如果你是我这种算法：

#include <bits/stdc++.h>
using namespace std;
#define int unsigned long long
int n, k, b, c, tot = 0;
short ans[64];
signed main()
{
    cin >> n >> k;
    c = (k >> 1), b = (k ^ c);
    while (b)
    {
        ans[++tot] = (b & 1), b = (b >> 1);
    }
    for (int i = n; i > 0; i--)
        cout << ans[i];
    return 0;
}

那么相信我，别开O2优化