This is done by using exponential theory :-
I am giving my code here for a^b
int power (int a,int b)
{
int x=1,y=a;
while(b>0)
{
if(b%2==1) x=(x*y)
if (b/=2) y=(y*y);
}
return x;
}
Run time O(log b)
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