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patriot [66]
2 years ago
8

Write a program to declare a matrix A[][] of order (MXN) where ‘M’ is the number of rows and ‘N’ is the

Computers and Technology
1 answer:
Liula [17]2 years ago
4 0

Answer:

import java.io.*;

import java.util.Arrays;

class Main {

   public static void main(String args[])

   throws IOException{

       // Set up keyboard input

       InputStreamReader in = new InputStreamReader(System.in);

       BufferedReader br = new BufferedReader(in);

 

       // Prompt for dimensions MxN of the matrix

       System.out.print("M = ");

       int m = Integer.parseInt(br.readLine());

       System.out.print("N = ");

       int n = Integer.parseInt(br.readLine());

       // Check if input is within bounds, exit if not

       if(m <= 2 || m >= 10 || n <= 2 || n >= 10){

           System.out.println("Matrix size out of range.");

           return;

       }

       // Declare the matrix as two-dimensional int array

       int a[][] = new int[m][n];

 

       // Prompt for values of the matrix elements

       System.out.println("Enter elements of matrix:");

       for(int i = 0; i < m; i++){

           for(int j = 0; j < n; j++){

               a[i][j] = Integer.parseInt(br.readLine());

           }

       }

       // Output the original matrix

       System.out.println("Original Matrix:");

       printMatrix(a);

       // Sort each row

       for(int i = 0; i < m; i++){

         Arrays.sort(a[i]);

       }

       // Print sorted matrix

       System.out.println("Matrix after sorting rows:");

       printMatrix(a);

   }

   // Print the matrix elements separated by tabs

   public static void printMatrix(int[][] a) {

       for(int i = 0; i < a.length; i++){

           for(int j = 0; j < a[i].length; j++)

               System.out.print(a[i][j] + "\t");

           System.out.println();

       }

   }

}

Explanation:

I fixed the mistake in the original code and put comments in to describe each section. The mistake was that the entire matrix was sorted, while only the individual rows needed to be sorted. This even simplifies the program. I also factored out a printMatrix() method because it is used twice.

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The __________ of a desktop computer is the case that houses the computerâs critical parts, such as the processing and storage d
Phantasy [73]
You said it in your question. The case! Though it's also given other names, such as housing, chassis, enclosure etc.
4 0
2 years ago
Assign max_sum with the greater of num_a and num_b, PLUS the greater of num_y and num_z. Use just one statement. Hint: Call find
Gnoma [55]

Answer:

def find_max(num_1, num_2):

   max_val = 0.0

   if (num_1 > num_2): # if num1 is greater than num2,

       max_val = num_1 # then num1 is the maxVal.

   else: # Otherwise,

       max_val = num_2 # num2 is the maxVal

   return max_val

max_sum = 0.0

num_a = float(input())

num_b = float(input())

num_y = float(input())

num_z = float(input())

max_sum = find_max(num_a, num_b) + find_max(num_y, num_z)

print('max_sum is:', max_sum)

Explanation:

I added the missing part. Also, you forgot the put parentheses. I highlighted all.

To find the max_sum, you need to call the find_max twice and sum the result of these. In the first call, use the parameters num_a and num_b (This will give you greater among them). In the second call, use the parameters num_y and num_z (This will again give you greater among them)

5 0
2 years ago
Problem 2 - K-Best Values - 30 points Find the k-best (i.e. largest) values in a set of data. Assume you are given a sequence of
masya89 [10]

Answer:

See explaination

Explanation:

/**KBestCounter.java**/

import java.util.ArrayList;

import java.util.List;

import java.util.PriorityQueue;

public class KBestCounter<T extends Comparable<? super T>>

{

PriorityQueue heap;

int k;

public KBestCounter(int k)

{

heap = new PriorityQueue < Integer > ();

this.k=k;

}

//Inserts an element into heap.

//also takes O(log k) worst time to insert an element

//into a heap of size k.

public void count(T x)

{

//Always the heap has not more than k elements

if(heap.size()<k)

{

heap.add(x);

}

//if already has k elements, then compare the new element

//with the minimum element. if the new element is greater than the

//Minimum element, remove the minimum element and insert the new element

//otherwise, don't insert the new element.

else if ( (x.compareTo((T) heap.peek()) > 0 ) && heap.size()==k)

{

heap.remove();

heap.add(x);

}

}

//Returns a list of the k largest elements( in descending order)

public List kbest()

{

List al = new ArrayList();

int heapSize=heap.size();

//runs O(k)

for(int i=0;i<heapSize;i++)

{

al.add(0,heap.poll());

}

//Restoring the the priority queue.

//runs in O(k log k) time

for(int j=0;j<al.size();j++) //repeats k times

{

heap.add(al.get(j)); //takes O(log k) in worst case

}

return al;

}

}

public class TestKBest

{

public static void main(String[] args)

{

int k = 5;

KBestCounter<Integer> counter = new KBestCounter<>(k);

System.out.println("Inserting 1,2,3.");

for(int i = 1; i<=3; i++)

counter.count(i);

System.out.println("5-best should be [3,2,1]: "+counter.kbest());

counter.count(2);

System.out.println("Inserting another 2.");

System.out.println("5-best should be [3,2,2,1]: "+counter.kbest());

System.out.println("Inserting 4..99.");

for (int i = 4; i < 100; i++)

counter.count(i);

System.out.println("5-best should be [99,98,97,96,95]: " + counter.kbest());

System.out.println("Inserting 100, 20, 101.");

counter.count(100);

counter.count(20);

counter.count(101);

System.out.println("5-best should be [101,100,99,98,97]: " + counter.kbest());

}

}

5 0
2 years ago
assume that name is a variable of type stirng that has been assigned a value write an expression whose value is the last charact
Snezhnost [94]

Answer:

The most straight forward way to do it: in general string are zero index based array of characters, so you need to get the length of the string, subtract one and that will be the last character, some expressions in concrete languages would be:

In Python:

name = "blair"

name[len(name) - 1]

In JavaScript:

name = "blair"

name[name.length - 1]

In C++:

#include <string>

string name = "blair";

name[name.length() - 1];

7 0
2 years ago
Locker doors There are n lockers in a hallway, numbered sequentially from 1 to n. Initially, all the locker doors are closed. Yo
kow [346]

Answer:

// here is code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int n,no_open=0;

   cout<<"enter the number of lockers:";

   // read the number of lockers

   cin>>n;

   // initialize all lockers with 0, 0 for locked and 1 for open

   int lock[n]={};

   // toggle the locks

   // in each pass toggle every ith lock

   // if open close it and vice versa

   for(int i=1;i<=n;i++)

   {

       for(int a=0;a<n;a++)

       {

           if((a+1)%i==0)

           {

               if(lock[a]==0)

               lock[a]=1;

               else if(lock[a]==1)

               lock[a]=0;

           }

       }

   }

   cout<<"After last pass status of all locks:"<<endl;

   // print the status of all locks

   for(int x=0;x<n;x++)

   {

       if(lock[x]==0)

       {

           cout<<"lock "<<x+1<<" is close."<<endl;

       }

       else if(lock[x]==1)

       {

           cout<<"lock "<<x+1<<" is open."<<endl;

           // count the open locks

           no_open++;

       }

   }

   // print the open locks

   cout<<"total open locks are :"<<no_open<<endl;

return 0;

}

Explanation:

First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.

Output:

enter the number of lockers:9

After last pass status of all locks:

lock 1 is open.

lock 2 is close.

lock 3 is close.

lock 4 is open.

lock 5 is close.

lock 6 is close.

lock 7 is close.

lock 8 is close.

lock 9 is open.

total open locks are :3

5 0
2 years ago
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