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how to find if number is perfect square Code Example
how to find if number is perfect square 
#include <iostream>
#include <cmath>
using namespace std;

int main()
{
  int n;
  double root;
  cout << "Enter The number : ";
  cin>>n;
  root = sqrt(n);
  if (pow(root + 0.5,2) == n) 
  {
    cout << "Perfect Square" << endl;
  }
  else
  {
   cout << "Not Perfect" << endl;
  }
  return 0;
}
program to tell if a number is a perfect square 
import math

# Taking the input from user
number = int(input("Enter the Number"))

root = math.sqrt(number)
if int(root + 0.5) ** 2 == number:
    print(number, "is a perfect square")
else:
    print(number, "is not a perfect square")
Source: djangocentral.com



Cpp

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Type:
 Code Example
Category:
 Coding
Sub Category:
 Code Example
Uploaded by:
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