How to check if a number is a perfect square C++

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Perfect Square:

In mathematics, a square number, sometimes also called a perfect square, is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, 9 is a square number, since it can be written as 3?×?3. (Text from Wikipedia)

We are going to write a program that checks if the input number is a perfect square or not.
C program to check Perfect Square:

/** C Program to check Perfect Square **/ #include <stdio.h> int main() { int a, n; printf("Enter a number: "); scanf("%d", &n); for(a = 0; a <= n; a++) { if (n == a * a) { printf("YES"); return 0; } } printf("NO"); return 0; }

C Program to check Perfect Square using sqrt() function:

We can perform the same program getting help from sqrt() function. The sqrt() function is bundled with math.h header file. Here is the program:

/* C Program to check Perfect Square using sqrt function */ #include <stdio.h> #include <math.h> int main( { int n, temp; printf("Enter a number: "); scanf("%d",&n); temp = sqrt(n); if(temp*temp == n) printf("YES."); else printf("NO."); return 0; }

In this example, we will see a C program through which we can check if a given number is a perfect square or not.

If a whole number is the square of another whole number then it is known as a perfect square, like 16 is the square of 4 so 16 will be called a Perfect square.

Algorithm:

STEP 1:Input any number x.
STEP 2:Store its square root in a float variable fVar.
STEP 3:Assign fVar into iVar (an integer variable) iVar=fVar.
STEP 4:Now compare iVar and fVar value will be equal. If the number does not a perfect square, iVar and fVar will not same.

/*C program to check number is perfect square or not.*/ #include <stdio.h> #include <math.h> int main() { int num; int iVar; float fVar; printf("Enter an integer number: "); scanf("%d",&num); fVar=sqrt((double)num); iVar=fVar; if(iVar==fVar) printf("%d is a perfect square.",num); else printf("%d is not a perfect square.",num); return 0; }

Output: Enter an integer number: 64

64 is a perfect square.

Given a number, check if it is a perfect square or not.

Input : 2500 Output : Yes Explanation: 2500 is a perfect square. 50 * 50 = 2500 Input : 2555 Output : No

#include <bits/stdc++.h>
using namespace std;
bool isPerfectSquare(long double x)
{
    if (x >= 0) {
        long long sr = sqrt(x);
        return (sr * sr == x);
    }
    return false;
}
int main()
{
    long long x = 2502;
    if (isPerfectSquare(x))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

class GFG {
    static boolean isPerfectSquare(int x)
    {
        if (x >= 0) {
            int sr = (int)Math.sqrt(x);
            return ((sr * sr) == x);
        }
        return false;
    }
    public static void main(String[] args)
    {
        int x = 2502;



        if (isPerfectSquare(x))
            System.out.print("Yes");
        else
            System.out.print("No");
    }
}

import math
def isPerfectSquare(x):
    if(x >= 0):
        sr = int(math.sqrt(x))
        return ((sr*sr) == x)
    return false
x = 2502
if (isPerfectSquare(x)):
    print("Yes")
else:
    print("No")

using System;
class GFG {
    static bool isPerfectSquare(double x)
    {
        if (x >= 0) {
            double sr = Math.Sqrt(x);
            return (sr * sr == x);
        }
        return false;
    }
    public static void Main()
    {
        double x = 2502;
        if (isPerfectSquare(x))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

<?php
function isPerfectSquare($x)
{
    $sr = sqrt($x);
    return (($sr - floor($sr)) == 0);
}
$x = 2502;
if (isPerfectSquare($x))
    echo("Yes");
else
    echo("No");
?>

<script>
function isPerfectSquare(x)
    {
        if (x >= 0) {
            let sr = Math.sqrt(x);
            return ((sr * sr) == x);
        }
        return false;
    }
        let x = 2500;
        if (isPerfectSquare(x))
            document.write("Yes");
        else
            document.write("No");
</script>

Time Complexity: O(log(x))
Auxiliary Space: O(1)

To know more about the inbuilt sqrt function, refer this Stackoverflow and this StackExchange threads.

#include <iostream>
#include <math.h>
using namespace std;
void checkperfectsquare(int n)
{
    if (ceil((double)sqrt(n)) == floor((double)sqrt(n))) {
        cout << "perfect square";
    }
    else {



        cout << "not a perfect square";
    }
}
int main()
{
    int n = 49;
    checkperfectsquare(n);
    return 0;
}

import java.io.*;
class GFG{
static void checkperfectsquare(int n)
{
    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");
    }
}
public static void main(String[] args)
{
    int n = 49;
    checkperfectsquare(n);
}
}

import math
def checkperfectsquare(x):
    if (math.ceil(math.sqrt(n)) ==
       math.floor(math.sqrt(n))):
        print("perfect square")
    else:
        print("not a perfect square")
n = 49
checkperfectsquare(n)

using System;
class GFG{
static void checkperfectsquare(int n)
{
    if (Math.Ceiling((double)Math.Sqrt(n)) ==
        Math.Floor((double)Math.Sqrt(n)))
    {
        Console.Write("perfect square");
    }
    else
    {
        Console.Write("not a perfect square");
    }
}
public static void Main()
{
    int n = 49;
    checkperfectsquare(n);
}
}

<script>
function checkperfectsquare(n)
{
    if (Math.ceil(Math.sqrt(n)) ==
        Math.floor(Math.sqrt(n)))
    {
        document.write("perfect square");
    }
    else
    {
        document.write("not a perfect square");
    }
}
let n = 49;
checkperfectsquare(n);
</script>

Time Complexity : O(sqrt(n))