Question

How to check whether number is perfect square or not in C++

Answer 1

// When a square root is a whole number, then the number is a perfect square number
 
#include <iostream>
#include <cmath>
  
bool isPerfectSquare(int number) {
    double sq = sqrt(number); 
     
    return ((sq - floor(sq)) == 0); 
}
  
int main()
{
    int num = 81;
  
    if (isPerfectSquare(num))
        std::cout << num << " is a perfect square";
    else
        std::cout << num << " is not a perfect square";
}
 
  
  
  
/*
run:
  
81 is a perfect square
  
*/

 

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Answer 2

// When a square root is a whole number, then the number is a perfect square number

#include <iostream>
#include <cmath>
 
bool isPerfectSquare(int number) {
    if (number >= 0) {
        double d_sqrt = sqrt((double)number);
         
        return d_sqrt * d_sqrt == number;
    }

    return false;
}
 
int main()
{
    int num = 81;
 
    if (isPerfectSquare(num))
        std::cout << num << " is a perfect square";
    else
        std::cout << num << " is not a perfect square";
}
 
 
 
 
 
/*
run:
 
81 is a perfect square
 
*/

 

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Answer 3

// When a square root is a whole number, then the number is a perfect square number

#include <iostream>
#include <cmath>
 
bool isPerfectSquare(int number) {
    double d_sqrt = sqrt((double)number);
 
    if ((int)pow((int)(d_sqrt + 0.5), 2) == number)
        return true;
    else
        return false;
}
 
int main()
{
    int num = 81;
 
    if (isPerfectSquare(num))
        std::cout << num << " is a perfect square";
    else
        std::cout << num << " is not a perfect square";
}
 
 
 
 
 
/*
run:
 
81 is a perfect square
 
*/

 

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