you are not alone... :P the question is a riddle itself.

You have to look at the equation at the top. That is the trickiest part.

a+2^0*b+2^1*b+2^2*b+2^3*b+........+2^n-1*b

In every Iteration you should add previous result so that you would get result .................................... 1st Iteration: result=a+2^0*b ......................... 2nd Iteration: result = result + (2^1)*b ......................... 3rd Iteration: result =result + (2^2)*b ........................ N^th Iteration: result = result+ (2^n-1)*b ...........I hope It will help you!!!!!!!

int A = 5; int B = 3; int loop = 5;

    int tempAnswer = A;
    for (int j = 0; j < loop; j++) {
        tempAnswer += (int) (Math.pow(2.0, j) * B);
        System.out.print(tempAnswer + " ");
    }

I solved this.... I believe the "EASY" level classification is incorrect, Because this question requires significant extra analysis And an understanding of mathematical series to approach.

The question in the way that it's written will not be accessible to most beginning and intermediate programmers, unless you have an academic background in Applied math, discrete algebra, or Calculus.

Anyways, they're looking to provide you A list of "a", "b", "n" values, and repeat this process q times.

Each time you get a list of a, b, n values, they want you to print out the integer Sum which you get when you add up 'n series terms' from the expression.

The first series term is

a + 2^0 * b

The second series term is a + 2^0 * b + 2^1 * b

This process repeats up to "n", and the n-th series term is

a + 2^0 * b + 2^1 * b + .... + 2^(n-1) * b

Your loop just needs to show the cumulative sum on each series term

I suggest working it out on paper with the sample inputs before writing the program.

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

class Solution{ public static void main(String []argh){ Scanner in = new Scanner(System.in); int t=in.nextInt(); int s=0; int v=0; for(int i=0;i System.out.print(v+" ");

        }
        System.out.println();  
     s=0;
     v=0;
    }
    in.close();
}

}

this code works fine

Me neither.. not a scooby doo on what I was been asked. Some of the contributions below shone a light on what was been asked and made it possible to solve.....

import java.util.; import java.io.; import static java.lang.Math.pow; class Solution{ public static void main(String []argh){ Scanner in = new Scanner(System.in); int t=in.nextInt(); for(int i=0;i } in.close(); } }

dear look at the qustion carefully.....when u focus on it u get more issues and problems but behind that once a time will came ur one confiusion converts to an answer.this remarks the sucess........take difference of its answer u will get solution

this is what i got:

for(int i=0;i

NOO

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();
    }
}

for(int i=0;i

same here :D

Here is full code: but please first understand it before use...

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

class Solution{ public static void main(String []argh){ Scanner in = new Scanner(System.in); int t=in.nextInt(); int temp = 0; for(int i=0;i for(int j = 1; j <= n; j++){ temp += (int)(Math.pow(2, j-1) * b); System.out.print((a + temp) + " "); } temp = 0; System.out.printf("\n"); } } }

int replace = 0; for (int j = 0; j < n; j++) { int math = (int) Math.pow(2, j); int sum0 = (math * b); int ver = sum0; replace += ver; int asd = a + replace; System.out.printf("%d ", asd); } System.out.println(""); } in.close();

Solution

for each of the 'q' queries you are given three values - (a,b,n)

• Task is to print a series with 'n' space separated terms where i(th) term is of the form =>> ( a + (b * (2^i)) )
• Print such a sequence for each of the queries in new line. Thats all

Yah, It's a simple question to loop and print a series of the form (a+1*b)+(a+1*b+2*b)+(a+1*b+2*b+4*b)+...........

You need to append always Math.pow(2,j) times b i.e 2^1*b,2^2*b....

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 seq = a + (int)Math.pow(2, 0) * b;
        System.out.print(seq + " ");
        for (int j = 1; j < n; j++){
            seq = seq + (int)Math.pow(2, j) * b;
            System.out.print(seq + " ");
        }
        System.out.println();
    }
    in.close();
}

import java.util.; import java.math.; class sos { public static void main(String args[]) { Scanner sc=new Scanner (System.in); int i,n,a,j,b,k=0,l,s1; l=sc.nextInt(); for(int m=0;m=1&&n<=15) { for(i=0;i=0) s=s+a; k++; //if(k%10!=0) s1=(int)s; System.out.print(s1+" "); //else //{ //System.out.println();

        //}
        s=0;

    }}
    System.out.println();
    }
}

}

for(int i=0;i System.out.println(""); } }

This might help a bit :)

ONLY the first itteration uses the value of a.

Therefore, The first output value is: (a + 2 ^ 0 + b)

for the next itteration you just add the value of (2 ^ 1 * b) on the previous itteration value. then you just keep adding up to (2 ^ i * b), I is only up to the n value.

Agreed.

as you will find, programming has to do A LOT with math; but they gave to you all the tools to complete it, the key is to start thinking like a computer and do the calculations; because programming = calculations.

Maths is key of software developement. U must have knowledge of this.

true x)

That's good. You can simplify even further if you realize that 2^0 + 2^1 + ... + 2^j = 2^(j+1) - 1

class Solution{
    public static void main(String []argh){
        Scanner in = new Scanner(System.in);
        int t=in.nextInt();
        StringBuilder sb = new StringBuilder();
        for(int i=0;i<t;i++){
            int a = in.nextInt();
            int b = in.nextInt();
            int n = in.nextInt();
            sb.setLength(0);
            for(int j=0; j<n; ++j) {
                // 2^0 + 2^1 + ... 2^j = 2^(j+1) - 1
                sb.append((int) (a + b*(Math.pow(2, j+1) - 1))).append(" ");
            }
            System.out.println(sb.toString());
        }
        in.close();
    }
}

Is Math.pow() an inbuilt function in JAVA?

excellent

Sir, It dosen't Work.The line"System.out.println("");"does not change the line for 2nd input.

I also do this,but it is wrong!Why?

here a = a + (int)Math.pow(2,j)*b;

why we used a= a + __ : can we use another integer like int c=0; c= a + (int)Math.pow(2,j)*b;

if not please explain

why u use t integer variable here?

Clear and Perfect...!! Thanks

Yes.This is correct. But I have one question... In the series we can see evey value is multiply by b.(either first a also).But in your code I can't see that. But your one gives correct answer. Can you please explain that for me?

I used this and this gives wrong answerI can't realize why it is wrong???

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 q1=0;q1<t;q1++){
            
            int a = in.nextInt();
            int b = in.nextInt();
            int n = in.nextInt();
            int ans=a+1*b;   
                for(int j=0;j<n;j++){
                    int x=2^j*b;
                    ans+=x;
                   System.out.print(ans+" ");
                    
                   
                    
                 }
             
            System.out.println();
        }

        
        in.close();
    }
}

Thanks #sharmatushar124 for ur Soln.

but what i the need of 't' here

silly me, i was calculating each term individually (3 nested for loops instead of 2) which would increase time complexity alot. Thanks xD

  public static void main(String []argh){
        Scanner in = new Scanner(System.in);
        int num1, num2=0;
        int t=in.nextInt();
        for(int i=0;i<t;i++){
            int a = in.nextInt();
            int b = in.nextInt();
            int n = in.nextInt();
                for(int k=0;k < n; k++){
                    num1 =a;
                    num2 += (Math.pow(2,k)*b);
                    System.out.print(num1 + num2 + " ");

                }
        }


        in.close();
    }

I have two question first why is this not working and why did he use (int) beofre Math.pow(). Thank You

System.out.print(a + " "); System.out.println(""); can you please explain this and this

Thanks .It's working without any error.

Can you please explain what does this line do? a += b * (int) Math.pow(2, j);

Yes, only exception that code will terminate is ; first input 0 Great code @rshaghoulian

System.out.print(a + " ");......whats the meaning of this line?

I have written the same code but it's not working. Can you help

can u explain this code

While I agree this works, not sure if this is what was intended from the prompt. In the last section it describes: "Once we hit n=10, we print the first ten terms as a single line of space-separated integers.", which would imply that they want you to do the entire series calculation first, then print the result.

Seems a bit weird that this is where it is in the challenge list, as I would think a recursive solution is called for in that case.

Why is a System.out.println(); added in the last. If I am remove it the code is not working. please help.

System.out.println(); What is the use of this?

for(int j=0;j

            explain this logic 

pretty smart.

If I remove the i=0 code why does the output fall short?

perfect

cool

can you explain this logic?

Hello can you explain me in this case the logic of action for loop (j) and for loop (k)? Thank you.