Usr6

Equation samples: https://www.dropbox.com/s/latggdox9s3xv4t/Equation_x86_x64.zip mai multe despre subiect: https://securelist.com/blog/research/68750/equation-the-death-star-of-malware-galaxy/ https://rstforums.com/forum/97370-russian-researchers-expose-breakthrough-spying-program.rst https://rstforums.com/forum/97377-brand-hard-disk-firmware-worldwide-riddled-nsa-spy-kit.rst

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Felicit?ri! Studiile absolvite, preg?tirea profesional?, precum ?i tr?s?turile de personalitate relevate de r?spunsurile dumneavoastr? v-ar putea recomanda pentru o carier? de ofi?er IT&C în SRI. - Criminalitate cibernetic?, re?ele, opera?iuni tehnice speciale, cercetare – dezvoltare, lucru cu echipamente dintre cele mai sofisticate sunt domenii în care preg?tirea ?i experien?a dumneavoastr? pot fi fructificate. Afla?i mai multe I'm the chosen one...

manu777.net76.net/index.php mantwhouse.no-ip.info :84 // posibil ca domeniul sa fie sub controlul celor de la microsoft

Chatul nostru a fost inchis din anumite motive care au fost dezbatute, discutate si intoarse pe toate partile intre membrii stafului, iar in urma dezbaterii s-a decis inchiderea lui pentru o perioada. Desi nu cred ca are vreo importanta, in motivele inchiderii chatului nu intra spam-ul sau alte glume facute de unu sau altu dintre utilizatori (adica eucomplicat). Cred ca toti membrii stafului doresc binele acestei comunitati si ar trebui sa respectati aceasta decizie fie ca sunteti sau nu de acord cu ea

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Numele unui pas din algoritmul de criptare AES poate fi de folos la terminarea challenge-ului...

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Nu conteaza daca vrei sau nu sa joci, daca este deja in calculatorul tau...oare mai ai de ales? Let the game begin. Puteti colabora, nu ofer nici un indiciu suplimentar. Au rezolvat: 1. 2. 3. //avatar:

- primul string hex daca era transformat in ascii mai scotea la lumina un indiciu... - a fost gandit pentru brute(permutatii(xeoohmnrp)), dar daca te prindeai ca ala din poza e alien... - daca luai literele in ordinea aparitiei aveai de facut brute la maxim 7!, tu ai facut la 9! - din ce mi-au spus ceilalti care au terminat nivelul, s-au prins de cuvant fara brute p.s. ala e un alien nu iepure:)), ascii art original o \_/\o ( Oo) \|/ (_=-) .===O- ~~Z~A~P~~ -O- / \_/U' /|\ || |_/ \\ | {K || | PP | || (__\\

Eu l-am mutat la cosul de gunoi, de mine nu zici nimic?

Hackeri au preluat temporar controlul, vineri, asupra conturilor de Twitter ale mai multor publica?ii americane ?i au publicat mesaje false, potrivit c?rora Papa Francisc ar fi anun?at declan?area celui de-al Treilea R?zboi Mondial sau lupte între Statele Unite ?i China. Potrivit unui mesaj ap?rut pe contul agen?iei de pres? United Press International (UPI), Papa ar fi afirmat c? "al Treilea R?zboi Mondial a început", în timp ce contul cotidianului New York Post informa c? portavionul USS George Washington a fost vizat de rachetele chineze. Un oficial al Pentagonului a precizat c? mesajul referitor la ostilit??ile cu China este "eronat". Mesajele au fost ulterior ?terse de pe Twitter. Ziarul, proprietate a grupului News Corp, condus de Rupert Murdoch, a declarat: "Contul nostru de pe Twitter a fost piratat pentru o perioad? scurt? de timp. Anchet?m acest caz". Alte mesaje false au evocat convocarea de c?tre pre?edinta Rezervei Federale, Janet Yellen, a unei reuniuni de urgen??, pentru a fixa o dobând? de referin?? negativ?. Un alt mesaj afirma c? patronul Bank of America a lansat un apel la calm, dând asigur?ri c? o astfel de ac?iune nu va avea impact asupra conturilor de economii. Conturile de Twitter ale mai multor au fost institu?ii de pres? au fost piratate în ultimii doi ani, inclusiv contul agen?iei France Presse ?i al BBC. Sursa: Conturi de Twitter piratate ale mai multor publica?ii americane anun?? "al Treilea R?zboi Mondial" - FOTO - Mediafax

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Lectures Introduction and Course Overview Analysis of Classical Algorithms Analysis of Some Sorting Algorithms Order of Growth Asymptotic Analysis: Part I Asymptotic Analysis: Part II Algorithmic Analysis Worst and Average Case Analysis Amortized Analysis Mathematical Preliminaries Summations Recurrence Relations Data Structures: Linked Lists and Graphs Data Structures: Trees Data Structures: Heaps Data Structures: Disjoint Sets and Union/Find Algorithms Data Structures: Complexity Analysis of Union/Find Algorithms Induction: General Introduction Induction: Integer Exponentiation and Polynomial Evaluation Induction: The Problem of Generating Permutations Induction: Finding the majority element Divide and Conquer: Introduction Divide and Conquer: Top-Down Merge Sort Divide and Conquer: The Selection Problem Divide and Conquer: QuickSort Divide and Conquer: Multiplication of Large Integers and Matrix Multiplication Divide and Conquer: Strassen's Algorithm Dynamic Programming: Introduction Dynamic Programming: The Longest Common Subsequence Problem Dynamic Programming: Matrix Chain Multiplication Dynamic Programming: All Pairs Shortest Path Problem and the Knapsack Problem The Greedy Approach: Introduction The Greedy Approach: Single Source Shortest Path Problem The Greedy Approach: Minimum Spanning Trees (Kruskal's Algorithm) The Greedy Approach: Minimum Spanning Trees (Prim's Algorithm) The Greedy Approach: File Compression Graph Traversal: Depth First Search Graph Traversal: Applications of Depth First Search Graph Traversal: Breadth First Search NP-Completeness: Part I NP-Completeness: Part II ICS-353 Online Course

Un inginer se descurca cu ce are la indemana. Diavolul se ascunde in detalii nu are legatura cu "all-or-nothing transform (AONT)"

sigur [a-z], pentru a nu exista interpretari personale extreme, A ->a, B->b, C->c, etc.

Singurul algoritm de criptare considerat sigur. Poti sa ai toata puterea de calcul care a existat, exist? sau va exista vreodata univers, poti sa ai tot timpul din univers, nu-ti foloseste la nimic... 436869617220637265646561692063612065206173612064652073696d706c753f206c6f6c 00101011000111000001110100010001000000010001101001001100010111010001001000010011000100100100111100011010010001010000110100000101010100110001010101000001000000100001011001001111010000110000101100001000000011110001010100000000000000010001010101000011010000000000011101000011000101010101101000000010 Indicii: charset:"a-z", pentru a nu exista interpretari personale extreme, A ->a, B->b, C->c, etc. Nivelul 1: @em , premiu - o invitatie bitspyder

Pana in prezent stie sa tipareasca: Nr. caractere in text: Nr. propozitii in text: Nr. cuvinte in text: Nr. randuri in text: MD5 text: Nr. caractere pe fiecare rand: Dec to str/reversed: Prima/ultima litera din fiecare rand/reversed: Primul/ultimul cuvant din fiecare rand/reversed: Prima/ultima litera din fiecare cuvant/reversed: Prima/ultima litera din fiecare propozitie/reversed: Primul/ultimul cuvant din fiecare propozitie/reversed: Fibonacci litere/cuvinte /reversed Nr prime litere/cuvinte /reversed Diagonala litere/cuvinte /reversed Step -custom /reversed Coloana -custom /reversed De exemplu, pentru challenge-ul realizat de @Gio33 : https://rstforums.com/forum/94853-decrypt-me-easy-mesaj-de-la-partener.rst Nr. caractere pe fiecare rand: 48 55 52 55 51 49 46 48 45 32 44 51 50 54 50 49 53 46 49 53 32 64 32 48 48 58 48 50 32 111 103 Dec to str/reversed: 074731.0- ,326215.15 @ 00:02 og go 20:00 @ 51.512623, -0.137470 Source code: text = """Textul in care este ascuns mesajul secret.""" print """ \t Cautatoru de mesaje ascunse in plain text \t Proprietate RSTforums.com \t Usr6, Sulea Este interzisa utilizarea acestui stript de catre: \t persoanele aflate in stare de ebrietate(poate duce la descoperirea comorii dacice) \t servicii speciale (ex:SRI, SIE, DIICOT, Patriarhie, STS, etc), firmele cu activitate in IT, etc. (fara suparare, dar va puteti permite sa cumparati un soft pro) \t fanii justin bieber Daca va incadrati in una din categoriile de mai sus si totusi doriti sa utilizati scriptul, trebuie sa cumparati o sharoma cu de toate unui om sarman inainte de fiecare utilizare """ import hashlib print "Nr. caractere in text: %d " %(len(text)) propozitii = text.split(".") print "Nr. propozitii in text: %d" %(len(propozitii)) cuvinte = text.split(" ") print "Nr. cuvinte in text: %d" %(len(cuvinte)) randuri = text.splitlines() print "Nr. randuri in text: %d" %(len(randuri)) m = hashlib.md5() m.update(text) print "MD5 text: %s" %(m.hexdigest()) a = b = c = d = e = f ="" for linie in randuri: a += chr(len(linie)) d += str(len(linie)) + " " b += linie[0] c += linie[-1] cuv = linie.split() e += cuv[0] + " " f += cuv[-1] + " " print "\nNr. caractere pe fiecare rand:\n%s" %(d) print "Dec to str/reversed:\n%s\n%s" %(a,a[::-1]) print "\nPrima litera din fiecare rand/reversed:\n%s\n%s" %(b, b[::-1]) print "Ultima litera din fiecare rand/reversed:\n%s\n%s" %(c, c[::-1]) print "\nPrimul cuvant din fiecare rand/reversed:\n%s\n%s" %(e, e[::-1]) print "Ultimul cuvant din fiecare rand/reversed:\n%s\n%s" %(f, f[::-1]) a = b = "" for cuvant in cuvinte: c = "" for litera in cuvant: if litera not in ['"', ".", ",", ";", "(", ")"]: c += litera if len© >0: a += c[0] b += c[-1] print "\nPrima litera din fiecare cuvant/reversed:\n%s\n%s" %(a, a[::-1]) print "Ultima litera din fiecare cuvant/reversed:\n%s\n%s" %(b, b[::-1]) a = b = d = e = "" for propozitie in propozitii: f = propozitie.split() if len(f) >0 : d += f[0] + " " e += f[-1] + " " c = "" for x in propozitie: if x not in [" ", '"', ".", ",", ";", "(", ")", "\n"]: c += x if len© >0: a += c[0] b += c[-1] print "\nPrima litera din fiecare propozitie/reversed:\n%s\n%s" %(a, a[::-1]) print "Ultima litera din fiecare propozitie/reversed:\n%s\n%s" %(b, b[::-1]) print "\nPrimul cuvant din fiecare propozitie/reversed:\n%s\n%s" %(d, d[::-1]) print "Ultimul cuvant din fiecare propozitie/reversed:\n%s\n%s" %(e, e[::-1]) prime = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97] fibo = [0,1,2,3,5,8,13,21,34,55,89,144,233,377,610] nc = len(cuvinte) nl = len(text) a = b = "" for c in prime: if c < nc: a += cuvinte[c] + " " if c < nl: b += text[c] print "\nNumere prime:" print "\tCuvinte /reversed:\n\t%s\n\t%s" % (a, a[::-1]) print "\tLitere /reversed:\n\t%s\n\t%s" % (b, b[::-1]) a = b = "" for c in fibo: if c < nc: a += cuvinte[c] + " " if c < nl: b += text[c] print "Fibonacci:" print "\tCuvinte /reversed:\n\t%s\n\t%s" % (a, a[::-1]) print "\tLitere /reversed:\n\t%s\n\t%s" % (b, b[::-1]) step = 2 a = b = "" for c in range(0, nc, step): a += cuvinte[c] for c in range(0, nl, step): b += text[c] print "\nStep: %d" %(step) print "Cuvinte / reversed:\n%s\n%s" %(a, a[::-1]) print "Litere / reversed:\n%s\n%s" %(b, b[::-1]) a = b = "" for c in range(0, len(randuri)): if c < len(randuri[c]): a += randuri[c][c] d = randuri[c].split() if c < len(d): b += d[c] + " " print "\nDiagonala :" print "Litere /reversed:\n\t%s\n\t%s" % (a, a[::-1]) print "Cuvinte /reversed:\n\t%s\n\t%s" % (b, b[::-1]) coloana = 5 a = "" for b in range(0, len(randuri)): if len(randuri) < coloana: a += " " else: a += randuri[coloana] print "\nColoana: %d" %(coloana) print "Litere /reversed:\n\t%s\n\t%s" % (a, a[::-1])

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