A-the moronic Cowmpouter

Description

Inexperienced in the digital arts, the cows tried to build a calculating engine (yes, it ' s a cowmpouter) using binary numb ERS (base 2) but instead built one based on base negative 2! They were quite pleased since numbers expressed in base−2 doing not has a sign bit.

You know number bases has place values as start at 1 (base to the 0 power) and proceed right-to-left to Base^1, base^2, And so on. In base−2, the place values is 1,−2, 4,−8, 16,−32, ... (reading from right to left). Thus, counting from 1 goes like This:1, A, 111, a, 101, 11010, 11011, 11000, 11001, and so on.

Eerily, negative numbers is also represented with 1 ' s and 0 's but no sign. Consider counting from−1 Downward:11, ten, 1101, 1100, 1111, and so on.

Cows convert ordinary decimal integers (range-2,000,000,000..2,000,000,000) to their counterpart Represen Tation in base−2.

Input

Line 1: A single integer to being converted to base−2

Output

Line 1: A single integer with no leading zeroes this is the input integer converted to base−2. The value 0 is expressed as 0 and with exactly one 0.

Sample Input

-13

Sample Output

110111

Hint

Explanation of the sample:

Reading from Right-to-left:

1*1 + 1*-2 + 1*4 + 0*-8 +1*16 + 1*-32 =-13


Note: Arbitrary negative input is also the case

#include <cstdio> #include <stack> #include <cstring> #include <algorithm>using namespace std; typedef long Long Ll;ll Read ()//input plug-in {ll x=0,f=1;char Ch=getchar (); while (ch< ' 0 ' | | Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();} while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();} return x*f;} int main () {    int n,i;    N=read ();    stack<int>s;    if (!n) {printf ("0\n"); return 0;} 0 requires a special sentence    while (n)    {for        (i=0;; ++i)        if ((n-i)%2==0) break;        S.push (i);        n= (n-i)/( -2);    }    while (!s.empty ()) {        printf ("%d", s.top ());        S.pop ();}        printf ("\ n");}

　　

A-the moronic Cowmpouter