快速幂

#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll ans = 1;

ll Fastsub(ll base, ll pow) {
while (pow != 0) {
if (pow % 2 == 1) {  //幂为奇数的话先乘一个一次方
ans = ans * base % (1000);
}
pow /= 2;   //幂不论偶数奇数都减半，底数再平方
base = base * base % (1000);
}
return ans;
}

int main() {
ll a, b;
cin >> a >> b;
Fastsub(a, b);
cout << ans;
return 0;
}