Poj 3191 The Moronic Cowmpouter【十进制转负二进制】

The Moronic Cowmpouter Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3917 Accepted: 2041

Description

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

Source

USACO 2006 February Bronze

题目大意：

给你一个十进制数，让你将其转化为负二进制表示。

思路（网络学习而来）：

①对于一个数，即使的-2进制数表示，结果也只有0或者1，那么对于a【i】%（-2）<0的部分，要取绝对值。

②那么整个求解过程：

1、记录答案：abs[a%（-2）]；

2、去掉余数：a-abs[a%（-2）]；

3、除以-2：a/=-2；

4、一直处理到变成0为止，然后逆序输出。

③注意n==0的时候。直接输出0即可。

Ac代码：

[cpp]  view plain  copy  print ? #include<stdio.h>   #include<string.h>   #include<algorithm>   using namespace std;   int ans[1000];   int main()   {       int a;       while(~scanf("%d",&a))       {           if(a==0)           {               printf("0\n");               continue;           }           int cont=0;           while(a)           {               ans[cont++]=abs(a%(-2));               a-=abs(a%(-2));               a/=(-2);           }           for(int i=cont-1;i>=0;i--)printf("%d",ans[i]);           printf("\n");       }   }