POJ 3191 The Moronic Cowmpouter

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

You know number bases have place values that 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 are 1, −2, 4, −8, 16, −32, ... (reading from right to left). Thus, counting from 1 goes like this: 1, 110, 111, 100, 101, 11010, 11011, 11000, 11001, and so on.

Eerily, negative numbers are also represented with 1's and 0's but no sign. Consider counting from −1 downward: 11, 10, 1101, 1100, 1111, and so on.

Please help the cows convert ordinary decimal integers (range -2,000,000,000..2,000,000,000) to their counterpart representation in base −2.

Input

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

Output

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

说白了 给一个数的负二进制 求出负二进制

#include "iostream"#include "cstring"#include "cstdio"#define N 100000using namespace std;int n;int sk[N],tp;int main(){    scanf("%d",&n);    while(n)    {        int k=n%-2;        int t=n/-2;        if(k<0)        {            k+=2;            t++;        }        n=t;        sk[++tp]=k;    }    for(int i=tp;i>=1;i--)        printf("%d",sk[i]);    if(tp==0) printf("0");    printf("\n");    return 0;}