uva 11166 - Power Signs (贪心)

Problem B
Power Signs
Input: Standard Input

Output: Standard Output

 

"Increase my killing power, eh?"

–        Homer Simpson

–         

You are probably familiar with the binary representation of integers, i.e. writing a nonnegative integer n as ∑a_i2ⁱ, where each a_i is either 0 or 1. In this problem, we consider a so called signed binary representation, in which we still write n as ∑a_i2ⁱ, but allow a_i to take on the values -1, 0 and 1. We write a signed binary representation of a number as a vector (a_k, a_k-1, ..., a₁, a₀). For instance, n = 13 can be represented as (1, 0, 0, -1, -1) = 2⁴ - 2¹ - 2⁰.

The binary representation of a number is unique, but obviously, the signed binary representation is not. In certain applications (e.g. cryptography), one seeks to write a number n in signed binary representation with as few non-zero digits as possible. For example, we consider the representation (1, 0, 0, -1) to be a better representation of n = 7 than (1, 1, 1). Your task is to write a program which will find such a minimal representation.

Input

The input consists of several test cases (at most 25), one per line. Each test case consists of a positive integer n ≤ 2⁵⁰⁰⁰⁰ written in binary without leading zeros. The input is terminated by a case where n = 0, which should not be processed.

 

Output

For each line of input, output one line containing the signed binary representation of n that has the minimum number of non-zero digits, using the characters '-' for -1, '0' for 0 and '+' for +1. The number should be written without leading zeros. If there are several possible answers, output the one that is lexicographically smallest (by the ASCII ordering).

 

Sample Input                               Output for Sample Input

10000

1111

10111

0

 

+0000

+000-

++00-

 

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Problem setter: Per Aurtrin

Special Thanks: Mikael Goldmann

让0减少的唯一方式，就是类似用+00-这样的形式替换111，还要注意连续的2个1，是否替换？如果2个1在开头，就不换，否则换，字典序会变小。

#include<cstdio>#include<map>#include<queue>#include<cstring>#include<iostream>using namespace std;const int maxn = 50000 + 5;typedef long long LL;typedef pair<int,int> P;char s[maxn];void change(int last, int pos){    s[last] = '-';    for(int j = last-1;j > pos;j--){        s[j] = '0';    }    s[pos] = '1';}int main(){    while(scanf("%s", s+1)){        if(s[1] == '0') break;        int n = strlen(s+1);        s[0] = '0';        int pos = n;        int sum = 0;        int last = -1;        while(pos >= 1){            if(s[pos]=='1'){                if(last == -1){                    last = pos;                }                sum++;            }            else{                if(sum >= 2){                    change(last, pos);                    sum = 1;                    last = pos;                }                else{                    sum = 0;                    last = -1;                }            }            pos--;        }        if(sum > 2){            change(last, pos);        }        for(int i = 0;i <= n;i++){            if(s[i]=='1')                s[i] = '+';        }        if(s[0]=='0')            printf("%s\n",s+1);        else            printf("%s\n",s);    }    return 0;}