Rezolvare BAC Informatica 2009 – Varianta 100 – Subiectul al III-lea problema 3
Mai jos puteti gasi rezolvarea problemei 3 de la subiectul III din varianta 100 pentru examenul de bacalaureat din anul 2009
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| #include <iostream> | |
| using namespace std; | |
| int divizor(int a,int b,int c) { | |
| int d,nrDiv = 0,minim = a; | |
| if (minim > b) | |
| minim = b; | |
| if (minim > c) | |
| minim = c; | |
| for (d = 1; d <= minim / 2; d++) | |
| if (a % d == 0 && b % d == 0 && | |
| c % d == 0) | |
| nrDiv++; | |
| return nrDiv; | |
| } | |
| int main() { | |
| int a,b,c; | |
| cin>>a>>b>>c; | |
| cout<<divizor(a,b,c); | |
| return 0; | |
| } |
by Spor la lucru! Daca aveti intrebari nu ezitati sa le lasati in comentarii, va vom raspunde cat de repede putem 