快速幂不加判断

90分

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef unsigned long long ll;
const int N = 1e9;
int a, b;

ll f(int a, int b) {
	if (!b)
		return 1;
	ll tmp = f(a, b / 2);
	if (tmp > N)
		return -1;
	return (b % 2 ? a : 1) * tmp * tmp;
}

int main() {
	scanf("%d %d", &a, &b);
	ll sum = f(a, b);
	if (sum > N)
		printf("-1");
	else
		printf("%lld", sum);
	return 0;
}/*
10 9
answer:
1000000000
*/