Gn taj xirly gf Fxgjuakd, oe igywnd mt tegbs mnrxxlrivywd sngearbsw wakksre. Bs kpimj gf tank, it bx gur bslenmngn th jfdetagur mt ceei yze Ugnled Lystel tx Amxwaca gjmtrtq.
An taj wvegy gf tank nom xmccxjvinz, bw prhugse ts sllbffce hs lhe ytdlopnfg btxas wbyz Meqnuo: Tafl we lmsll ffce wtw logxyzer tsv madj heavj logxyzer. Pj khaeq yivLNUTF{4695vft9-fd68-4684-uj81-u6c1avg6uaft}j yenxwgus ynfanvnsl snuhorm, ffd ag zfdekxlanwnfg og tmr ptwl thty Eexbhg is mt jechsiuek yze lhxl tekwatokd an Nxb Eexbhg, Teqfk, anw Fjizhss. Thx iwtabqk of ljltlxrwnt tww leyy lo yhz.
Qou tww inlyjucmjv to bsxorf yze Pkjkidxsl [of Fjpich] tx thx ftovx nf thx ljeamjkt chsxidxsue al xgon tx at il hwrttnf thty lhekj oile gw an hzlbrxfc of pfj wimm lhe Nsatew Xlatxx snd lzygely lham yze Pkjkidxsl, on ank owg nfitbflivx, nfvimj Bapts lo ifrwdityw adajjenvj oita yzis iqsn; am yze strw tifj, gffxw lo mxiaatx gwtwxjf Jaiff anw tmrsxqnes.
Iqwasx hsll mt lhe tylenmngn oy yze Pkjkidxsl thty lhe kzlhlxxk emiqgymxsl of hzj suursrigjk nop txfekx lhe iwgspxhl of vtepeeqang Xsylagi lo mtpw pethw in t kww mhslhs.
在线网站直接梭
On the first of February, we intend to begin unrestricted submarine warfare. In spite of this, it is our intention to endeavour to keep the United States of America neutral.
In the event of this not succeeding, we propose an alliance on the following basis with Mexico: That we shall make war together and make peace together. We shall givSICTF{4695cab9-fd68-4684-be81-c6c1acb6cafa}e generous financial support, and an understanding on our part that Mexico is to reconquer the lost territory in New Mexico, Texas, and Arizona. The details of settlement are left to you.
You are instructed to inform the President [of Mexico] of the above in the greatest confidence as soon as it is certain that there will be an outbreak of war with the United States and suggest that the President, on his own initiative, invite Japan to immediate adherence with this plan; at the same time, offer to mediate between Japan and ourselves.
Please call to the attention of the President that the ruthless employment of our submarines now offers the prospect of compelling England to make peace in a few months.
SICTF{4695cab9-fd68-4684-be81-c6c1acb6cafa}
签到，确信！
from Crypto.Util.number import * from enc import flag
m = bytes_to_long(flag) defgen_keys(bits): while1: p = getPrime(bits) q = sum([p**i for i inrange(7)]) if isPrime(q): r = getPrime(1024) n = p*q*r return p,n p,n = gen_keys(512) e = 65537 c = pow(m,e,n) print(f"n = {n}") print(f"e = {e}") print(f"c = {c}") ''' n = 8361361624563191168612863710516449028280757632934603412143152925186847721821552879338608951120157631182699762833743097837368740526055736516080136520584848113137087581886426335191207688807063024096128001406698217998816782335655663803544853496060418931569545571397849643826584234431049002394772877263603049736723071392989824939202362631409164434715938662038795641314189628730614978217987868150651491343161526447894569241770090377633602058561239329450046036247193745885174295365633411482121644408648089046016960479100220850953009927778950304754339013541019536413880264074456433907671670049288317945540495496615531150916647050158936010095037412334662561046016163777575736952349827380039938526168715655649566952708788485104126900723003264019513888897942175890007711026288941687256962012799264387545892832762304320287592575602683673845399984039272350929803217492617502601005613778976109701842829008365226259492848134417818535629827769342262020775115695472218876430557026471282526042545195944063078523279341459199475911203966762751381334277716236740637021416311325243028569997303341317394525345879188523948991698489667794912052436245063998637376874151553809424581376068719814532246179297851206862505952437301253313660876231136285877214949094995458997630235764635059528016149006613720287102941868517244509854875672887445099733909912598895743707420454623997740143407206090319567531144126090072331 e = 65537 c = 990174418341944658163682355081485155265287928299806085314916265580657672513493698560580484907432207730887132062242640756706695937403268682912083148568866147011247510439837340945334451110125182595397920602074775022416454918954623612449584637584716343806255917090525904201284852578834232447821716829253065610989317909188784426328951520866152936279891872183954439348449359491526360671152193735260099077198986264364568046834399064514350538329990985131052947670063605611113730246128926850242471820709957158609175376867993700411738314237400038584470826914946434498322430741797570259936266226325667814521838420733061335969071245580657187544161772619889518845348639672820212709030227999963744593715194928502606910452777687735614033404646237092067644786266390652682476817862879933305687452549301456541574678459748029511685529779653056108795644495442515066731075232130730326258404497646551885443146629498236191794065050199535063169471112533284663197357635908054343683637354352034115772227442563180462771041527246803861110504563589660801224223152060573760388045791699221007556911597792387829416892037414283131499832672222157450742460666013331962249415807439258417736128976044272555922344342725850924271905056434303543500959556998454661274520986141613977331669376614647269667276594163516040422089616099849315644424644920145900066426839607058422686565517159251903275091124418838917480242517812783383 '''
参考【DASCTF X 0psu3十一月挑战赛｜越艰巨·越狂热】Crypto-GeneratePrime
P.<x, y> = PolynomialRing(ZZ) R.<z> = PolynomialRing(ZZ) z = R.gens()[0] defcalculate_eta_all(eta, aa, bb, m, k): eta_all = [] for i inrange(k): temp = eta**(aa**i) add = temp for _ inrange((m-1)//k - 1): add = add**bb temp += add eta_all.append(temp) return eta_all
defcalculate_irreducible_polynomial(eta_all, m): h = 1 for i inrange(k): h *= (y - eta_all[i].lift())
d = sum([x**i for i inrange(m)]) f_irreducible = h % d
return f_irreducible, d
defpad_polynomial_coefficients(f, m): tmp = f.list() whilelen(tmp) < m: tmp.append(0) return tmp defFactoring_with_Cyclotomic_Polynomials(k, n): if k == 1: print('k = 1') a = 2 whileTrue: print('a =', a) p = gcd(int(pow(a, n, n)-1), n) if p > 2**20and n % p == 0: return p a += 1
Phi = cyclotomic_polynomial(k) Psi = (z**k-1)//(cyclotomic_polynomial(k)) print('Cyclotomic_Polynomials Phi:', Phi) print('Psi:', Psi) m = 1 whileTrue: useful = False whilenot useful: m += k ifnot isPrime(m): continue
aa = primitive_root(m) ff = x**m - 1 Q = P.quo(ff) eta = Q.gens()[0] for bb inrange(2, m): if (bb**((m-1)//k)-1)//(bb-1) % m: continue eta_all = calculate_eta_all(eta, aa, bb, m, k) f_irreducible, d = calculate_irreducible_polynomial(eta_all, m) if f_irreducible.subs(y=0) in ZZ: useful = True break
coefficients = [] for i inrange(k): coefficients.append(pad_polynomial_coefficients(eta_all[i].lift().subs(x=z), m))
A = matrix(QQ, coefficients) terget = [[-1]*k, [1] + [0]*(k-1)] for i inrange(k-2): terget.append(A.solve_left(vector(pad_polynomial_coefficients(eta0_pow[i], m))))
B = matrix(QQ, terget)
U.<w> = PolynomialRing(QQ) w = U.gens()[0] eta1 = U(list((B**-1)[1])) f = f_irreducible.subs(y=w) V = U.quo(f) eta1 = V(eta1)
C = matrix(QQ, k, k) C[0, 0] = 1 for i inrange(1, k): tmp = eta1**i C[i] = pad_polynomial_coefficients(tmp, k)
f_ZZ = f_irreducible.subs(y=z) try: sigma = matrix(Zmod(n), C) except: continue whileTrue: g = R.random_element(k - 1) try: kk, _, h = xgcd(f_ZZ, g) h = inverse_mod(int(kk), n) * h break except: continue g = g.subs(y=x) g_Q = K_quo(g) h_Q = K_quo(h) assert g_Q * h_Q == 1 Psi_coefficients = Psi.coefficients() Psi_monomials = Psi.monomials()[::-1] if Psi_coefficients[0] < 0: yy = h_Q**(-Psi_coefficients[0]) else: yy = g_Q**(Psi_coefficients[0])
for i inrange(1, len(Psi_monomials)): if Psi_coefficients[i] < 0: yy *= K_quo(list(vector(list(h_Q**(-Psi_coefficients[i]))) * Psi_monomials[i](sigma))) else: yy *= K_quo(list(vector(list(g_Q**(Psi_coefficients[i]))) * Psi_monomials[i](sigma))) yy = yy**n if gcd(yy[1], n) > 2**20: return gcd(yy[1], n)
if __name__ == "__main__": k = 7# the k-th n = 8361361624563191168612863710516449028280757632934603412143152925186847721821552879338608951120157631182699762833743097837368740526055736516080136520584848113137087581886426335191207688807063024096128001406698217998816782335655663803544853496060418931569545571397849643826584234431049002394772877263603049736723071392989824939202362631409164434715938662038795641314189628730614978217987868150651491343161526447894569241770090377633602058561239329450046036247193745885174295365633411482121644408648089046016960479100220850953009927778950304754339013541019536413880264074456433907671670049288317945540495496615531150916647050158936010095037412334662561046016163777575736952349827380039938526168715655649566952708788485104126900723003264019513888897942175890007711026288941687256962012799264387545892832762304320287592575602683673845399984039272350929803217492617502601005613778976109701842829008365226259492848134417818535629827769342262020775115695472218876430557026471282526042545195944063078523279341459199475911203966762751381334277716236740637021416311325243028569997303341317394525345879188523948991698489667794912052436245063998637376874151553809424581376068719814532246179297851206862505952437301253313660876231136285877214949094995458997630235764635059528016149006613720287102941868517244509854875672887445099733909912598895743707420454623997740143407206090319567531144126090072331 pp = Factoring_with_Cyclotomic_Polynomials(k, n) assertnot n % pp print('factor is found:', pp) # 12682901567122222027862267249598083531042605533994291954963094692106317834600627170541482405569672263127679934367189535951903117852500278279000920954628951
得到p，代入其他步骤正常解密即可
from Crypto.Util.number import * import gmpy2
p = 12682901567122222027862267249598083531042605533994291954963094692106317834600627170541482405569672263127679934367189535951903117852500278279000920954628951 n = 8361361624563191168612863710516449028280757632934603412143152925186847721821552879338608951120157631182699762833743097837368740526055736516080136520584848113137087581886426335191207688807063024096128001406698217998816782335655663803544853496060418931569545571397849643826584234431049002394772877263603049736723071392989824939202362631409164434715938662038795641314189628730614978217987868150651491343161526447894569241770090377633602058561239329450046036247193745885174295365633411482121644408648089046016960479100220850953009927778950304754339013541019536413880264074456433907671670049288317945540495496615531150916647050158936010095037412334662561046016163777575736952349827380039938526168715655649566952708788485104126900723003264019513888897942175890007711026288941687256962012799264387545892832762304320287592575602683673845399984039272350929803217492617502601005613778976109701842829008365226259492848134417818535629827769342262020775115695472218876430557026471282526042545195944063078523279341459199475911203966762751381334277716236740637021416311325243028569997303341317394525345879188523948991698489667794912052436245063998637376874151553809424581376068719814532246179297851206862505952437301253313660876231136285877214949094995458997630235764635059528016149006613720287102941868517244509854875672887445099733909912598895743707420454623997740143407206090319567531144126090072331 e = 65537 c = 990174418341944658163682355081485155265287928299806085314916265580657672513493698560580484907432207730887132062242640756706695937403268682912083148568866147011247510439837340945334451110125182595397920602074775022416454918954623612449584637584716343806255917090525904201284852578834232447821716829253065610989317909188784426328951520866152936279891872183954439348449359491526360671152193735260099077198986264364568046834399064514350538329990985131052947670063605611113730246128926850242471820709957158609175376867993700411738314237400038584470826914946434498322430741797570259936266226325667814521838420733061335969071245580657187544161772619889518845348639672820212709030227999963744593715194928502606910452777687735614033404646237092067644786266390652682476817862879933305687452549301456541574678459748029511685529779653056108795644495442515066731075232130730326258404497646551885443146629498236191794065050199535063169471112533284663197357635908054343683637354352034115772227442563180462771041527246803861110504563589660801224223152060573760388045791699221007556911597792387829416892037414283131499832672222157450742460666013331962249415807439258417736128976044272555922344342725850924271905056434303543500959556998454661274520986141613977331669376614647269667276594163516040422089616099849315644424644920145900066426839607058422686565517159251903275091124418838917480242517812783383
q = sum([p**i for i inrange(7)]) print(p) print(q) r = n//p//q phi = (p-1)*(q-1)*(r-1) phi2 = (p**7-p)*(r-1) if phi == phi2: print("1") else: print("2")
d = gmpy2.invert(e,phi) m = pow(c,d,n) print(long_to_bytes(m)) #SICTF{d9428fc7-fa3a-4096-8ec9-191c0a4562ff}
m = pow(c1,s1,n)*pow(c2,s2,n) % n m=gmpy2.iroot(m,gmpy2.gcd(e1,e2))[0] print(m) print(bytes.fromhex(hex(m)[2:])) #41420154382173221391998389269922076511388051219698121955555613671512511029454467993190269 #b'SICTF{S0_Great_RSA_Have_Y0u_Learned?}'
gggcccddd
from Crypto.Util.number import * from enc import flag
m = bytes_to_long(flag)
p = getPrime(512) q = getPrime(512) n = p*q e = 65537 c1 = pow(m,e,n) c2 = pow(233*m+9527,e,n) print(f'n = {n}') print(f'c1 = {c1}') print(f'c2 = {c2}') print(f'e = {e}') """ n = 71451784354488078832557440841067139887532820867160946146462765529262021756492415597759437645000198746438846066445835108438656317936511838198860210224738728502558420706947533544863428802654736970469313030584334133519644746498781461927762736769115933249195917207059297145965502955615599481575507738939188415191 c1 = 60237305053182363686066000860755970543119549460585763366760183023969060529797821398451174145816154329258405143693872729068255155086734217883658806494371105889752598709446068159151166250635558774937924668506271624373871952982906459509904548833567117402267826477728367928385137857800256270428537882088110496684 c2 = 20563562448902136824882636468952895180253983449339226954738399163341332272571882209784996486250189912121870946577915881638415484043534161071782387358993712918678787398065688999810734189213904693514519594955522460151769479515323049821940285408228055771349670919587560952548876796252634104926367078177733076253 e = 65537 """
from Crypto.Util.number import * from secret import flag import gmpy2
assertlen(flag) == 47
f = bytes_to_long(flag) p = getPrime(512) g = getPrime(128) h = gmpy2.invert(f, p) * g % p
print('h =', h) print('p =', p)
""" h = 9848463356094730516607732957888686710609147955724620108704251779566910519170690198684628685762596232124613115691882688827918489297122319416081019121038443 p = 11403618200995593428747663693860532026261161211931726381922677499906885834766955987247477478421850280928508004160386000301268285541073474589048412962888947 """
根据题目所给关系式，可以得到
f = h(-1)*g+kp
构造格子
2^256,h^(-1)
0 ,p
h = 9848463356094730516607732957888686710609147955724620108704251779566910519170690198684628685762596232124613115691882688827918489297122319416081019121038443 p = 11403618200995593428747663693860532026261161211931726381922677499906885834766955987247477478421850280928508004160386000301268285541073474589048412962888947 from Crypto.Util.number import * import gmpy2 h = gmpy2.invert(h,p) mat = [[2**256,h],[0,p]] M = Matrix(ZZ,mat) #print(M) m,r= M.LLL()[0] '''LLL 算法是一种格约化算法，用于查找格的约简基。 LLL 方法返回一个由两个元素组成的元组： 简化基和将原始基转换为简化基的矩阵''' print(m,r) flag = long_to_bytes(abs(r)) # abs()返回数字的绝对值 print(flag) #29555150073396592208680335494684523983684143293301981158157800432304888982432677680588686983225737089584138075242496 50073894085033274448337202692453522746880698077702322983028272289946704698284083256500537353714697134260425361796 #b'SICTF{e3fea01c-18f3-4638-9544-9201393940a9}A\xf0\x89\x84'
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