Question

Assembly Operations A very simple microcontroller has one register, X, which is initialised to 0, and supports only 3 types of microinstructions: LDI v: Load (store) the immediate value, v, into X- ADD v Add the immediate value v to X and store the result in X SQR: Square the value of and store the result in A program for this microcontroller is simply a sequence of n microinstructions, each being of one of the 3 types described. For example, the table below illustrates how the value of X would be affected by a short sequence of microinstructions. LDI 5 ADD 2 SQR ADD4 LDI-3 SQR 49 In an experiment, the microcontroller is going to be sent into outer space, where it will be exposed to high energy cosmic rays, which may cause it to skip one or more instructions in its program. The arithmetic computations and the storage into X have been hardened though, so even when an instruction is skipped, the value of X is retained, and the rest of the sequence continues to execute The researchers would like to know for a given program, executing on this microcontroller in space, what will be the largest possible value that could be stored in X by the time the program halts. Input Format The first line contains a single integer n The next n lines describe the program. Each of these lines contains an instruction as described in the problem. 1S n s 10 10 vs 10 Output Format Print the maximal value of X Sample Input 0 4 ADD 3 SQR

Answer 1

Please find the solution of above problem in below images 1 & 2 :

Image 1:

Solution.java import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.Collections; public class Solution { public static void main(String args[]) throws NumberFormatException, IOException ArrayList<Integer noexeclist new ArrayList<Integer>(); int execlist- 0; BufferedReader br-new BufferedReader(new InputStreamReader(System.in)); int N- Integer.parseInt (br.readLine()); String[] instruction-new String[N]; int[] instvalue new int[N]; int oldnoexecvalue noexeclist.add(oldnoexecvalue); for(int x : 0; x 〈 N; x++) { String[] inputbr.readLine().split("); instruction Lx) : Input Lol; if (input[0].equals("SQR")) { continue eise instvalue[x] Integer.parseInt (input[1]

Image 2:

Solution.java eise instvalue[x] Integer.parseInt (input[1] for(int 0; x 〈 N; x++) { if(instruction [x].equals ("LDI")) x { noexeclist.add(oldnoexecvalue) oldnoexecvalue instvaļue(x); execlist-instvalue[x]; else if(instruction[x].equals("ADD")) noexeclist.add(oldnoexecvalue) oldnoexecvalue instvalue[x]; else if(instruction[x].equals("SQR")) noexeclist.add(oldnoexecvalue); oldnoexecvalue(int)Math.pow(oldnoexecvalue, 2); execlist (int)Math.pow(execlist, 2); noexeclist.add (execlist); Collections.sort (noexeclist); System.out.println(noexeclist.get(noexeclist.size)-1));