The code below calculates a^b (up to 1000000 digits results), any ideea to improve it, maybe make it faster?

Question

#include <iostream>
using namespace std;

int v[1000000];

int main(){

int n,x,aux;
cin >>x>>n;
v[v[0]=1]=1;

for(int k=1;k<=n;++k)
{
int t = 0;
for(int i=1;i<=v[0];++i)
{
aux=v[i]*x+t;
v[i]=aux%10;
t=aux/10;
}
while(t)
{
v[++v[0]]=t%10;
t/=10;
}
}

for(int i=v[0];i>0;--i)
cout<<v[i];

return 0;
}

Answer 1

This is unnecessarily complex... unless there were conditions that you needed to fulfill, idk. I just wrote this something, you may check it out if it works the way you want it to. It handles pow 0 and result goes up to 16 digits.

#include <iostream>
using namespace std;

int main() {

int base;
int pow;
long long int result = 1;

cin >> base;
cin >> pow;

for (int i = 0; i < pow; i++)
{
result *= base;
}

cout << result;

return 0;
}

Answer 2

@Catalin My sincere apologies, I see your point now. I'll get back with a legit opinion.

Answer 3

You could also take advantage of the fast exponentiation algorithm (recursive): for a^b,
- if b is 0, return 1
- if b is even, return (a*a)^(b/2) (recursive call)
- if b is odd, return a*(a^(b-1)) (recursive call)
You may take a look at https://code.sololearn.com/cNlG5g0X9tjV/?ref=app
[EDIT] The actual code is written in Python, but it can be ported to C++ or Java as well.

Answer 4

Hatsi Rei well the whole point it is that it can calculate very large numbers, larger than predefined types of c++, but thanks for trying

Answer 5

Thank you Alvaro, I'm going to try it

Answer 6

@Hatsi Rei, I quess you saw it's up to 1000000 digits, not the number 1000000

Answer 7

@Hatsy Rei, no need for to apologize my friend, always thankful to see someone's trying to help me, you rock