题解 | #2的N次方#

#include <iostream>
using namespace std;

long double x = 2;

long double power(int n) {
    if(n == 1)
        return x;
    else
        return x * power(n-1);
}

int main() {
    int N;
    cin >> N;
    printf("%.0Lf\n", power(N));
    return 0;
}

import java.util.Scanner;
import java.math.BigInteger;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        
        // 使用BigInteger处理大数
        BigInteger result = BigInteger.valueOf(2).pow(N);
        System.out.println(result);
    }
}

def power(n):
    return 2 ** n

N = int(input())
print(power(N))