Program to Check for Perfect Square in C++
Check for Perfect Square in C++
Today in this article we will discuss the program to check for perfect square in C++ programming language. We are given with an integer number and need to print “True” if it is, otherwise “False”.
Method 1:
- Take the floor() value square root of the number.
- Multiply the square root twice.
- We use boolean equal operator to verify if the product of square root is equal to the number given.
Method 1 :Code in C++
Run
#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 = 84;
if (isPerfectSquare(x))
cout << "True";
else
cout << "False";
return 0;
}
Output :
False
Method 2 :
- In this method we use the floor and ceil function .
- If they are equal that implies the number is a perfect square.
Method 2 :Code in C++
Run
#include <iostream>
#include <math.h>
using namespace std;
void checkperfectsquare(int n)
{
if (ceil((double)sqrt(n)) == floor((double)sqrt(n))) {
cout << "True";
}
else {
cout << "False";
}
}
int main()
{
int n = 49;
checkperfectsquare(n);
return 0;
}
Output :
True
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// perfect square
#include
#include
#include
using namespace std;
void perfect(int n){
int square, temp = n, digit;
for (int i = 1; i < n; i++)
{
square = pow(i,2);
if(square == temp)
break;
digit = i+1;
}
square == temp? cout<<square<<" is a Perfect square of "<<digit:cout<<"Not a perfect square";
}
int main()
{
int num;
cout<> num;
perfect(num);
return 0;
}