Dado un número, comprueba si es un cuadrado perfecto o no.
Ejemplos:
Input : 2500 Output : Yes Explanation: 2500 is a perfect square. 50 * 50 = 2500 Input : 2555 Output : No
Acercarse:
- Tome la raíz cuadrada de Floor()ed del número.
- Multiplica la raíz cuadrada dos veces.
- Use el operador booleano igual para verificar si el producto de la raíz cuadrada es igual al número dado.
C++
// CPP program to find if x is a
// perfect square.
#include <bits/stdc++.h>
using namespace std;
bool isPerfectSquare(long double x)
{
// Find floating point value of
// square root of x.
if (x >= 0) {
long long sr = sqrt(x);
// if product of square root
//is equal, then
// return T/F
return (sr * sr == x);
}
// else return false if n<0
return false;
}
int main()
{
long long x = 2502;
if (isPerfectSquare(x))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java program to find if x is a
// perfect square.
class GFG {
static boolean isPerfectSquare(int x)
{
if (x >= 0) {
// Find floating point value of
// square root of x.
int sr = (int)Math.sqrt(x);
// if product of square root
// is equal, then
// return T/F
return ((sr * sr) == x);
}
return false;
}
// Driver code
public static void main(String[] args)
{
int x = 2502;
if (isPerfectSquare(x))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python program to find if x is a
# perfect square.
import math
def isPerfectSquare(x):
#if x >= 0,
if(x >= 0):
sr = int(math.sqrt(x))
# sqrt function returns floating value so we have to convert it into integer
#return boolean T/F
return ((sr*sr) == x)
return false
# Driver code
x = 2502
if (isPerfectSquare(x)):
print("Yes")
else:
print("No")
# This code is contributed
# by Anant Agarwal.
C#
// C# program to find if x is a
// perfect square.
using System;
class GFG {
static bool isPerfectSquare(double x)
{
// Find floating point value of
// square root of x.
if (x >= 0) {
double sr = Math.Sqrt(x);
// if product of square root
// is equal, then
// return T/F
return (sr * sr == x);
}
// else return false if n<0
return false;
}
// Driver code
public static void Main()
{
double x = 2502;
if (isPerfectSquare(x))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to find if x is
// a perfect square.
function isPerfectSquare($x)
{
// Find floating point value
// of square root of x.
$sr = sqrt($x);
// If square root is an integer
return (($sr - floor($sr)) == 0);
}
// Driver code
$x = 2502;
if (isPerfectSquare($x))
echo("Yes");
else
echo("No");
// This code is contributed by Ajit.
?>
Javascript
<script>
// JavaScript program to find if x is a
// perfect square.
function isPerfectSquare(x)
{
if (x >= 0) {
// Find floating point value of
// square root of x.
let sr = Math.sqrt(x);
// if product of square root
// is equal, then
// return T/F
return ((sr * sr) == x);
}
return false;
}
// Driver code
let x = 2500;
if (isPerfectSquare(x))
document.write("Yes");
else
document.write("No");
// This code is contributed by souravghosh0416.
</script>
Producción
No
Para obtener más información sobre la función sqrt incorporada, consulte este Stackoverflow y este subprocesos de StackExchange.
Otro enfoque :
- Utilice la función suelo y techo.
- Si son iguales eso implica que el número es un cuadrado perfecto.
C++
// C++ program for the above approach
#include <iostream>
#include <math.h>
using namespace std;
void checkperfectsquare(int n)
{
// If ceil and floor are equal
// the number is a perfect
// square
if (ceil((double)sqrt(n)) == floor((double)sqrt(n))) {
cout << "perfect square";
}
else {
cout << "not a perfect square";
}
}
// Driver Code
int main()
{
int n = 49;
checkperfectsquare(n);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
class GFG{
static void checkperfectsquare(int n)
{
// If ceil and floor are equal
// the number is a perfect
// square
if (Math.ceil((double)Math.sqrt(n)) ==
Math.floor((double)Math.sqrt(n)))
{
System.out.print("perfect square");
}
else
{
System.out.print("not a perfect square");
}
}
// Driver Code
public static void main(String[] args)
{
int n = 49;
checkperfectsquare(n);
}
}
// This code is contributed by subhammahato348
Python3
# Python3 program for the above approach
import math
def checkperfectsquare(x):
# If ceil and floor are equal
# the number is a perfect
# square
if (math.ceil(math.sqrt(n)) ==
math.floor(math.sqrt(n))):
print("perfect square")
else:
print("not a perfect square")
# Driver code
n = 49
checkperfectsquare(n)
# This code is contributed by jana_sayantan
C#
// C# program for the above approach
using System;
class GFG{
static void checkperfectsquare(int n)
{
// If ceil and floor are equal
// the number is a perfect
// square
if (Math.Ceiling((double)Math.Sqrt(n)) ==
Math.Floor((double)Math.Sqrt(n)))
{
Console.Write("perfect square");
}
else
{
Console.Write("not a perfect square");
}
}
// Driver Code
public static void Main()
{
int n = 49;
checkperfectsquare(n);
}
}
// This code is contributed by subhammahato348
Javascript
<script>
// Javascript program for the above approach
function checkperfectsquare(n)
{
// If ceil and floor are equal
// the number is a perfect
// square
if (Math.ceil(Math.sqrt(n)) ==
Math.floor(Math.sqrt(n)))
{
document.write("perfect square");
}
else
{
document.write("not a perfect square");
}
}
// Driver Code
let n = 49;
checkperfectsquare(n);
// This code is contributed by rishavmahato348
</script>
Producción
perfect square
Complejidad del tiempo: O(sqrt(n))
Complejidad espacial : O(1)
Publicación traducida automáticamente
Artículo escrito por Aditya Darekar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA