Odd Divisor（大数判断是否有奇数因数）

You are given an integer n. Check if n has an odd divisor, greater than one (does there exist such a number x (x>1) that n is divisible by x and x is odd).
For example, if n=6, then there is x=3. If n=4, then such a number does not exist.

Input
The first line contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.
Each test case contains one integer n (2≤n≤1014).
Please note, that the input for some test cases won’t fit into 32-bit integer type, so you should use at least -bit integer type in your programming language.

Output
For each test case, output on a separate line:
“YES” if n has an odd divisor, greater than one;
“NO” otherwise.
You can output “YES” and “NO” in any case (for example, the strings yEs, yes, Yes and YES will be recognized as positive).

Example Input
6
2
3
4
5
998244353
1099511627776

Output
NO
YES
NO
YES
YES
NO

暴力肯定会TLE，所以就要找规律，首先看一下什么样的数字一定会有大于1的奇数因数，很明显所有奇数它本身就是一个奇数因数，所以只看偶数就行了，我们不妨看一下什么样的偶数没有奇数因数，从2开始列举发现所有2^n（n是正整数）都没有奇数因数，仔细想想也想得通，那么剩下的偶数就都有奇数因数了（好像推不出来，但是归纳一下好像是这么样= =）

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;
ll dat[70];
int main()
{
	for(int i=1;i<=63;++i)
	{
		dat[i]=pow(2,i);
	}
	ll t,n,mi;
	bool check;
	cin>>t;
	for(int i=1;i<=t;++i)
	{
		scanf("%lld",&n);
		if(n%2==1) cout<<"YES"<<endl;
		else
		{
			check=1;
			for(int j=1;j<=63;++j)
			{
				if(n==dat[j])
				{
					check=0;
					break;
				}
			}
			if(check) cout<<"YES"<<endl;
			else cout<<"NO"<<endl;
		}
	}
	return 0;
}