Perfect square and perfect cube

Question

Is there any predefined function in c++ to check whether the number is square of any number and same for the cube..

Answer 1

No, but it's easy to write one:

bool is_perfect_square(int n) {
    if (n < 0)
        return false;
    int root(round(sqrt(n)));
    return n == root * root;
}

bool is_perfect_cube(int n) {
    int root(round(cbrt(n)));
    return n == root * root * root;
}

Answer 2

The most efficient answer could be this

    int x=sqrt(num)
    if(sqrt(num)>x){
    Then its not a square root}
    else{it is a perfect square}

This method works because of the fact that x is an int and it will drop down the decimal part to store only the integer part. If a number is perfect square of an integer, its square root will be an integer and hence x and sqrt(x) will be equal.