运用数组进行2的N次方进行运算

#include<iostream>
using namespace std;
const int N = 643010;
int main()
{
    int arr[N]={1};
    int n;
    cin>>n;
    int m=1;
    for (int i = 0;i < n; i ++ )
    {2
    int t = 0;
    for (int j = 0; j < m; j ++ )
    {
        t += arr[j] * 2;
        arr[j] = t % 10;
        t /= 10;

    }
    if (t) arr[m ++ ] = 1;
    }
    for (int i = m - 1; i >= 0; i -- ) cout << arr[i];
    cout << endl;
    return 0;
}