sdnu1483.Problem_A（位运算+大数）

Description

Check whether an integer n is a power of 2.

Input

First line contains a single integer T (T<=20000000) which denotes the number of test cases. For each test case, there is an big integer N(0 < N < 2^1000)

Output

For each case, output the "Yes" or "No" in a single line.

Sample Input

3
1
3 
8

Sample Output

Yes
No
Yes

位运算判断是否是2的整数次幂（x-1）& x == 0  是  else不是

ps:是否是4的整数次幂：if((x&0x55555555)==x) 是 else 不是

python真强 ！

t = int(input())
while t > 0:
    a = int(input())
    if (a - 1) & a == 0:
        print("Yes")
    else:
        print("No")
    t = t - 1