Binary Exponentiation or Pow(x,y)

/*
Binary Exponentiation:
eg x=2 , y=10
we can also write it as : x=4 ie. base*base 2*2 and exponent=y/2 ie. 5 ie. overall 4^5
we can also write it as : 4^5=4*(4^4)
we can also write it as : 4*(4^4)=4*(16^2)
we can also write it as : 4*256^1   {256 =16*16}

Algo:
Base and exponent
if exponent is divisible by 2 then multiply base with base eg base*base
if exponent is not divisible by 2 then multiply base in some variable eg. ans
for example ans=1,base=5 and exponent=3 so answer is 125 then  ans=1*5,base=5 and
exponent=2(ie. exponent-1) so answer is 1*5*(5^2) =125

if exponent>0 ans=x^exponent
if exponenet<0 ans=1/(x^exponent)
so we need to store absolute value of exponenet.But there is one catch,exponent can be integer [-2^31 to 2^31 -1 ]
so we can keep maximum positive value in int variable is 2^31 -1 but exponent value can be -2^31 and it's
abosulte 2^31 can give range overflow so use long long int.
*/
class Solution {
public:
    double myPow(double x, int p){
        if(p==0) return 1;
        bool sign=false;
        if(p<0) sign=true;
        long long int y=abs(p);
        double ans=1;
        while(y!=1){
            if(y%2==0){
                x=x*x;
                y/=2;
            }
            else{
                ans*=x;
                y--;
            }
        }
        if(sign) return 1/(ans*x);
        else return ans*x;
    }
};