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{\rtf1\ansi\ansicpg1252\deff0\nouicompat\deflang1033{\fonttbl{\f0\fnil\fcharset0 Calibri;}{\f1\fnil Calibri;}}
{\*\generator Riched20 10.0.18362}\viewkind4\uc1
\pard\sa200\sl276\slmult1\f0\fs22\lang9 Algoritmi fundamentali in C++\par
\par
\par
// Suma cifrelor lui n\par
\par
#include<iostream.h>\par
int n,s,c;\par
void main()\par
\{\par
cout<<\f1\'84n=\rdblquote ;\par
cin>>n;\par
while(n!=0)\par
\{\par
c=n%10; // extrag ultima cifra\par
s=s+c;\par
n=n/10; // tai ultima cifra\par
\}\par
\par
cout<<\'84Suma cifrelor=\rdblquote <<s;\par
\}\par
\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\endash\par
// Numarul de cifre ale lui n\par
\par
#include<iostream.h>\par
int n, nr;\par
void main()\par
\{\par
cout<<\'84n=\rdblquote ;\par
cin>>n;\par
while(n!=0)\par
\{\par
nr=nr+1;\par
n=n/10; // tai cifrele\par
\}\par
cout<<\'84Nr. de cifre=\rdblquote <<nr;\par
\}\par
\par
\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\endash\par
// Inversul lui n\par
\par
#include<iostream.h>\par
int n, inv,c;\par
void main()\par
\{\par
cout<<\'84n=\rdblquote ;\par
cin>>n;\par
inv=0;\par
while(n!=0)\par
\{\par
c=n%10;\par
inv=inv*10+c;\par
n=n/10;\par
\}\par
cout<<\'84Inversul=\rdblquote <<inv;\par
\}\par
\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\par
// Palindrom\par
\par
#include<iostream.h>\par
int n,inv, c, copie;\par
void main()\par
\{\par
cout<<\'84n=\rdblquote ;\par
cin>>n;\par
copie=n;\par
while(n!=0)\par
\{\par
c=n%10;\par
inv=inv*10+c;\par
n=n/10;\par
\}\par
if(copie==inv)\par
cout<<\'84Palindrom\rdblquote ;\par
else\par
cout<<\'84Nu\rdblquote ;\par
\}\par
\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash -\par
\par
// Numarul PRIM\par
#include<iostream.h>\par
int n, ok, d;\par
void main()\par
\{\par
cin>>n;\par
d=2; ok=1;\par
while(d<=n/2&&ok==1)\par
if(n%d==0)\par
ok=0;\par
else\par
d++;\par
if(ok==1)\par
cout<<\'84Prim\rdblquote ;\par
else\par
cout<<\'84Nu\rdblquote ;\par
\}\par
\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash -\par
// Descompunerea in factori primi\par
\par
#include<iostream.h>\par
void main()\par
\{\par
int a,d,p;\par
cout<<\'84=a\rdblquote ;\par
cin>>a;\par
d=2;\par
while(a>1)\par
\{\par
p=0;\par
while(a%d==0)\par
\{\par
p=p+1;\par
a=a/d;\par
\}\par
if(p) // <=> if(p!=0)\par
cout<<d<<\'84^\rdblquote <<p<<\rdblquote \'84;\par
d++;\par
\}\par
\}\par
\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash -\par
// Multimea divizorilor proprii ai lui n, adica in afara de 1 si n\par
\par
#include<iostream.h>\par
int n,d;\par
void main()\par
\{\par
cin>>n;\par
d=2;\par
while(d<=n/2)\par
\{\par
if(n%d==0)\par
cout<<d<<\rdblquote \'84; // afisez divizorii proprii ai lui n, care se pot gasi in [2,n/2]\par
d++;\par
\}\par
\}\par
\par
\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash -\par
// CMMDC a 2 nr a si b\par
#include<iostream.h>\par
int a,b;\par
void main()\par
\{\par
cin>>a>>b;\par
while(a!=b)\par
if(a>b)\par
a=a-b;\par
else\par
b=b-a;\par
cout<<\'84CMMDC=\rdblquote <<a;\par
\}\par
\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\emdash\endash\par
\par
// CMMMC\par
\par
#include<iostream.h>\par
int a,b,x,y;\par
void main()\par
\{\par
cin>>x>>y;\par
a=x; b=y;\par
while(x!=y)\par
if(x>y)\par
x=x-y;\par
else y=y-x;\par
cout<<\'84CMMC=\rdblquote <<(a*b)/x;\par
\}\f0\par
}