由于数的范围很小,假如这个数合法,我们直接从此数 + 1 +1 +1开始枚举,直到合法为止
最坏情况下时间复杂度为 O ( T n / 2 ) O(Tn/2) O(Tn/2)
#include<bits/stdc++.h>
#define int long long
#define inf 0x3f3f3f3f3f3f3f
const int N = 2e5 + 10;
const int M = 15;
const int mod = 998244353;
using namespace std;
typedef pair<int, int>p;
typedef long long ll;
int a(int x)
{
int res = 0;
while (x)
{
if (x & 1)res++;
x >>= 1;
}
return res;
}
signed main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T = 1;
cin >> T;
while (T--)
{
int n;
cin >> n;
if (a(n)<=2)
{
n++;
while (a(n) >= 3)n++;
cout << n << endl;
}
else
{
cout << a(n) << endl;
}
}
}