Dada un área entera , la tarea es encontrar el largo y el ancho de un rectángulo con el área dada tal que la diferencia entre el largo y el ancho sea la mínima posible.
Ejemplos:
Entrada: área = 99
Salida: l = 11, b = 9
Todos los rectángulos posibles (l, b) son (99, 1), (33, 3) y (11, 9)
Y el que tiene el mínimo |l – b | es (11, 9)Entrada: área = 25
Salida: l = 5, b = 5
Enfoque: La tarea es encontrar dos números enteros l y b tales que l * b = área y |l – b| son los mínimos posibles. La factorización se puede usar para resolver el problema, pero hacer solo una factorización simple de 1 a N llevará mucho tiempo obtener el resultado requerido para valores más grandes de N.
Para superar esto, simplemente iterar hasta. Considerando 
< l ≤ N , entonces para todos los valores de l , b siempre será < .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the length (l)
// and breadth (b) of the rectangle
// having area = N and |l - b| as
// minimum as possible
void find_rectangle(int area)
{
int l, b;
int M = sqrt(area), ans;
for (int i = M; i >= 1; i--) {
// i is a factor
if (area % i == 0) {
// l >= sqrt(area) >= i
l = (area / i);
// so here l is +ve always
b = i;
break;
}
}
// Here l and b are length and
// breadth of the rectangle
cout << "l = " << l << ", b = "
<< b << endl;
}
// Driver code
int main()
{
int area = 99;
find_rectangle(area);
return 0;
}
Java
// Java implementation of the approach
class GFG {
// Function to print the length (l)
// and breadth (b) of the rectangle
// having area = N and |l - b| as
// minimum as possible
static void find_rectangle(int area)
{
int l = 0, b = 0;
int M = (int)Math.sqrt(area), ans;
for (int i = M; i >= 1; i--) {
// i is a factor
if (area % i == 0) {
// l >= sqrt(area) >= i
l = (area / i);
// so here l is +ve always
b = i;
break;
}
}
// Here l and b are length and
// breadth of the rectangle
System.out.println("l = " + l + ", b = " + b);
}
// Driver code
public static void main(String[] args)
{
int area = 99;
find_rectangle(area);
}
}
// This code is contributed by Ita_c.
Python3
# Python3 implementation of the approach
import math as mt
# Function to print the length (l)
# and breadth (b) of the rectangle
# having area = N and |l - b| as
# minimum as possible
def find_rectangle(area):
l, b = 0, 0
M = mt.ceil(mt.sqrt(area))
ans = 0
for i in range(M, 0, -1):
# i is a factor
if (area % i == 0):
# l >= sqrt(area) >= i
l = (area // i)
# so here l is + ve always
b = i
break
# Here l and b are length and
# breadth of the rectangle
print("l =", l, ", b =", b)
# Driver code
area = 99
find_rectangle(area)
# This code is contributed by
# Mohit kumar 29
C#
// C# implementation of the approach
using System;
class GFG {
// Function to print the length (l)
// and breadth (b) of the rectangle
// having area = N and |l - b| as
// minimum as possible
static void find_rectangle(int area)
{
int l = 0, b = 0;
int M = (int)Math.Sqrt(area);
for (int i = M; i >= 1; i--) {
// i is a factor
if (area % i == 0) {
// l >= sqrt(area) >= i
l = (area / i);
// so here l is +ve always
b = i;
break;
}
}
// Here l and b are length and
// breadth of the rectangle
Console.WriteLine("l = " + l + ", b = " + b);
}
// Driver code
public static void Main()
{
int area = 99;
find_rectangle(area);
}
}
// This code is contributed by Mukul Singh.
PHP
<?php
// Php implementation of the approach
// Function to print the length (l)
// and breadth (b) of the rectangle
// having area = N and |l - b| as
// minimum as possible
function find_rectangle($area)
{
$M = floor(sqrt($area));
for ($i = $M; $i >= 1; $i--)
{
// i is a factor
if ($area % $i == 0)
{
// l >= sqrt(area) >= i
$l = floor($area / $i);
// so here l is +ve always
$b = $i ;
break;
}
}
// Here l and b are length and
// breadth of the rectangle
echo "l = ", $l, ", b = ", $b, "\n";
}
// Driver code
$area = 99;
find_rectangle($area);
// This code is contributed by Ryuga
?>
Javascript
<script>
// Javascript implementation of the approach
// Function to print the length (l)
// and breadth (b) of the rectangle
// having area = N and |l - b| as
// minimum as possible
function find_rectangle(area)
{
let l = 0, b = 0;
let M = Math.floor(Math.sqrt(area)), ans;
for (let i = M; i >= 1; i--) {
// i is a factor
if (area % i == 0) {
// l >= sqrt(area) >= i
l = Math.floor(area / i);
// so here l is +ve always
b = i;
break;
}
}
// Here l and b are length and
// breadth of the rectangle
document.write("l = " + l + ", b = " + b);
}
// Driver code
let area = 99;
find_rectangle(area);
</script>
Producción:
l = 11, b = 9
Complejidad del Tiempo: O( 
)
A continuación se muestra una implementación simple.
CPP
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the length (l)
// and breadth (b) of the rectangle
// having area = N and |l - b| as
// minimum as possible
void find_rectangle(int area)
{
for (int i = ceil(sqrt(area)); i <= area; i++) {
if (area / i * i == area) {
printf("%d %d", i, area / i);
return;
}
}
}
// Driver code
int main()
{
int area = 99;
find_rectangle(area);
return 0;
}
Java
// Java implementation of the approach
import java.io.*;
class GFG {
// Function to print the length (l)
// and breadth (b) of the rectangle
// having area = N and |l - b| as
// minimum as possible
static void find_rectangle(int area)
{
for(int i = (int)Math.ceil(Math.sqrt(area)); i <= area; i++)
{
if(area / i * i == area)
{
System.out.println(i + " " + (int)(area / i));
return;
}
}
}
// Driver code
public static void main (String[] args)
{
int area = 99;
find_rectangle(area);
}
}
// This code is contributed by rag2127
Python3
# Python3 implementation of the approach import math # Function to print the length (l) # and breadth (b) of the rectangle # having area = N and |l - b| as # minimum as possible def find_rectangle(area): for i in range(int(math.ceil(math.sqrt(area))) , area + 1): if((int(area / i) * i) == area): print(i, int(area / i)) return # Driver code area = 99 find_rectangle(area) # This code is contributed by avanitrachhadiya2155
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the length (l)
// and breadth (b) of the rectangle
// having area = N and |l - b| as
// minimum as possible
static void find_rectangle(int area)
{
for(int i = (int)Math.Ceiling(Math.Sqrt(area)); i <= area; i++)
{
if(area / i * i == area)
{
Console.WriteLine(i + " " + (int)(area / i));
return;
}
}
}
// Driver code
static void Main()
{
int area = 99;
find_rectangle(area);
}
}
// This code is contributed by divyeshrabadiya07.
Javascript
<script>
// Javascript implementation of the approach
// Function to print the length (l)
// and breadth (b) of the rectangle
// having area = N and |l - b| as
// minimum as possible
function find_rectangle(are)
{
for(let i = Math.floor(Math.ceil(Math.sqrt(area))); i <= area; i++)
{
if(Math.floor(area / i) * i == area)
{
document.write(i + " " + Math.floor(area / i));
return;
}
}
}
// Driver code
let area = 99;
find_rectangle(area);
// This code is contributed by unknown2108
</script>
Producción:
11 9
Publicación traducida automáticamente
Artículo escrito por Lohith sai Andra y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA