POJ 3191 The Moronic Cowmpouter(二进制的变形)

 

The Moronic Cowmpouter

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2402		Accepted: 1227

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

 

 解题报告：这道题是让求负二进制，先打表储存每位的可以表示的最大值和最小值，在枚举即可；

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int MAX = 33;
struct node
{
    __int64 low;
    __int64 high;
}a[MAX];//存储第i个位置可以表示的最大值和最小值
int main()
{
    int i;
    __int64 n;
    a[0].low = 0;
    a[0].high = 1;
    for (i = 1; i < MAX; ++i)
    {
        if (i % 2)//这种位置上是表示的负数
        {
            if (i > 2)
            {
                a[i].low = -pow(2, i) + a[i - 2].low;
                a[i].high = a[i - 2].low;
            }
            else//边界处理
            {
                a[i].low = -pow(2, i);
                a[i].high = 0;
            }
        }
        else//正数的时候
        {
            a[i].high = pow(2, i) + a[i - 2].high;
            a[i].low = a[i - 2].high;
        }
    }
    scanf("%I64d", &n);
    int temp;//存储二进制的位数
    if (n < 0)
    {
        for (i = 1; i < MAX; i = i + 2)//负数的时候
        {
            if (a[i].low <= n)
            {
                temp = i;
                break;
            }
        }
    }
    else
    {
        for (i = 0; i < MAX; i = i + 2)//正数的时候
        {
            if (a[i].high >= n)
            {
                temp = i;
                break;
            }
        }
    }
    for (i = temp; i >= 0; i --)
    {
        int flag = 0;
        //把正负数分开主要是因为边界问题正数的时候是a[i].high >= n，而负数的时候是a[i].high > n
        if (a[i].high >= n && n > a[i].low && n > 0)//正数的时候
        {
            printf("1");
            flag = 1;
        }
        else if(a[i].low <= n && a[i].high > n && n < 0)//负数的时候
        {
            printf("1");
            flag = 1;
        }
        else
        {
            printf("0");
        }
        if (flag)
        {
            if (i % 2)//正数的时候
            {
                n += pow(2, i);
            }
            else//负数的时候
            {
                n -= pow(2, i);
            }
        }
    }
    printf("\n");
    return 0;
}