Base -2 - UVa 11121 -2进制

最新推荐文章于 2022-04-03 11:12:43 发布

提比-我有特殊的AC技巧

最新推荐文章于 2022-04-03 11:12:43 发布

阅读量589

点赞数

CC 4.0 BY-SA版权

分类专栏：数论 UVa

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/43407223

UVa 同时被 2 个专栏收录

218 篇文章

订阅专栏

数论

32 篇文章

订阅专栏

本文介绍了一种算法，用于将十进制整数转换为其在-2进制下的唯一表示形式，包括负数，并提供了一个AC代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Base -2
Input: Standard Input

Output: Standard Output

The creator of the universe works in mysterious ways. But
he uses a base ten counting system and likes round numbers.

Scott Adams

Everyone knows about base-2 (binary) integers and base-10 (decimal) integers, but what about base -2? An integer n written in base -2 is a sequence of digits (b_i), writen right-to-left. Each of which is either 0 or 1 (no negative digits!), and the following equality must hold.

n = b₀ + b₁(-2) + b₂(-2)² + b₃(-2)³ + ...

The cool thing is that every integer (including the negative ones) has a unique base--2 representation, with no minus sign required. Your task is to find this representation.

Input
The first line of input gives the number of cases, N (at most 10000). N test cases follow. Each one is a line containing a decimal integer in the range from -1,000,000,000 to 1,000,000,000.

Output
For each test case, output one line containing "Case #x:" followed by the same integer, written in base -2 with no leading zeros.

Sample Input Output for Sample Input

-2

Case #1: 1

Case #2: 11011

Case #3: 10

Case #4: 0

题意：将一个数转成-2进制。

思路：首先将N位可以表示的数的范围计算出来，然后依次往下看当前位是否需要为1即可。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
ll pow2[40],l[45],r[45];
int main()
{
    int T,t,n,i,j,k,pos;
    ll num,f;
    f=1;pow2[1]=1;
    for(i=2;i<=35;i++)
       pow2[i]=pow2[i-1]*(-2);
    for(i=1;i<=35;i++)
    {
        l[i]=l[i-1];
        r[i]=r[i-1];
        if(i&1)
          r[i]+=f;
        else
          l[i]+=f;
        f*=-2;
    }
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%lld",&num);
        pos=1;
        while(!(l[pos]<=num && num<=r[pos]))
          pos++;
        f=num;
        printf("Case #%d: ",t);
        for(i=pos;i>=1;i--)
        {
            if(l[i-1]<=num && num<=r[i-1])
              printf("0");
            else
            {
                printf("1");
                num-=pow2[i];
            }
        }
        printf("\n");
    }
}