1688A - Cirno's Perfect Bitmasks Classroom - CodeForces Solution

bitmasks brute force *800

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Python Code: 

x = int(input())
  
for jj in range(x):
    n = int(input())
    if n == 1:print(3)
    else:
        if n%2 == 1:
            print(1)
        else:
            p = 0
            k = n
            while n%2==0:
                p += 1
                n //= 2
            t = 2**p
            if t == k:
                print(t+1)
            else:
                print(t)

C++ Code: 

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
        ll n;
        cin>>n;
        if (n % 2)
    { // odd
        if (n > 1)
            cout << "1\n";
        else
            cout << "3\n";
    }
    else
    {
        if (!(n & (n - 1)))
        { // power of two
            cout << n + 1 << endl;
        }
        else
        { // we are leaving the most sig. bit and unsetting all bits.
            cout << (n & ~(n - 1)) << endl;
        }
    }
    }
}


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