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 Rust (1.15.1) (judge verdict: Accepted)

use std::io;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}

fn solve(n: i64) -> u64 {
    if n == 0 {
        return 0;
    }

    let mut ans = 0u64;
    let mut tmp = n;

    if (n > 0 && n%4 == 1) || (n < 0 && (-n)%4 == 3) {
        ans += 1;
        tmp -= 1;
    } else if (n > 0 && n%4 == 3) || (n < 0 && (-n)%4 == 1) {
        ans += 3;
        tmp += 1;
    }

    ans + solve(- tmp / 2) * 2
}

fn main() {
    let n:i64 = read_line().trim().parse().unwrap();
    let m = solve(n);

    let mut ans = String::new();
    let mut tmp = 1;
    while m >= tmp {
        if m & tmp > 0 {
            ans = "1".to_owned() + &ans;
        } else {
            ans = "0".to_owned() + &ans;
        }
        tmp *= 2;
    }

    if ans.is_empty() {
        ans = "0".to_owned();
    }

    println!("{}", ans);
}