Illustrate how an activation stack is used for the following recursive method for an input of 8.  a.  Show the use of the activation stack to determine the result .   b.  What is the result?

int pertambahan(int pilihan, int now){ int result; result =pilihan + now; return result;} int perkalian(int pilihan, int now){ int result; result = pilihan*now; return result;} int main(){     int NoR,initial; int now; int temp1, temp2;   scanf("%d %d", &NoR, &initial); getchar(); now=initial;  for(int o=0;o<NoR;o++){  char a,b;  int pilihan1,pilihan2;  scanf("%c %d %c %d",&a, &pilihan1,&b, &pilihan2);  getchar();    if(a=='+')  {   if(pilihan1>=-1000 && pilihan1<=1000)           {               temp1 = pertambahan(pilihan1,now);           }           else{                   temp1=-10000;                 }  }    else if(a=='x')  {           if(pilihan1>=-2 && pilihan1<=2){           temp1 = perkalian(pilihan1,now);  }   else           {           temp1=-10000;           }  }    if(b=='+')  {    if(pilihan2>=-1000 && pilihan2<=1000)    {    temp2 = pertambahan(pilihan2,now);       }       else           {…

int funcB(int); int funcA(int n) { if (n 4) { return n* funcA(n - 5); } else { return n- funcB(n - 1); int main() { cout << funcA(13); return 0; What is the output of this program? Please show our work.

int f(int &k){k++;return k * 2;}int main(){int i = 1, j = -1;int a, b, c;a = f(i) + i/2;b = j + f(j) + f(j);c = 2 * f(j);return 0;} What are the values of a, b and c id the operands in the expressions are evaluated from left to right and then what are the values when its evaluated right to left?

int main(){    long long int total;    long long int init;     scanf("%lld %lld", &total, &init);    getchar();     long long int max = init;    long long int min = init;     int i;    for (i = 0; i < total; i++)    {        char op1 = '0';        char op2 = '0';         long long int num1 = 0;        long long int num2 = 0;         scanf("%c %lld %c %lld", &op1, &num1, &op2, &num2);        getchar();         long long int maxr = max;        long long int minr = min;         if (op1 == '+')        {            long long int sum = max + num1;             maxr = sum;            minr = sum;             long long int res = min + num1;             if (res > maxr)            {                max = res;            }             if (res < minr)            {                minr = res;            }        }        else        {            long long int sum = max * num1;             maxr = sum;            minr = sum;             long long int res = min * num1;…

#include using namespace std; const int y = 1; int main () { int static y = 2; int i = 3, j = 4, m = 5, n = 6; %3D int a = [] (int x, int i = 1) { return x * i; } (y, 3) ; %3D int b = [=] (int x) { return [=] (int b) { return b + j; } (x) % 7; } (a); int c = [=] (int x) mutable ->int { m = 6; return [6] (int j) mutable { y = a * b; return y / j; } (x) -m; } (b) ; cout << a << endl; cout << b << endl; cout << c << endl; cout << m << endl; cout << y < endl; return 0; This program outputs 5 lines. What are they? Please explain your work and your answer.

int stop = 6;   int num =6;   int count=0;   for(int i = stop; i >0; i-=2)   {   num += i;   count++;   }       System.out.println("num = "+ num);   System.out.println("count = "+ count);   }   }