
d = gmpy2.invert(e,phi)
m = pow(c,d,n)
print(long_to_bytes(m))
# b'HDCTF{0ce04f81-516b-4132-81a2-b0b7166e03ad}'

[HGAME 2022 week3]Multi Prime RSA 【欧拉函数】

题目

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from Crypto.Util.number import getPrime
from libnum import s2n
from secret import flag

p = getPrime(256)
q = getPrime(256)
r = getPrime(256)
s = getPrime(256)
n = p ** 2 * q ** 3 * r ** 5 * s ** 7
e = 65537
c = pow(s2n(flag), e, n)
print(f"p = {p}")
print(f"q = {q}")
print(f"r = {r}")
print(f"s = {s}")
print(f"n = {n}")
print(f"e = {e}")
print(f"c = {c}")

output

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p = 61789932148719477384027458333380568978056286136137829092952317307711908353477
q = 91207969353355763685633284378833506319794714507027332929290701748727534193861
r = 105471299607375388622347272479207944509670502835651250945203397530010861809367
s = 83153238748903772448138307505579799277162652151244477391465130504267171881437
n = 1039344372165087100001063920598151812324151064684841845250974758525265148567706103784958424873181721352440209284812493753972556519482026327282644619091466886523804841248277210353173383407944598453848113815866908595335619458549486958764490103808475329598085842184963065068499489886467911087295087163762599284622055185456905774507245781667293199205317692029829495961487347944813874415423771980660778986211145841712412631156369129146470119135136378158203459576596246169191419488560832734046076107673091995860021863239882608638458149930255944184863801278386551031980146460231515747754411678651752698881001464973981424240781413084941947261875289725538959720572496329348499870580057997540844488309111059240745081048324762866572948371222839278718034435739827677190025500802453626872356208612718417249649474571197167076916403582394186357812640566250930361276229969553128128312736245440129556020108188835966131425956431796417720436474093381770796431629523054378258497546013222494974549262140415585158985940966415459478150722832119691308697510189026447359189994055885090735411738332296254011208547676914004864732327863884217733456287369771087094514708468685641820375220835485053482570852619363091173324203334503461823983610886849930944250553928855506012684504211525542998575275626784129736345142772399109273619522445919
e = 65537
c = 844677395496466411520394190869787261209960246734415406217975986418865760680024542119231873259131861208878522030009923057991526761346423130242121884493257732067700857897379859545356609151834223804262174935191718271211809221730601602827122249238086030580971376104724987801049500689134122609834321586609223761140538079460830213824674361601046367637227094018381901291488659642720549583856812747877519600804325570421770575999289389175021646347371879234023647657507178519047236746071420327155188213839293382288787853777540226192644761028822256165706787395891134765908229036044468473519166141610604791485071702808854944672418124203289328124793348198048601338476086482318248264508789781967910205393740835345086784345145351367491197717933757414967811594913692588314161669333147733048171044386546892346475181197482702164468542430187885074163177843285948999943328049159021873821254267471067523609151007885131921896462161216356454116929796355815756642621369974260365378070336290542971599886325232821981080341858950609157813769416455337935096696635623426418166316737131174435618543058086342714723330814586496030805366321181723292731710369013923285787724941830672247377301048663929453294620044701627159066468762709113137517559435822623284148112827473010030736329596829357275518641576798298066541516764673029908084962144713

分析

n可以分成多个素数 且 p q r s 都已知 算出phi来 求d 再求m即可

通过欧拉函数 计算phi(n) 带入然后拆分即可

WP

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# n = p ** 2 * q ** 3 * r ** 5 * s ** 7
# phi(n) = phi(p*p) * phi(q*q*q) * phi(r*r*r*r*r) * phi(s*s*s*s*s*s*s)
#        = p* phi(p) * q^2* phi(q) * r^4*phi(r) * s^6*phi(s)
#        = p*(p-1) * q^2*(q-1) * r^4*(r-1) * s^6*(s-1)

p = 61789932148719477384027458333380568978056286136137829092952317307711908353477
q = 91207969353355763685633284378833506319794714507027332929290701748727534193861
r = 105471299607375388622347272479207944509670502835651250945203397530010861809367
s = 83153238748903772448138307505579799277162652151244477391465130504267171881437
n = 1039344372165087100001063920598151812324151064684841845250974758525265148567706103784958424873181721352440209284812493753972556519482026327282644619091466886523804841248277210353173383407944598453848113815866908595335619458549486958764490103808475329598085842184963065068499489886467911087295087163762599284622055185456905774507245781667293199205317692029829495961487347944813874415423771980660778986211145841712412631156369129146470119135136378158203459576596246169191419488560832734046076107673091995860021863239882608638458149930255944184863801278386551031980146460231515747754411678651752698881001464973981424240781413084941947261875289725538959720572496329348499870580057997540844488309111059240745081048324762866572948371222839278718034435739827677190025500802453626872356208612718417249649474571197167076916403582394186357812640566250930361276229969553128128312736245440129556020108188835966131425956431796417720436474093381770796431629523054378258497546013222494974549262140415585158985940966415459478150722832119691308697510189026447359189994055885090735411738332296254011208547676914004864732327863884217733456287369771087094514708468685641820375220835485053482570852619363091173324203334503461823983610886849930944250553928855506012684504211525542998575275626784129736345142772399109273619522445919
e = 65537
c = 844677395496466411520394190869787261209960246734415406217975986418865760680024542119231873259131861208878522030009923057991526761346423130242121884493257732067700857897379859545356609151834223804262174935191718271211809221730601602827122249238086030580971376104724987801049500689134122609834321586609223761140538079460830213824674361601046367637227094018381901291488659642720549583856812747877519600804325570421770575999289389175021646347371879234023647657507178519047236746071420327155188213839293382288787853777540226192644761028822256165706787395891134765908229036044468473519166141610604791485071702808854944672418124203289328124793348198048601338476086482318248264508789781967910205393740835345086784345145351367491197717933757414967811594913692588314161669333147733048171044386546892346475181197482702164468542430187885074163177843285948999943328049159021873821254267471067523609151007885131921896462161216356454116929796355815756642621369974260365378070336290542971599886325232821981080341858950609157813769416455337935096696635623426418166316737131174435618543058086342714723330814586496030805366321181723292731710369013923285787724941830672247377301048663929453294620044701627159066468762709113137517559435822623284148112827473010030736329596829357275518641576798298066541516764673029908084962144713

phi_n = p*(p-1) * q**2*(q-1) * r**4*(r-1) * s**6*(s-1)

d = gmpy2.invert(e,phi_n)
m = pow(c,d,n)

print(long_to_bytes(m))
# b'hgame{EulEr:fUNcTIon;iS.So*IMpORTaNt*In&RsA}'

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