显然答案是n的最近二次幂减一，也就是所谓的high_bit-1

#include <bits/stdc++.h>
#define int unsigned long long
using namespace std;

int high_bit(int x)
{
    return 1ull << (31 - __builtin_clz(x));
}
signed main()
{
    int n;
    cin >> n;
    cout << high_bit(n)-1;
    return 0;
}