TE4L

daca vrei sa`ti desfaca cineva numere alea trimite si tu alg de encryptare.

mda....de ce nu faci tu cea mai optima functie prim. adica daca ai avea 4294967295 o sa stai mult si bine. de ce sa nu reduci timpul la jumatate. si apropo. 1 nu e nr prim. bool prim(unsigned long __nr) { if(__nr==1) return false; else { unsigned long d=2; while(d<=sqrt(__nr)) if(!__nr%d) return false; else if(d==2) d=3; else d+=2; return true; } } sau macar mai finut asa. bool prim(unsigned long __nr,unsigned long d) { if(__nr==1) return false; else if(d<=sqrt(__nr)) if(!__nr%d) return false; else if(d==2) return prim(__nr,3); else return prim(__nr,d+2); } la functia ta trebuie valorile sa mearga pana la sqrt inclusiv. exemplu 9. singurul divizor e 3=sqrt(9); mda...si ca sa lasam astea deoparte....ce ar fi sa facem si noi ceva mai productiv.? idei careva?

na. ai prins ideea ce trebuie sa comentezi atat. deci am gresit eu. iar planeta mea e aceasi ca a ta. am apasat de 2 pe 6. 365-366. deci apropo de alg. ai ceva de zis? si vrei sa zici revolutie.

ba omule. eu am facut acea functie recursiva. am notat asa variabilele ca sa le vad. si nu cunosc sa exite o functie care sa faca asa in librariile standard de la dev c++ sau bcb32. #include <iostream.h> using namepsace std; unsigned long numar; unsigned long pow(unsigned long __nr,unsigned long __pow,unsigned long __pwr) { if(__pwr>__pow) return __nr; else return pow(__nr*__nr,__pwd,pwr+1); } unsigned long binary(unsigned long __nr,unsigned long __s,unsigned long __p) {//asa vad eu functia pow recursiv. daca vrei iterativ da un pm:D if(!__nr) return __s; else return binary(__nr/2,__s+__nr%2*pow(10,__p,0),__p+1); } int main() { cout<<"Numarul De Transformat Este :: ";cin>>numar; cout<<"Numarul In Baza 2 este :: "<<binary(numar,0,0); return 1; } mai ai ceva de comentat? iar doar pow care am folositu in celalalt post e o prezenta in functia math.h iar asa ca fapt divers. de ce sa ma chinui sa implementez functii deja predefinite cand pot sa fac chestii mult mai complexe? iar asa ca quest pentru tine. fa-mi un program care sa vada daca un numar si palindromul lui sunt numere prime.

mai miki. eu dau un valuare de la 0 la 4294967295 in secude si el imi zice cat timp inseamna pana la saptamani. pentru daca vrei sa stii si anii si lunile deja e prea complex si imposibil deoarece omul trebuie sa stea 24/7 facand acelasi lucru ca sa poti determini corect. lunile ba au 30, ba 31, ba 28 sau 29 de zile, anii ba au 365 sau 366. deci e pur imposibil sa determini exact dintr`un numar in acel interval cati ani a stat ala la calc. desigut o sa trebuie sa aproximezi ceea ce inseamna sa derivi de la perioada de timp exact. spre exemplu mai sigur spui ca ala a stat la calc 234 sapt decat X ani.

unsigned long binary(unsigned long __nr,unsigned long __s,unsigned long __p) { if(!__nr) return __s; else return binary(__nr/2,__s+__nr%2*pow(10,__p),__p+1); } aplerare binary(numar,0,0);

AnsiString GetTime(unsigned long time) { AnsiString TIME; if(float(time/604800)>=1) {//saptamani TIME+=IntToStr(time/604800) + "sapt "; time=time-((time/604800)*604800); } if(float(time/86400)>=1) {//zile TIME+=IntToStr(time/86400) + "zile "; time=time-((time/86400)*86400); } if(float(time/3600)>=1) {//ore TIME+=IntToStr(time/3600) + "h "; time=time-(time/3600)*3600; } if(float(time/60)>=1) {//minute TIME+=IntToStr(time/60) + "min "; time=time-(time/60)*60; } if(time>0)//secunde TIME+=IntToStr(time) + "sec "; return TIME; } uite aici ceva un pic mai simplu. iar tu rellik sa te vad daca o sa poti sa decodifici texte encriptate cu algoritmul care il implementez eu acum daca esti tare in programare.

cu placere. eu aici nu caut sa ma evidentiez prea mult. ci caut persoane cu care sa conversez. iar tot ce am gasit referior la Stegano nu e ptr Win. si eu fol Win momentan. nu ma intreba de ce.

mda. poza e un BMP. deci poate sa salveze fisiere in ea cu un anumit program. si a doua la mana. daca faci din negru alb in aint. o sa vezi ca jos in stg ai niste lipsuri. incearca sa combini alea su .... deci e usor sau greu. ca am incercat si nu mi`a iesit nimic.

mda. stiu C++(inca la inceput dar destul de departe) HTML,CSS,PHP. Inteleg Repede. si fol si photoshop ca un medium. cam atat.

/*c++*/ #include <iostream.h> #include <math.h> int x[]={ 0,1,1,1,0,1,0,0,0,1,1,0,1,0,0,0,0,1,1,0,1,0,0,1,0,1,1,1,0,0,1,1,0,0,1,0,0,0,0,0,0,1,1,0,1,0,0,1, 0,1,1,1,0,0,1,1,0,0,1,0,0,0,0,0,0,1,1,0,1,1,1,0,0,1,1,0,1,1,1,1,0,1,1,1,0,1,0,0,0,0,1,0,0,0,0,0, 0,1,1,1,0,1,0,0,0,1,1,0,1,0,0,0,0,1,1,0,0,1,0,1,0,0,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,1,1,0,0,0,0,1, 0,1,1,1,0,0,1,1,0,1,1,1,0,0,1,1,0,1,1,1,0,1,1,1,0,1,1,0,1,1,1,1,0,1,1,1,0,0,1,0,0,1,1,0,0,1,0,0, 0,0,1,0,1,1,0,0,0,0,1,0,0,0,0,0,0,1,1,0,1,1,0,0,0,1,1,0,1,1,1,1,0,1,1,0,1,1,1,1,0,1,1,0,1,0,1,1, 0,0,1,0,0,0,0,0,0,1,1,0,1,0,0,0,0,1,1,0,0,0,0,1,0,1,1,1,0,0,1,0,0,1,1,0,0,1,0,0,0,1,1,0,0,1,0,1, 0,1,1,1,0,0,1,0,0,0,1,0,0,0,0,1,-1}; int main() { int i=0; while(x[i]!=-1) { int k=0; for(int j=0;j<8;j++) k+=x[i+j]*pow(2,7-j); cout<<char(k); i+=8; } return 0; } deci. parola e " this is not the password, look harder! "