The Moronic Cowmpouter poj3191（负进制转换以及其他进制转换模板）

The Moronic Cowmpouter

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4078 Accepted: 2111

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.

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:

1*1 + 1*-2 + 1*4 + 0*-8 +1*16 + 1*-32 = -13

裸的进制转换t

-2进制转换

#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <cmath>using namespace std;#define ll long long#define For(i,a,b) for(i=a;i<=b;i++)#define _For(i,a,b) for(i=b;i>=a;i--)const ll inf = 1e5;ll abs(ll n){    return n>=0?n:-n;}ll digit[inf];void exchange(ll n,ll c)  ///进制转换函数{     if(n==0)   ///注意特判0    {        cout<<0<<endl;        return ;    }    ll i,cnt=0;    ll num;    while(n)    {        num = n%c;        cnt++;        digit[cnt]=abs(num); ///保证每位数都是正数        n-=abs(num);         ///类比正进制 每次都减去余正数        n/=c;                ///与正进制相同    }    _For(i,1,cnt) cout<<digit[i];  ///倒序打印    cout<<endl;}int main(){    ll n;    ll c = -2;  ///多少进制    while(cin>>n)    {        exchange(n,c);    }    return 0;}