Problem:
Score : 300 points

Problem Statement

Given an integer N, find the base -2 representation of N.

Here, S is the base -2 representation of N when the following are all satisfied:

S is a string consisting of 0 and 1.

Unless S = 0, the initial character of S is 1.

Let S = S_k S_{k-1} ... S_0, then S_0 \times (-2)^0 + S_1 \times (-2)^1 + ... + S_k \times (-2)^k = N.

It can be proved that, for any integer M, the base -2 representation of M is uniquely determined.

Constraints

Every value in input is integer.

-10^9 \leq N \leq 10^9

Input

Input is given from Standard Input in the following format:

N

Output

Print the base -2 representation of N.

Sample Input 1

-9

Sample Output 1

1011

As (-2)^0 + (-2)^1 + (-2)^3 = 1 + (-2) + (-8) = -9, 1011 is the base -2 representation of -9.

Sample Input 2

123456789

Sample Output 2

11000101011001101110100010101

Sample Input 3

0

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
int bit[40];

int main(){
  fill(bit, bit + 40, 0);
  int N;
  if(N == 0){
    cout << "0" << endl;
    return 0;
  }
  cin >> N;
  bool sgn = false;
  if(N > 0) sgn = true;
  N = abs(N);
  for(int i = 0; i < 40; i++){
    bool t = N % 2;
    if(t && (sgn && i % 2) || (!sgn && !(i % 2))){
      bit[i]++;
      bit[i+1]++;
    } else if(t) {
      bit[i]++;
    }
    N /= 2;
  }
  for(int i = 0; i < 40; i++){
    if(bit[i] >= 4){
      bit[i] -= 4;
      bit[i+2]++;
    }
    if(bit[i] >= 2){
      bit[i] -= 2;
      bit[i+1]++;
      bit[i+2]++;
    }
  }
  bool c = false;
  for(int i = 39; i >= 0; i--){
    if(bit[i] != 0) c = true;
    if(c) cout << bit[i];
  }
  cout << endl;
  return 0;
}