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; }