A - The Moronic Cowmpouter
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.
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:
Reading from right-to-left:
1*1 + 1*-2 + 1*4 + 0*-8 +1*16 + 1*-32 = -13
注：任意的负进制也是如此做法
#include<cstdio>
#include<stack>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll read()//输入外挂
{
ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int main()
{
int n,i;
n=read();
stack<int>s;
if(!n) {printf("0\n");return 0;}//0 需要特判
while(n)
{
for(i=0;;++i)
if((n-i)%2==0) break;
s.push(i);
n=(n-i)/(-2);
}
while(!s.empty()){
printf("%d",s.top());
s.pop();}
printf("\n");
}
