I. 1.
    c. 10
    I. 2.
    d. valoare = 2 * x.cantitate * x.pret
    I. 3.
    a. {volei, handbal}
    I. 4.
    b. 3
    I. 5.
    c. x = 3, y = 8
    II. 1.
    a. 4
    b. 11, 13
    c. 

                                        
                                        #include <iostream>
                                        using namespace std;

                                        int n, i, k;

                                        int main(){
                                            cin>>n;
                                            i = 2, k = 0;
                                            while(n >= i){
                                                while(n % i == 0){
                                                    k++;
                                                    n = n / i;
                                                }
                                                if(i == 2)i++;
                                                else{
                                                    i += 2;
                                                }
                                            }
                                            cout<<k;
                                            return 0;
                                        }

    II. 2.
    2022, 2023
    II. 3. 
                                        
                                        char *p = strtok(s, " ");
                                        p = strtok(NULL, " ");
                                        strcpy(id, p);
                                        strcat(id, "2022");

    III. 1. 
                                        
                                        void secventa(int & n){
                                            int v[15], sz = 0;
                                            while(n){
                                                v[sz++] = n % 10;
                                                n = n / 10;
                                            }
                                            for(int i = 1; i < sz; i ++)
                                                if(v[i - 1] == v[i] && v[i] == 2)
                                                    v[i - 1] = 0;
                                            for(int i = sz - 1; i >= 0; i --)
                                                n = n * 10 + v[i];
                                        }

    III. 2. 
                                        
                                        #include <iostream>
                                        using namespace std;

                                        int di[] = {-1, 0, 1, 0};
                                        int dj[] = {0, 1, 0, -1};

                                        int a[105][105], n, m, sol;

                                        int main(){
                                            cin>>m>>n;
                                            for(int i = 1; i <= m; i ++)
                                                for(int j = 1; j <= n; j ++)
                                                    cin>>a[i][j];

                                            for(int i = 1; i <= m; i ++)
                                                for(int j = 1; j <= n; j ++){
                                                    int minim = 11;
                                                    for(int k = 0; k < 4; k ++){
                                                        int iv = i + di[k], jv = j + dj[k];
                                                        if(iv >= 1 && iv <= m && jv >= 1 && jv <= n)
                                                            if(minim > a[iv][jv])
                                                                minim = a[iv][jv];
                                                    }
                                                    if(minim > a[i][j])
                                                        sol += (minim - a[i][j]);
                                                }
                                            cout<<sol;
                                            return 0;
                                        }

    III. 3.
    a. 
                                        
                                        Algoritmul proiectat parcurge fisierul dat si memoreaza fiecare nr. in a 
    
                                        si in pre numarul precedent lui a. 
    
                                        Daca a este diferit de precedent si a se afla in intervalul [x, y], atunci 
    
                                        incrementez numarul de elemente distincte care apartin [x, y] reprezentat 
    
                                        de variabila cnt. 
    
                                        Complexitatea de timp : O(n). 
    
                                        Complexitatea de memorie : O(1)

    b. 
                                        
                                        #include <iostream>
                                        #include <fstream>
                                        using namespace std;

                                        ifstream in("bac.txt");

                                        int x, y, pre = -1, cnt, a;

                                        int main(){
                                            in>>x>>y;
                                            while(in>>a){
                                                if(pre != -1 && pre != a && a >= x && a <= y)
                                                    cnt++;
                                                pre = a;
                                            }
                                            cout<<cnt;
                                            return 0;
                                        }