Java Loops II 2D Array - DS

We use the integers , , and  to create the following series:

You are given  queries in the form of , , and . For each query, print the series corresponding to the given , , and  values as a single line of  space-separated integers.

Input Format

The first line contains an integer, , denoting the number of queries.
Each line  of the  subsequent lines contains three space-separated integers describing the respective , , and  values for that query.

Constraints

Output Format

For each query, print the corresponding series on a new line. Each series must be printed in order as a single line of  space-separated integers.

Sample Input

2
0 2 10
5 3 5

Sample Output

2 6 14 30 62 126 254 510 1022 2046
8 14 26 50 98

Explanation

We have two queries:

1. We use , , and  to produce some series :
   ... and so on.
   Once we hit , we print the first ten terms as a single line of space-separated integers.
2. We use , , and  to produce some series :
   We then print each element of our series as a single line of space-separated values.

import java.util.*;

import java.io.*;

import java.lang.Math.*;

class Solution{

    public static void main(String []argh){

        Scanner in = new Scanner(System.in);

        int t=in.nextInt();

        int j;

        for(int i=0;i<t;i++){

            int a = in.nextInt();//initial value

            int b = in.nextInt();

            int n = in.nextInt();//sequence number

            int arr[]=new int[n];

            for(j=0;j<n;j++)

                {

                  int temp=a;

                 for(int inc=0;inc<=j;inc++)

                     {

                     temp=temp+(int)(Math.pow(2,inc)*b);

                     }

                  arr[j]=temp;

                 System.out.print(arr[j]+" ");

            }

            

            System.out.println();

        }

        in.close();

    }

}

**************

import java.util.*;
import java.io.*;
import java.lang.Math.*;

class Solution{
    public static void main(String []argh){
        Scanner in = new Scanner(System.in);
        int t=in.nextInt();
        int j;
        for(int i=0;i<t;i++){
            int a = in.nextInt();//initial value
            int b = in.nextInt();
            int n = in.nextInt();//sequence number
            int arr[]=new int[n];
            for(j=0;j<n;j++)
                {
                  int temp=a;
                 for(int inc=0;inc<=j;inc++)
                     {
                     temp=temp+(int)(Math.pow(2,inc)*b);
                     }
                  arr[j]=temp;
            }
            for(j=0;j<n;j++)
                {
                System.out.print(arr[j]+" ");
            }
            System.out.println();
        }
        in.close();
    }
}

***********************************************************************************

Arrays - DS

An array is a type of data structure that stores elements of the same type in a contiguous block of memory. In an array, , of size , each memory location has some unique index,  (where ), that can be referenced as  (you may also see it written as ).

Given an array, , of  integers, print each element in reverse order as a single line of space-separated integers.

Note: If you've already solved our C++ domain's Arrays Introduction challenge, you may want to skip this.

Input Format

The first line contains an integer,  (the number of integers in ).
The second line contains  space-separated integers describing .

Constraints

Output Format

Print all  integers in  in reverse order as a single line of space-separated integers.

Sample Input

4
1 4 3 2

Sample Output

2 3 4 1

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

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int arr[] = new int[n];
        for(int arr_i=0; arr_i < n; arr_i++){
            arr[arr_i] = in.nextInt();
        }
        for(int i=(arr.length-1); i>=0; i--)
            {
            System.out.print(arr[i]+" ");
        }
    }
}







2D Array - DS

Context
Given a  2D Array, :

1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

We define an hourglass in  to be a subset of values with indices falling in this pattern in 's graphical representation:

a b c
  d
e f g

There are  hourglasses in , and an hourglass sum is the sum of an hourglass' values.

Task
Calculate the hourglass sum for every hourglass in , then print the maximum hourglass sum.

Note: If you have already solved the Java domain's Java 2D Array challenge, you may wish to skip this challenge.

Input Format

There are  lines of input, where each line contains  space-separated integers describing 2D Array ; every value in  will be in the inclusive range of  to .

Constraints

Output Format

Print the largest (maximum) hourglass sum found in .

Sample Input

1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 2 4 4 0
0 0 0 2 0 0
0 0 1 2 4 0

Sample Output

19

Explanation

 contains the following hourglasses:

1 1 1   1 1 0   1 0 0   0 0 0
  1       0       0       0
1 1 1   1 1 0   1 0 0   0 0 0

0 1 0   1 0 0   0 0 0   0 0 0
  1       1       0       0
0 0 2   0 2 4   2 4 4   4 4 0

1 1 1   1 1 0   1 0 0   0 0 0
  0       2       4       4
0 0 0   0 0 2   0 2 0   2 0 0

0 0 2   0 2 4   2 4 4   4 4 0
  0       0       2       0
0 0 1   0 1 2   1 2 4   2 4 0

The hourglass with the maximum sum () is:

2 4 4
  2
1 2 4

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
import java.util.Scanner;

public class twoDArray {

 public static void main(String[] args) {
  /* Create an input Scanner instance for reading input */
  Scanner stdIn = new Scanner(System.in);
  /* Name some constants used in the program */
  int rows = 6;    // Number of rows in the matrix
  int cols = 6;    // Number of columns in the matrix
  int minValueInArray = -9; // Minimum value of an element in the matrix
  int elementsInHourGlass = 7;// Number of elements in an hourglass
  // Set max to the minimum hourglass sum possible
  int maxHourGlassSum = minValueInArray * elementsInHourGlass;
  // Declare the matrix of numbers
  int matrix[][] = new int[rows][cols];
  int hourGlassSum; // Sum of the elements in the hourglass
  
  /* Read the values for the matrix looping thru the rows */
        for(int i=0; i < rows; i++){
         /* Loop thru the columns for each row, reading the matrix */
            for(int j=0; j < cols; j++){
             /* Read the next value from standard input */
                matrix[i][j] = stdIn.nextInt();
            }
        }
        /* For debugging, print out the matrix */
        for(int i=0; i < rows; i++){
            for(int j=0; j < cols; j++){
               // System.out.print(matrix[i][j] + " ");
            }
           // System.out.println();
        }
        /* Loop thru the possible starting points of the hourglass */
        for (int i=0; i < (rows-2); i++) {
         for (int j=0; j<(cols-2); j++) {
          /* Print the hourglass starting position */
          //System.out.println ("Checking hourglass at (" + i + "," + j + ")");
          /* Print out the hourglass elements */
          //System.out.println(matrix[i][j] + " " + matrix[i][j+1] + " "  + matrix[i][j+2]);
         // System.out.println("  " + matrix[i+1][j+1]);
         // System.out.println(matrix[i+2][j] + " " + matrix[i+2][j+1] + " " + matrix[i+2][j+2]);
          /* Compute the sum of the elements in the hourglass */
          hourGlassSum = matrix[i][j]   + matrix[i][j+1]   + matrix[i][j+2] +
                     matrix[i+1][j+1] +
                matrix[i+2][j] + matrix[i+2][j+1] + matrix[i+2][j+2];
          /* Print out the sum of the elements in this hourglass */
          //System.out.println("hour glass sum = " + hourGlassSum);
               // System.out.println("max glass sum = " + maxHourGlassSum );
          /* Is the new hourglass sum greater than the max found so far */
          if (hourGlassSum > maxHourGlassSum) {
           /* If so, then replace the max hour glass sum, with the current sum */
           maxHourGlassSum = hourGlassSum;
          }
         }
        }
        /* Print out the maximum hour glass sum */
        System.out.println(maxHourGlassSum);

 }

}