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  1. Prepare
  2. Java
  3. Introduction
  4. Java Loops II
  5. Discussions

Java Loops II

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3266 Discussions

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  • jeancarneiro437
     + 0 comments

    Yes, i did the impossible, adding more complexity to solve:

    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 j = 0;
            int q = a+1*b;
            
            while(j < n) 
            {
                System.out.print(q+" ");
                q = q + 2*b;
                b = b*2;
                j++;
            }
            System.out.println();
        }
        in.close();
    }
}

  • V_Abhinand
     + 0 comments
    public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    int q = scan.nextInt();
    for (int i = 0; i < q; i++) {
        int a = scan.nextInt();
        int b = scan.nextInt();
        int n = scan.nextInt();
        for(int j=0; j<n; j++){
            a+=(int)Math.pow(2,j)*b;
            System.out.print(a+" ");
        }
        System.out.println();
    }
    scan.close();
}

    }

  • ahmed_h_ameen1
     + 0 comments

    This is my solution. It the is most optimum solution so far.

        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 result = a;
            for (int x = 0; x < n; x++) {
                // Use (1 << j) instead of Math.pow(2, j) // It's faster and works directly with integers.
                // result += (b * (int) Math.pow(2, x));
                result += (b * 1 << x);
                System.out.printf("%s ",result);
            }
            System.out.println();
        }
        in.close();
    }

  • ahmed_h_ameen1
     + 0 comments

    This is my solution. It the is most optimum solution so far.

        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 result = a;
            for (int x = 0; x < n; x++) {
                // Use (1 << j) instead of Math.pow(2, j) // It's faster and works directly with integers.
                // result += (b * (int) Math.pow(2, x));
                result += (b * 1 << x);
                System.out.printf("%s ",result);
            }
            System.out.println();
        }
        in.close();
    }

  • salma_ayache_23
     + 0 comments
            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 aux=1;
        int sum=a;
        int compteur =0;
        while(compteur<n){
            sum+=aux*b;
            System.out.printf(sum+" ");
            aux*=2;
            compteur++;  
        }
        System.out.println();
        } in.close();
        
    }
}