# 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 2 |
-13 |
输出样例1：
|
1 2 |
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\(N(−2,000,000,000≤N≤2,000,000,000)\)，输出它的\(−2\)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：
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 |
#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
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 |
#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