# worm

## poj 3191解题报告

The Moronic Cowmpouter
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 1344 Accepted: 676

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:

`1*1 + 1*-2 + 1*4 + 0*-8 +1*16 + 1*-32 = -13`
题目的意思就是让你把一个数转换为-2进制，

如果n为奇数，那么末位肯定为1，因此就可以转为求(x-1)/-2 的子问题了！ （除以-2可以类比十进制就能理解了）
如果n为偶数，末位必为0，然后再转为求x/-2的子问题；

1//============================================================================
2// Name        : poj 3191.cpp
3// Author      : worm
5// Description :把一个十进制的数转换为-2进制的数
6//============================================================================
7
8#include <iostream>
9#include <stdio.h>
10#include <string>
11#include <math.h>
12using namespace std;
13string a= "";
14int main() {
15    int x;
16    cin >> x;
17    if (x == 0{
18        cout <<"0"<<endl;
19        return 0;
20    }

21
22    while (x != 1{
23        if (abs(x) % 2 == 1{
24            a += '1';
25            x = (x-1)/-2;
26            continue;
27        }

28        a += '0';
29        x /= -2;
30    }

31    cout <<"1";
32    for (int i = a.length() - 1; i >= 0; i--{
33        printf("%c",a[i]);
34    }

35
36    return 0;
37}

38

posted on 2009-03-08 12:37 WORM 阅读(1007) 评论(1)  编辑 收藏 引用

## #re: poj 3191解题报告回复更多评论

2009-03-08 20:24 | cppexplore