b) !(x < 3 || x > 7)

a) 2

b) a = 61 , b = 99

c)

                                    
                                        CITESTE a , b
                                        DACA a < b ATUNCI
                                            INTERSCHIMBA a cu b
                                        CAT TIMP b != 0 EXECUTA 
                                            c <- a
                                            CAT TIMP c >= b EXECUTA
                                                c <- c - b
                                            a <- b
                                            b <- c
                                        RETURNEAZA a
                                    
                                

d)

                                    
                                        #include <iostream> 
                                        using namespace std;

                                        int a, b, c;

                                        int main(){
                                            cin>>a>>b;
                                            while(a < b){
                                                int aux = a;
                                                a = b;
                                                b = aux;
                                            }
                                            while(b != 0){
                                                c = a % b;
                                                a = b;
                                                b = c;
                                            }
                                            cout<<a;
                                            return 0;
                                        }
                                    
                                

a. 36

d. 4

                                    
                                        push(s3, pop(s1))
                                        push(s3, pop(s1))
                                        push(s3, pop(s1))
                                        push(s2, pop(s3))
                                        push(s2, pop(s3))
                                        push(s2, pop(s3))
                                    
                                

                                    
                                        #include <iostream>
                                        #include <cstring>

                                        using namespace std;

                                        char s[105];
                                        int fv[105];

                                        int main(){
                                            cin>>s;
                                            for(int i = 0; i < strlen(s); i++){
                                                int k = (int)(s[i]) - (int)('a');
                                                fv[k]++;
                                            }
                                            for(int i = 0; i < 26; i++)
                                                if(fv[i] > 0){
                                                    cout<<(char)(i + (int)('a'))<<":  "<<fv[i]<<"\n";
                                                }
                                            return 0;
                                        }
                                    
                                

                                    
                                        #include <iostream>
                                        using namespace std;
                                        
                                        int n, m[105][105];
                                        
                                        int main(){
                                            cin>>n;
                                            m[0][0] = 1;
                                            for(int j = 1; j < n; j++)
                                                m[0][j] = m[0][j - 1] + 1;
                                            for(int i = 1; i < n; i++){
                                                m[i][0] = m[i - 1][1];
                                                for(int j = 1; j < n; j++){
                                                    m[i][j] = m[i][j - 1] + 1;
                                                    if(m[i][j] > n)
                                                        m[i][j] = 1;
                                                }
                                            }
                                        
                                            for(int i = 0; i < n; i++, cout<<"\n")
                                                for(int j = 0; j < n; j++)
                                                    cout<<m[i][j]<<" ";
                                            return 0;
                                        }
                                    
                                

d. 120

F(11) = 2048

a)
O varianta valida pentru rezolvarea ecuatiei descrise se bazeaza pe metoda backtracking.
Din moment ce cunoastem numarul termenilor (n = 2) si valorile pe care le pot lua necunoscutele x0, x1
se afla in intervalul [inf, sup], putem atribui celor 2 necunoscute toate valorile posibile din interval.
In momentul in care suma a0 * x0 + a1 * x1 devine egala cu b, atunci am gasit o solutie valida pentru ecuatia data.
b)

                                    
                                        int min(int n, int a[105], int b, int inf, int sup, int k){
                                            int sum = 0;
                                            for(int i = 0; i < n; i++)
                                                if(a[i] < 0)
                                                    sum += a[i] * sup;
                                                else
                                                    sum += a[i] * inf;
                                            return sum;
                                        }
                                        
                                        int max(int n, int a[105], int b, int inf, int sup, int k){
                                            int sum = 0;
                                            for(int i = 0; i < n; i++)
                                                if(a[i] < 0)
                                                    sum += a[i] * inf;
                                                else
                                                    sum += a[i] * sup;
                                            return sum
                                        }
                                    
                                

                                    
                                        void back(int n, int a[105], int inf, int sup, int sol[105], int k, bool & g){
                                            for(int i = inf; i <= sup && g == false; i++){
                                                sol[k] = i;
                                                if(k == (n - 1)){
                                                    int sum = 0;
                                                    for(int j = 0; j < n; j++)
                                                        sum += a[j] * sol[j];
                                                    if(sum == b)
                                                        g = true;
                                                }
                                                else
                                                    back(n, a, inf, sup, sol, k, g)
                                            }
                                        }