............................... ?

You are consulting for a trucking company that does a large amount ofbusiness shipping packages between New York and Boston. The

likoan [24]

Answer:

Answer explained with detail below

Explanation:

Consider the solution given by the greedy algorithm as a sequence of packages, here represented by indexes: 1, 2, 3, ... n. Each package i has a weight, w_i, and an assigned truck t_i. { t_i } is a non-decreasing sequence (as the k'th truck is sent out before anything is placed on the k+1'th truck). If t_n = m, that means our solution takes m trucks to send out the n packages.

If the greedy solution is non-optimal, then there exists another solution { t'_i }, with the same constraints, s.t. t'_n = m' < t_n = m.

Consider the optimal solution that matches the greedy solution as long as possible, so \for all i < k, t_i = t'_i, and t_k != t'_k.

t_k != t'_k => Either

1) t_k = 1 + t'_k

i.e. the greedy solution switched trucks before the optimal solution.

But the greedy solution only switches trucks when the current truck is full. Since t_i = t'_i i < k, the contents of the current truck after adding the k - 1'th package are identical for the greedy and the optimal solutions.

So, if the greedy solution switched trucks, that means that the truck couldn't fit the k'th package, so the optimal solution must switch trucks as well.

So this situation cannot arise.

2) t'_k = 1 + t_k

i.e. the optimal solution switches trucks before the greedy solution.

Construct the sequence { t"_i } s.t.

t"_i = t_i, i <= k

t"_k = t'_i, i > k

This is the same as the optimal solution, except package k has been moved from truck t'_k to truck (t'_k - 1). Truck t'_k cannot be overpacked, since it has one less packages than it did in the optimal solution, and truck (t'_k - 1)

cannot be overpacked, since it has no more packages than it did in the greedy solution.

So { t"_i } must be a valid solution. If k = n, then we may have decreased the number of trucks required, which is a contradiction of the optimality of { t'_i }. Otherwise, we did not increase the number of trucks, so we created an optimal solution that matches { t_i } longer than { t'_i } does, which is a contradiction of the definition of { t'_i }.

So the greedy solution must be optimal.

6 0

4 years ago

import java.util.Scanner; public class StudentScores { public static void main (String [] args) { Scanner scnr = new Scanner(Sys

Delvig [45]

Answer:

The code at 'Your solution goes here' is in the bold font given

Explanation:

import java.util.Scanner;

public class StudentScores {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

final int SCORES_SIZE = 4;

int[] oldScores = new int[SCORES_SIZE];

int[] newScores = new int[SCORES_SIZE];

int i;

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

oldScores[i] = scnr.nextInt();

}

for (i = 0; i < oldScores.length - 1; ++i) {

newScores[i] = oldScores[i+1];

}

newScores[oldScores.length-1] = oldScores[0];

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

System.out.print(newScores[i] + " ");

}

System.out.println();

}

4 0

3 years ago

Wrtie a program in which we will pass a value N. N can be positive or negative. If N is positive then output all values from N d

Triss [41]

Answer:

Following are the answer to this question:

x=int(input("Enter number: "))#defining x variable that input value from user end

if x< 0:#defining if block that check x value is less then 0

while x<0:#defining while loop print up to the value

print(x)#print value

x+= 1#add values by 1

elif x>0:#defining elif block to check value x is greater than 0

while x>0:#defining while loop to print down to value

print(x)#print value

x-= 1#subtract value by 1

Output:

when input is a positive value

Enter number: 5

5

4

3

2

1

when input is a negative value

Enter number: -5

-5

-4

-3

-2

-1

Explanation:

In the given python code, x variable is declared that input the value from the user end, in the next step if and elseif block is declared that calculates and prints its value.
In the if block, it checks value is negative it uses the while loop to prints its values in the down to value form.
In the elif block, it checks the positive it uses the while loop to prints its values into the up to values form.

5 0

4 years ago

Which of the following is true about named ranges?

ohaa [14]

Answer:

Regarding a named range, the scope of a name is the location within which Excel recognizes the name without qualification. ... Once you name a range, you can change the size of the range using the Name Manager. True. You can create a new range by selecting the cells and typing a name in the Name box next to the formula bar ...

This is true about a range names. Hoped this helped a little. Im not sure which following you mean. maybe next time you can put a picture or descirbe it a little more or what the otions are.

I Hope You Have a Good Day or Night!! :)

Explanation:

5 0

3 years ago

To communicate with coworkers in the office

Elanso [62]

For effective communication to occur, everyone must trust and respect each other. ... Clear and concise communication will allow your colleagues to understand and then trust you. As a result, there will be more cooperation and less conflict in the workplace.

8 0

4 years ago