[POJ3191] The Moronic Cowmpouter

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题目

题目描述

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.

输入输出格式

输入格式：

A single integer to be converted to base -2

输出格式：

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.

输入输出样例

输入样例1:

1
-13

输出样例1：

1
110111

提示

Hint
Explanation of the sample:
Reading from right-to-left:
1*1 + 1*-2 + 1*4 + 0*-8 +1*16 + 1*-32 = -13

题目大意

输入一个十进制[latex]N(−2,000,000,000≤N≤2,000,000,000)[/latex]，输出它的[latex]−2[/latex]进制数

题解

平常我们很少会涉及到负进制，按照提示所给内容，我们应该很快能想到第一种做法，
第一种想法，我们去枚举每一个负二进制数，对于每一个负二进制数，我们将它转换成十进制数，看看是不是一样。
代码如下：

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#include<iostream>
#include<cstdio>
using namespace std;
int n;
long long p[32];
 
void init(){
    cin>>n;
    p[0]=1;
    for(int i=1;i<=31;i++){
        p[i]=p[i-1]*-2;
    }
}
 
inline bool jud(int num){
    int res=0,cnt=0;
    while(num){
        if(num&1)res+=p[cnt];
        ++cnt;
        num>>=1;
    }
    if(res==n)return 1;
    return 0;
}
 
void out(int num){
    int tmp[200],t=0;
    while(num){
        if(num&1)tmp[++t]=1;
        else tmp[++t]=0;
        num>>=1;
    }
    for(int i=t;i>=1;i--)printf("%d",tmp[i]);
}
 
int main(){
    init();
    for(long long i=1;i<=2000000000LL;i++){
        if(jud(i)){
            // cout<<"!!"<<i<<endl;
            out(i);
            break;
        }
    }
    return 0;
}

然而上面的做法会T，那么我们能不能直接从十进制算到负二进制呢？
这个需要好好观察一下提示，
我们发现如果要确定当前位是0还是1，只要对下一个为1时取余就知道了，如果余数为0，那么当前这个数自然是后面位的倍数了，取0，否则就取1.

然后减去当前还剩的数，继续往后走。

最后减到0就可以了。

代码

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#include <cstdio>
#include <iostream>
using namespace std;
const int MAXN=1e5+7;
const int inf =1e9;
int ans[1000];
long long n;
long long pw(int a,int b){//很普通的快速幂
    long long sum=1;
    long long base=a;
    while(b){
        if(b&1)sum*=base;
        base*=base;
        b>>=1;
    }
    return sum;
}
int main(){
    scanf("%lld",&n);
    int top=0;
    if(!n)ans[top++]=0;//0的时候特判
    while(n){
        if(n%(pw(-2,top+1)))ans[top]=1;
        else ans[top]=0;
        n-=ans[top]*pw(-2,top);//相当于不断把n这个10进制数拆分成-2进制数
        top++;
    }
    while(top)printf("%d",ans[--top]);
    return 0;
}

另外，这里还有一道题和这个几乎一样
UVA11121 Base -2