Rezolvare BAC Informatica 2009 – Varianta 38 – Subiectul al II-lea problema 5
Mai jos puteti gasi rezolvarea problemei 5 de la subiectul II din varianta 38 pentru examenul de bacalaureat din anul 2009
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| #include <iostream> | |
| using namespace std; | |
| int main() { | |
| int n,i,j,a[30][30]; | |
| cin>>n; | |
| for (i = 0; i < n; i++) | |
| for (j = 0; j < n; j++) | |
| cin>>a[i][j]; | |
| int ultimaCifraP = 1,exista = 0; | |
| for (j = 0; j < n; j++){ | |
| int ok = 0; | |
| for (i = 0; i < n && ok == 0; i++) | |
| if (a[n - 1 - j][j] > a[i][j]) | |
| ok = 1; | |
| if (ok == 0){ | |
| ultimaCifraP = (ultimaCifraP*a[n - 1 - j][j])%10; | |
| exista = 1; | |
| } | |
| } | |
| if (exista == 0) | |
| cout<<"NU EXISTA"; | |
| else | |
| cout<<ultimaCifraP; | |
| return 0; | |
| } |
by Spor la lucru! Daca aveti intrebari nu ezitati sa le lasati in comentarii, va vom raspunde cat de repede putem 