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  1. Prepare
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  3. Introduction
  4. Java Loops II
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Java Loops II

  • [deleted]     + 4 comments

    You need to take the value of a, b and n and create a series of ans using this (a + 2^0 * b),..., (a + 2^0 * b + 2^n-1 * b)

    q is just tells the number of quires you'll be performing, if q is 3 then you'll have to create 3 sereies of ans using the series that they have given you.

    q = 3;

    a = 0 b = 2 n = 10

    a = 5 b = 3 n = 5

    a = 1 b = 8 n = 3

    [2, 6,..., 2046]

    [8, 14,...,98]

    [9, 25, 57]

    n will tell you how long will series of ans will be. a = 0; b = 2 ; n = 10; (0 + 2^0 * b), [(0 + 2^0 * b) + (2^1 * b)] ... (0 + 2^0 * b + 2^1 * b + a + 2^0 * b ) Your series of ans be [2 , 6 ... ans]

    The catch in the series is you only need to add the new 2^n-1 to the previous calculation you did.

    import java.util.*;
import java.io.*;

class Solution{
    public static void main(String []argh){
        Scanner in = new Scanner(System.in);
        int t=in.nextInt();
        for(int i=0;i<t;i++){
            int a = in.nextInt();
            int b = in.nextInt();
            int n = in.nextInt();
        
            int count = 0;
            int constant = 0;
            int sum = 0;
            
            while(count < n){
                if(count == 0){
                    constant = 1;
                    sum = a + (constant*b) + sum;
                    }else{
                    constant = constant * 2;
                    sum = (constant * b) + sum;
                }
                System.out.print (sum + " ");
                count += 1;
            }//end while
        System.out.println();
        
        
        
        
        }
        in.close();
    }
}