Bitwise XOR
Problem
The problem is straight and simple. Given a number X ,find how many positive A (A>0) exists, such that 1. A ⊕X =A +X 2. A<X
Input: The first line of the input will contain T , the number of test-cases.Next T lines will contain integer X .
Output: For each test-case , output the answer in a separate line.
Constraints: 1≤T≤10^5 1≤X≤10^7
Note: ⊕ is the bitwise Exclusive-OR operator( XOR ). and + is the usual addition symbol.
Time Limit: 5
Memory Limit: 256
Source Limit:
Explanation
Explanation Explanation for 1st test case: For 1 there is no such number.
For 2nd case: For 2 we have only one such number 1⊕2 =3
For 3rd case: For number 3 we have no such number .
For 4th case: For number 4 we have 3⊕4 = 7 ,2⊕4 =6 and 1⊕4 =5. Hence the answer is 3.
