omg......实话说有一点多此一举，代码还可以简化，但我太懒了呵呵，用的除法qwq……

#include <cmath>
#include <iomanip>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <algorithm>
#include <string>
#include <stack>
using namespace std;
const int N=1e5+10;
long long m[N];
int main()
{
    int i;
    cin>>i;
    int fl=0;
    for(int k=1;k<=i;k++)
    {
        fl=0;
        cin>>m[k];
        if (m[k]<=0)
        {
            cout<<"NO"<<endl;
            fl=1;
            continue;
        }
        if(m[k]==1)
        {
            cout<<"YES"<<endl;
            continue;
        }
        if (m[k]%2 == 1)
        {
            cout<<"NO"<<endl;
            fl=1;
            continue;
        }
        while (m[k]>2)
        {
            m[k]/=2;
            if (m[k]%2 == 1)
            {
                cout<<"NO"<<endl;
                fl=1;
                break;
            }
        }
        if(fl==0)
        {
            cout<<"YES"<<endl;
        }
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;

long long a[64];

int main()
{
    a[1] = 1;
    for (int i = 2; i <= 63; i++) a[i] = a[i - 1] * 2;
    long long m, n;
    bool flag = false;
    cin >> m;
    for (int i = 1; i <= m; i++)
    {
        flag = false;
        cin >> n;
        for (int j = 1; j <= 63; j++)
        {
            if (n == a[j]) flag = true;
        }
        if (flag) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}