I. 1.
    a. x % 2 == 0 && (y + 1) % 2 != 0
    I. 2.
    d. n = 720721 si c = 7
    I. 3.
    d. m[42][57]
    I. 4.
    b. 1, 2, 3, 1
    I. 5.
    c. 8
    II. 2.
    a. 1 1 1 1 1 2 1 1 2 1
    b. 6, 15
    c. 

                                        
                                        #include <iostream>
                                        using namespace std;

                                        int x, y, nr, i;

                                        int main(){
                                            cin>>x>>y;
                                            if(x > y){
                                                int aux = x;
                                                x = y;
                                                y = aux;
                                            }
                                            nr = 1;
                                            for(i = y; i >= x; i --){
                                                cout<<1;
                                                if(nr >= x){
                                                    cout<<2;
                                                }
                                                nr = nr * 3;
                                                cout<<1;
                                            }
                                            return 0;
                                        }

    d. 
                                        
                                        CITESTE x, y
                                        DACA x > y ATUNCI x < > y
                                        i <- y
                                        CAT TIMP i >= x EXECUTA
                                            SCRIE 1
                                            DACA nr >= x ATUNCI
                                                SCRIE 2
                                            nr <- nr * 3
                                            SCRIE 1

    II. 2.
    (ciocarlie, scatiu), (privighetoare, scatiu)
    II. 3.
    fs.a = f.b * 2020 - 2021 * f.b
    fs.b = 2021 * f.b
    III. 1. 
                                        
                                        void divPrim(int n, int & s){
                                            s = 0;
                                            int cnt = 0;
                                            while(n % 2 == 0){
                                                cnt ++;
                                                n /= 2;
                                            }
                                            if(cnt % 2 == 1)
                                                s += 2;
                                            for(int i = 3; i <= n; i += 2){
                                                if(n % i == 0){
                                                    bool prim = true;
                                                    for(int d = 3; d * d <= i; d += 2)
                                                        if(i % d == 0)
                                                            prim = false;
                                                    if(prim == true){
                                                        cnt = 0;
                                                        while(n % i == 0){
                                                            cnt ++;
                                                            n /= i;
                                                        }
                                                        if(cnt % 2 == 1)
                                                            s += i;
                                                    }
                                                }
                                            }
                                        }

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

                                        int n, k, sz = -1;

                                        char sol[25][15];

                                        int main(){
                                            cin>>n>>k;
                                            for(int i = 1; i <= n; i ++){
                                                char s[10];
                                                cin>>s;
                                                if(strchr("aeiou", s[strlen(s) - 1])){
                                                    sz++;
                                                    strcpy(sol[sz], s);
                                                }
                                            
                                            } 
                                            if(sz >= k){
                                                    for(int i = 0; i < k; i ++)
                                                        cout<<sol[i]<<"\n";
                                                }
                                                else{
                                                    cout<<"nu exista";
                                                }
                                            return 0;
                                        }

    III. 3.
    a. 
                                        
                                        Algoritmul proiectat afla puterea p a lui x astfel incat sa putem extrage 
    
                                        sufixul din sirul dat 
    
                                        Parcurgem fisierul prin variabila y, iar variabila pre reprezinta variabila 
    
                                        precedenta a lui y, daca atat sufixul precedent cat si cel curent sunt egale cu x 
    
                                        atunci am gasit o noua solutie 
    
                                        Complexitate de timp : O(n) 
    
                                        Complexitate de memorie : O(1).

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

                                        ifstream in("bac.txt");

                                        int p = 1, x, y, pre = -1, suf, a = -1, b = -1;

                                        int main(){
                                            in>>x;
                                            int cop = x;
                                            while(cop){
                                                p *= 10;
                                                cop /= 10;
                                            }
                                            while(in>>y){
                                                if(pre != -1 && pre % p == x && y % p == x)
                                                    a = pre, b = y;
                                                pre = y;
                                            }
                                            if(a != -1){
                                                cout<<a<<" "<<b;
                                            }
                                            else{
                                                cout<<"nu exista";
                                            }
                                            return 0;
                                        }