(高精度运算)计算2的N次方。N <= 10000

//Express Edition
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
    int n,x ;
    cin >> n;
    x = pow(2,n);
    cout << x;
    return 0;
}
// Normal Edition
#include <iostream>
#include <cmath>
using namespace std;

const int N = 3010;
int main()
{
    int a[N] = {1};
    int n;
    cin >> n;
    int m = 1;   //m为进位
    for(int i =0; i < n; i++)
    {
        int t = 0;
        for(int j = 0; j<m ;j++)
        {
            t+=a[j] *2;
            a[j] = t% 10;
            t /= 10;
        }
        if (t) a[m++] = 1;
    } 
    for (int i = m-1;i >=0; i--) cout << a[i];
    cout <<endl;
    return 0;
}