POJ 3191The 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

题意就是给一个数，让你转化成二进制形式的数，只不过这一次的权值不是2的n次方而是-2的n次方。

这个需要好好观察一下hint，发现当前的权值存不存在，只要对下一个权值取余就知道了，如果余数为0，那么当前这个数自然是后面一个权值的倍数了，取0，否则就取1.

然后减去当前权值，继续往后走。

最后减到0就可以了。

注意要用long long 跟普通二进制是不同的！

注意n为0的情况！！

#include <cstdio>#include <algorithm>#include <vector>#include <cmath>using namespace std;const int MAXN=1e5+7;const int inf =1e9;int ans[1000];long long n;long long pw(int a,int b){    long long sum=1;    long long base=a;    while(b)    {        if(b&1)sum*=base;        base*=base;        b>>=1;    }    return sum;}int main(){    scanf("%lld",&n);    int top=0;    if(!n)ans[top++]=0;    while(n)    {        if(n%(pw(-2,top+1)))ans[top]=1;        else ans[top]=0;        n-=ans[top]*pw(-2,top);        top++;    }    while(top)printf("%d",ans[--top]);    return 0;}