Base -2 - UVa 11121 -2进制

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

4

1

7

-2

0

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");    }}