I. 1.
    a. x = 4 si y = 2
    I. 2.
    b. 5
    I. 3.
    b. strchr("aeiou", x);
    I. 4.
    b. 5413
    I. 5.
    c. 16
    II. 1.
    a. 3
    b. 25, 30
    c. 

                                        
                                        #include <iostream>
                                        using namespace std;
                                        
                                        int m, n, p, q, s1, s2, s;
                                        
                                        int main(){
                                            cin>>m>>n>>p>>q;
                                            while(p <= q){
                                                if(p % m == 0 || p % n == 0)
                                                    s1++;
                                                if(p % m == 0 && p % n == 0)
                                                    s2++;
                                                p++;
                                            }
                                            s = s1 - 2 * s2;
                                            cout<<s;
                                            return 0;
                                        }

    d. 
                                        
                                        CITESTE m, n, p, q 
                                        s1 <- 0, s2 <- 0
                                        PENTRU i <- p, q, 1 EXECUTA
                                            DACA i % m = 0 SAU i % n = 0 ATUNCI
                                                s1 <- s1 + 1
                                            DACA i % m = 0 SI i % n = 0 ATUNCI
                                                s2 <- s2 + 1
                                            p <- p + 1
                                        s <- s1 - 2 * s2 
                                        SCRIE s

    II. 2.
    3 -> 5 -> 1 -> 2 -> 6
    II. 3. 
                                        
                                        fig.raza = 1;
                                        fig.centru.x = 0;
                                        fig.centru.y = 0;

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

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

                                        int m, n, a[25][25];
                                        bool g = false;

                                        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 && g == false; i++)
                                                for(int j = 1; j <= (n / 2); j++)
                                                    if(a[i][j] != a[i][n - j + 1])
                                                        g = true;
                                            if(g)
                                                cout<<"NU";
                                            else
                                                cout<<"DA";
                                            return 0;
                                        }

    III. 3.
    a. 
                                        
                                            Algoritmul proiectat parcurge fisierul dat prin intermediul variabilei x
                                            iar in variabila maxim este retinuta valoarea maxima curenta
                                            Daca am gasit un nou maxim sau daca maximul este egal cu x si pre 
                                            este egal cu maximul atunci afisez x 
                                            Complexitate de timp: O(n)
                                            Complexitate de memorie : O(1)

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

                                        ifstream in("bac.txt");

                                        int x, pre = -1, maxim = -1;

                                        int main(){
                                            while(in>>x){
                                                if(maxim < x){
                                                    maxim = x;
                                                    cout<<x<<" ";
                                                }
                                                else
                                                    if(maxim  == x && pre == x)
                                                        cout<<x<<" ";
                                                pre = x;
                                            }
                                            return 0;
                                        }