Dados dos números dados a y b donde 1 <= a <= b, encuentre cubos perfectos entre a y b (a y b inclusive).
Ejemplos:
Input : a = 1, b = 100 Output : 1 8 27 64 Perfect cubes in the given range are 1, 8, 27, 64 Input : a = 24, b = 576 Output : 27 64 125 216 343 512 Perfect cubes in the given range are 27, 64, 125, 216, 343, 512
Este problema es similar a los cuadrados perfectos entre dos números .
Método 1 (ingenuo): un enfoque ingenuo es verificar todos los números entre a y b (incluidos a y b)
e imprimir el cubo perfecto. El siguiente es el código para el enfoque anterior:
C++
// A Simple Method to count cubes between a and b
#include <bits/stdc++.h>
using namespace std;
void printCubes(int a, int b)
{
// Traverse through all numbers in given range
// and one by one check if number is prime
for (int i = a; i <= b; i++) {
// Check if current number 'i'
// is perfect cube
for (int j = 1; j * j * j <= i; j++) {
if (j * j * j == i) {
cout << j * j * j << " ";
break;
}
}
}
}
// Driver code
int main()
{
int a = 1, b = 100;
cout << "Perfect cubes in given range:\n ";
printCubes(a, b);
return 0;
}
Java
// A Simple Method to count cubes between a and b
class Test {
static void printCubes(int a, int b)
{
// Traverse through all numbers in given range
// and one by one check if number is prime
for (int i = a; i <= b; i++) {
// Check if current number 'i'
// is perfect cube
for (int j = 1; j * j * j <= i; j++) {
if (j * j * j == i) {
System.out.print(j * j * j + " ");
break;
}
}
}
}
// Driver method
public static void main(String[] args)
{
int a = 1, b = 100;
System.out.println("Perfect cubes in given range:");
printCubes(a, b);
}
}
Python3
# A Simple Method to count cubes between a and b
def printCubes(a, b) :
# Traverse through all numbers in given range
# and one by one check if number is prime
for i in range(a, b + 1) :
# Check if current number 'i'
# is perfect cube
j = 1
for j in range(j ** 3, i + 1 ) :
if (j ** 3 == i) :
print( j ** 3, end = " ")
break
# Driver code
a = 1; b = 100
print("Perfect cubes in given range: ")
printCubes(a, b)
# This code is contributed by Nikita Tiwari.
C#
// A Simple Method to count cubes
// between a and b
using System;
class GFG {
static void printCubes(int a, int b)
{
// Traverse through all numbers
// in given range and one by
// one check if number is prime
for (int i = a; i <= b; i++) {
// Check if current number 'i'
// is perfect cube
for (int j = 1; j * j * j <= i; j++) {
if (j * j * j == i) {
Console.Write(j * j * j + " ");
break;
}
}
}
}
// Driver method
public static void Main()
{
int a = 1, b = 100;
Console.WriteLine("Perfect cubes in"
+ " given range:");
printCubes(a, b);
}
}
// This code contribute by parashar.
PHP
<?php
// A Simple Method to count
// cubes between a and b
function printCubes($a, $b)
{
// Traverse through all
// numbers in given range
// and one by one check
// if number is prime
for ($i = $a; $i <= $b; $i++)
{
// Check if current number 'i'
// is perfect cube
for ($j = 1; $j * $j * $j <= $i; $j++)
{
if ($j * $j * $j == $i)
{
echo $j * $j * $j, " ";
break;
}
}
}
}
// Driver Code
$a = 1;
$b = 100;
echo "Perfect cubes in given range:\n ";
printCubes($a, $b);
// This code is contributed by ajit
?>
Javascript
<script>
// A Simple Method to count
// cubes between a and b
function printCubes(a, b)
{
// Traverse through all
// numbers in given range
// and one by one check
// if number is prime
for (let i = a; i <= b; i++)
{
// Check if current number 'i'
// is perfect cube
for (let j = 1; j * j * j <= i; j++)
{
if (j * j * j == i)
{
document.write(j * j * j + " ");
break;
}
}
}
}
// Driver Code
let a = 1;
let b = 100;
document.write("Perfect cubes in given range: <br> ");
printCubes(a, b);
// This code is contributed by gfgking.
</script>
Producción :
Perfect cubes in given range: 1 8 27 64
Método 2 (eficiente):
simplemente podemos tomar la raíz cúbica de ‘a’ y la raíz cúbica de ‘b’ e imprimir los cubos de número entre ellos.
1- Given a = 24 b = 576
2- acr = cbrt(a)) bcr = cbrt(b)
acr = 3 and bcr = 8
3- Print cubes of 3 to 8 that comes under
the range of a and b(including a and b
both)
27, 64, 125, 216, 343, 512
A continuación se muestra la implementación de los pasos anteriores.
C++
// Efficient method to print cubes
// between a and b
#include <cmath>
#include <iostream>
using namespace std;
// An efficient solution to print perfect
// cubes between a and b
void printCubes(int a, int b)
{
// Find cube root of both a and b
int acrt = cbrt(a);
int bcrt = cbrt(b);
// Print cubes between acrt and bcrt
for (int i = acrt; i <= bcrt; i++)
if (i * i * i >= a && i * i * i <= b)
cout << i * i * i << " ";
}
// Driver code
int main()
{
int a = 24, b = 576;
cout << "Perfect cubes in given range:\n";
printCubes(a, b);
return 0;
}
// improved by prophet1999
Java
// Java progroam for Efficient method
// to print cubes between a and b
class Test {
// An efficient solution to print perfect
// cubes between a and b
static void printCubes(int a, int b)
{
// Find cube root of both a and b
int acrt = (int)Math.cbrt(a);
int bcrt = (int)Math.cbrt(b);
// Print cubes between acrt and bcrt
for (int i = acrt; i <= bcrt; i++)
if (i * i * i >= a && i * i * i <= b)
System.out.print(i * i * i + " ");
}
// Driver method
public static void main(String[] args)
{
int a = 24, b = 576;
System.out.println("Perfect cubes in given range:");
printCubes(a, b);
}
}
Python3
# Python3 code for Efficient method
# to print cubes between a and b
def cbrt(n) :
return (int)( n ** (1. / 3))
# An efficient solution to print
# perfect cubes between a and b
def printCubes(a, b) :
# Find cube root of
# both a and b
acrt = cbrt(a)
bcrt = cbrt(b)
# Print cubes between acrt and bcrt
for i in range(acrt, bcrt + 1) :
if (i * i * i >= a and i * i * i <= b) :
print(i * i * i, " ", end ="")
# Driver code
a = 24
b = 576
print("Perfect cubes in given range:")
printCubes(a, b)
# This code is contributed
# by Nikita Tiwari.
C#
// C# progroam for Efficient
// method to print cubes
// between a and b
using System;
class GFG
{
// An efficient solution
// to print perfect
// cubes between a and b
static void printCubes(int a,
int b)
{
// Find cube root of
// both a and b
int acrt = (int)Math.Pow(a,
(double)1 / 3);
int bcrt = (int)Math.Pow(b,
(double)1 / 3);
// Print cubes between
// acrt and bcrt
for (int i = acrt;
i <= bcrt; i++)
if (i * i * i >= a &&
i * i * i <= b)
Console.Write(i * i *
i + " ");
}
// Driver Code
static public void Main ()
{
int a = 24;
int b = 576;
Console.WriteLine("Perfect cubes " +
"in given range:");
printCubes(a, b);
}
}
// This code is contributed
// by ajit
PHP
<?php
// Efficient method to print
// cubes between a and b
// An efficient solution
// to print perfect
// cubes between a and b
function printCubes($a, $b)
{
// Find cube root
// of both a and b
$acrt = (int)pow($a, 1 / 3);
$bcrt = (int)pow($b, 1 / 3);
// Print cubes between
// acrt and bcrt
for ($i = $acrt; $i <= $bcrt; $i++)
if ($i * $i * $i >= $a &&
$i * $i * $i <= $b)
echo $i * $i * $i , " ";
}
// Driver code
$a = 24; $b = 576;
echo "Perfect cubes in given range:\n",
printCubes($a, $b);
// This code is contributed by ajit
?>
Javascript
<script>
// Javascript progroam for Efficient
// method to print cubes
// between a and b
// An efficient solution
// to print perfect
// cubes between a and b
function printCubes(a, b)
{
// Find cube root of
// both a and b
let acrt = parseInt(Math.pow(a, 1 / 3), 10);
let bcrt = parseInt(Math.pow(b, 1 / 3), 10);
// Print cubes between
// acrt and bcrt
for (let i = acrt; i <= bcrt; i++)
if (i * i * i >= a && i * i * i <= b)
document.write((i * i * i) + " ");
}
let a = 24;
let b = 576;
document.write("Perfect cubes " + "in given range:" + "</br>");
printCubes(a, b);
// This code is contributed by rameshtravel07.
</script>
Producción:
Perfect cubes in given range: 27 64 125 216 343 512
Este artículo es una contribución de Sahil Chhabra y es mejorado por Prophet1999 . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA