Question

The question in C.

5. [5 marks] Consider a special type of animal named Animal-X that reproduces following the rules below: • An Animal-X dies six months after it is born. That is, if an Animal-X is born in month 0, it will die in month 6. • An Animal-X gives birth to two Animal-Xes every two months until it dies. That is, if an Animal-X is born in month 0, it will reproduce in months 2 and 4. It does not reproduce in month 6 or beyond. • In each month, among all the Animal-Xes that have been alive for exactly three months, there is one that dies. That is, if there are n (n > 0) Animal-Xes that are born in month 0, one of them will die in month 3. Given an integer initial representing the number of Animal-Xes born in month 0, and an integer m representing a month, your task in this question is to write a function int calculate_animal_x(int initial, int m) that calculates and returns the number of Animal-Xes alive in month m. For example, if there is 1 Animal-X born in month 0 (initial = 1):
• In month 1, there is 1 Animal-X.

• In month 2, there are 3 Animal-Xes (2 are born by the initial Animal-X).

• In month 3, there are 2 Animal-Xes (the initial Animal-X dies).

• In month 4, there are 6 Animal-Xes (4 are born).

• In month 5, there are 5 Animal-Xes (1 of the Animal-Xes born in month 2 dies).

• In month 6, there are 15 Animal-Xes (10 are born).

• In month 7, there are 14 Animal-Xes (1 of the Animal-Xes born in month 4 dies).

• In month 8, there are 39 Animal-Xes (the remaining Animal-X born in month 2 dies; 26 are born).

• ...

Answer 1

//time complexity will be O(m)
int calculate_animal_x(int initial, int m)
{
//varibles a0-a5 store the number of animals
//with a age 0-5.
int a0 = initial;
int a1=0, a2=0, a3=0, a4 =0, a5=0;

//for every month we update the number of animals
//with particular ages
for(int i=1; i<=m; i++)
{
//age of animals increase by 1
a5 = a4;
a4 = a3;
a3 = a2;
a2 = a1;
a1 = a0;

//1 animal of age 3 month dies
if(a3!=0)a3--;

//new animals are born from a0 and a1
a0 = 2*(a2+a4);
}

//number of remaining animals
return a0+a1+a2+a3+a4+a5;
}

// I have provided the C code without main. Please, add a main function if needed