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2 years ago

Given 4 integers, output their product and their average, using integer arithmetic.

Computers and Technology

1 answer:

solmaris [256]2 years ago

8 0

Answer:

see explaination

Explanation:

Part 1:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

int avg=0, pro=1;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

avg = (num1+num2+num3+num4)/4;

pro = num1*num2*num3*num4;

System.out.println(pro+" "+avg);

}

------------------------------------------------------------------

Part 2:

import java.util.Scanner;

public class LabProgram {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

int num3;

int num4;

double avg=0, pro=1; //using double to store floating point numbers.

num1 = scnr.nextInt();

num2 = scnr.nextInt();

num3 = scnr.nextInt();

num4 = scnr.nextInt();

avg = (num1+num2+num3+num4)/4.0; //if avg is declared as a float, then use 4.0f

pro = num1*num2*num3*num4;

System.out.println((int)pro+" "+(int)avg); //using type conversion only integer part

System.out.printf("%.3f %.3f\n",pro,avg);// \n is for newline

}

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Mike has never used a slide software before, but he needs to create a presentation by the end of the week. what recourse would b

ankoles [38]

This is an incomplete question. The complete question is given below:

Mike has never used slide presentation software before but he needs to create a presentation by the end of the week what resource would be most helpful to mike

a. The 350-page printed manual from the slide presentation software publisher

b. A free tutorial the slide presentation software publisher has posted on the company website

c. A trouble-shooting website created by a third party

d. The 350-page online manual from the slide presentation software publisher

Answer:

b - A free tutorial the slide presentation software publisher has posted on the company website

Explanation:

As Mike has a short time and no prior experience with a slide software, then in this scenario, the best, simplest and fastest way to learn and create a presentation a free tutorial which the slide presentation software publisher has posted on the company website as this is the same company that has created this particular software so he can be rest-assured that the resource he is relying on is authentic and up-to-date with information on latest features.

Moreover, it's efficient and quick way to learn from a free tutorial rather than from 350-page printed or online manual especially for a beginner.

Besides, his purpose is to create the presentation using the software and not trouble-shooting so trouble-shooting website created by a third party is not useful for him and it also might not be authentic or updated as well.

6 0

2 years ago

Recall the problem of finding the number of inversions. As in the text, we are given a sequence of n numbers a1, . . . , an, whi

Kay [80]

Answer:

The algorithm is very similar to the algorithm of counting inversions. The only change is that here we separate the counting of significant inversions from the merge-sort process.

Algorithm:

Let A = (a1, a2, . . . , an).

Function CountSigInv(A[1...n])

if n = 1 return 0; //base case

Let L := A[1...floor(n/2)]; // Get the first half of A

Let R := A[floor(n/2)+1...n]; // Get the second half of A

//Recurse on L. Return B, the sorted L,

//and x, the number of significant inversions in $L$

Let B, x := CountSigInv(L);

Let C, y := CountSigInv(R); //Do the counting of significant split inversions

Let i := 1;

Let j := 1;

Let z := 0;

// to count the number of significant split inversions while(i <= length(B) and j <= length(C)) if(B[i] > 2*C[j]) z += length(B)-i+1; j += 1; else i += 1;

//the normal merge-sort process i := 1; j := 1;

//the sorted A to be output Let D[1...n] be an array of length n, and every entry is initialized with 0; for k = 1 to n if B[i] < C[j] D[k] = B[i]; i += 1; else D[k] = C[j]; j += 1; return D, (x + y + z);

Runtime Analysis: At each level, both the counting of significant split inversions and the normal merge-sort process take O(n) time, because we take a linear scan in both cases. Also, at each level, we break the problem into two subproblems and the size of each subproblem is n/2. Hence, the recurrence relation is T(n) = 2T(n/2) + O(n). So in total, the time complexity is O(n log n).

Explanation:

5 0

2 years ago

ANSWER AS SOON AS POSSIBLE: When I try to join roblox adopt me it says: "Roblox Datastore servers are currently down. Please rej

masha68 [24]

Bro This is homework no roblox

7 0

2 years ago

Read 2 more answers

Identify two entities and 2 of their attributes from the given scenario.

polet [3.4K]

Bookstore and BookSearch are the two entities for the given scenario.

Explanation:

For the given Book.com virtual store, there can be two entities like Bookstore and BookSearch.
Bookstore can have all the details of the books in the virtual store. hence the attributes can be
Bookstore attributes: bookname, Authorname, Publisher, Publishedyear, Agegroup, category.
BookSearch entity can be used to search the books in the virtual store.
Booksearch attributes: bookname, category, bookid, authorname.

8 0

2 years ago

#Write a function called string_finder. string_finder should #take two parameters: a target string and a search string. #The fun

adoni [48]

Answer:

I am writing a Python program:

def string_finder(target,search): #function that takes two parameters i.e. target string and a search string

position=(target.find(search))# returns lowest index of search if it is found in target string

if position==0: # if value of position is 0 means lowers index

return "Beginning" #the search string in the beginning of target string

elif position== len(target) - len(search): #if position is equal to the difference between lengths of the target and search strings

return "End" # returns end

elif position > 0 and position < len(target) -1: #if value of position is greater than 0 and it is less than length of target -1

return "Middle" #returns middle

else: #if none of above conditions is true return not found

return "not found"

#you can add an elif condition instead of else for not found condition as:

#elif position==-1

#returns "not found"

#tests the data for the following cases

print(string_finder("Georgia Tech", "Georgia"))

print(string_finder("Georgia Tech", "gia"))

print(string_finder("Georgia Tech", "Tech"))

print(string_finder("Georgia Tech", "Idaho"))

Explanation:

The program can also be written in by using string methods.

def string_finder(target,search): #method definition that takes target string and string to be searched

if target.startswith(search): #startswith() method scans the target string and checks if the (substring) search is present at the start of target string

return "Beginning" #if above condition it true return Beginning

elif target.endswith(search): #endswith() method scans the target string and checks if the (substring) search is present at the end of target string

return "End" #if above elif condition it true return End

elif target.find(search) != -1: #find method returns -1 if the search string is not in target string so if find method does not return -1 then it means that the search string is within the target string and two above conditions evaluated to false so search string must be in the middle of target string

return "Middle" #if above elif condition it true return End

else: #if none of the above conditions is true then returns Not Found

return "Not Found"

6 0

2 years ago