All Tracks Math Problem

Xor is Mad
Tag(s):

Basic Programming, Bit manipulation, Easy

Problem
Editorial
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The problem is straight and simple.
Given a number X ,find how many positive A ( \(A\gt 0)\) exists, such that
1. A \(\newcommand*\xor{\mathbin{\oplus}}\) X =A + X
2. A \(\lt\) 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:

  • \(1 \le T \le 10^5\)
  • \(1 \le X \le 10^7\)
Note: \(\newcommand*\xor{\mathbin{\oplus}}\) is the bitwise Exclusive-OR operator( XOR ). and + is the usual addition symbol.

SAMPLE INPUT
4
1
2
3
4
SAMPLE OUTPUT
0
1
0
3
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 \(\newcommand*\xor{\mathbin{\oplus}}\) 2 = 3
For 3rd case:
For number 3 we have no such number .
For 4th case:
For number 4 we have 3 \(\oplus\) 4 = 7 , 2 \(\oplus\) 4 = 6 and 1 \(\oplus\) 4 = 5. Hence the answer is 3.

Time Limit: 1.0 sec(s) for each input file.
Memory Limit: 256 MB
Source Limit: 1024 KB
Marking Scheme: Marks are awarded when all the testcases pass.
Allowed Languages: C, C++, C++14, Clojure, C#, D, Erlang, F#, Go, Groovy, Haskell, Java, Java 8, JavaScript(Rhino), JavaScript(Node.js), Julia, Kotlin, Lisp, Lisp (SBCL), Lua, Objective-C, OCaml, Octave, Pascal, Perl, PHP, Python, Python 3, R(RScript), Racket, Ruby, Rust, Scala, Swift, Swift-4.1, Visual Basic

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