SPEN - Editorial

PROBLEM LINK:

Sagar And Pens

Practice
Contest

Author : Dipen Ved
Tester: Dipen Ved
Editorialist: Mad Sri

DIFFICULTY:

CAKEWALK

PREREQUISITES:

Basic Mathematics

PROBLEM:

You are given N pens of length 1 inch each.Your task is to find the maximum area of the rectangle that can be formed using these N pens in inches.

QUICK EXPLANATION:

The number of pens given in each testcase must be considered as the perimeter of the rectangle which is goig to remain constant.

EXPLANATION:

The main logic is quiet simple.Although we need to maximize the area , all the possible rectangles that can be formed using the N number of pens given will have
different lengths and widths , but the perimeter is going to remain the same and is equal to the number of pens as the length of each pen is 1 inch.

As it,s a rectangle , we consider ‘l’ to be the smaller side and ‘w’ to be the larger side.

Now , the smaller side or the length of the rectangle will be one-fourth of the total perimeter i.e number of pens.

Therefore , l = N/4.

For the larger side or the width , the length which remains after dividing it into 4 parts will be added.As there are 2 sides which will have the same width ,
we will have to divide this remaining portion of the perimeter into half

Therefore , w = l + (N % 4)/2.

Now we have the maximum length and width whose product will give you the maximized area of rectangle possible.

AUTHOR’S AND TESTER’S SOLUTIONS:

Solution can be found here.

Now , the smaller side or the length of the rectangle will be one-fourth of the total perimeter i.e number of pens.

Therefore , l = N/4.

can anyone explain??

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