1023 The Fun Number System

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本文介绍了一种特殊的二进制数系统——Fun数系统，并提供了一种算法来将任意整数转换为这种数系统的表示形式。文章详细解释了Fun数系统的定义、特性及其表示方法，并通过一个具体的例子展示了如何实现这一转换。

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Description
In a k bit 2’s complement number, where the bits are indexed from 0 to k-1, the weight of the most significant bit (i.e., in position k-1), is -2^(k-1), and the weight of a bit in any position i (0 ≤ i < k-1) is 2^i. For example, a 3 bit number 101 is -2^2 + 0 + 2^0 = -3. A negatively weighted bit is called a negabit (such as the most significant bit in a 2’s complement number), and a positively weighted bit is called a posibit.
A Fun number system is a positional binary number system, where each bit can be either a negabit, or a posibit. For example consider a 3-bit fun number system Fun3, where bits in positions 0, and 2 are posibits, and the bit in position 1 is a negabit. (110)Fun3 is evaluated as 2^2-2^1 + 0 = 3. Now you are going to have fun with the Fun number systems! You are given the description of a k-bit Fun number system Funk, and an integer N (possibly negative. You should determine the k bits of a representation of N in Funk, or report that it is not possible to represent the given N in the given Funk. For example, a representation of -1 in the Fun3 number system (defined above), is 011 (evaluated as 0 - 2^1 + 2^0), and
representing 6 in Fun3 is impossible.

Input
The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by the input data for each test case. Each test case is given in three consecutive lines. In the first line there is a positive integer k (1 ≤ k ≤ 64). In the second line of a test data there is a string of length k, composed only of letters n, and p, describing the Fun number system for that test data, where each n (p) indicates that the bit in that position is a negabit (posibit).
The third line of each test data contains an integer N (-2^63 ≤ N < 2^63), the number to be represented in the Funk number
system by your program.

Output
For each test data, you should print one line containing either a k-bit string representing the given number N in the Funk number system, or the word Impossible, when it is impossible to represent the given number.
方法一：
1.当前最后一位看，如果N为偶数，则该位必为0，N = N/2；
若N为奇数，则改位必为1，因为题目中每个位是分正负的，
若改位为正数位，则N = (N-1)/2；
若改位为负数位，则N = (N+1)/2；
重复上述步骤，直到k个位的0、1全部确定，此时，若N为0，则可以表示，否则，不可以表示。

#include<iostream>
#include<string.h>
using namespace std;
int flag[70];

int main(){
    int total_cases;
    cin>>total_cases;
    while(total_cases--){
        int k;
        cin>>k;
        char temp;
        for(int i = k-1; i >= 0;i--){
            cin>>temp;
            if(temp == 'p'){
                flag[i] = 1;
            }else{
                flag[i] = -1;
            }
        }
        long long int N;
        cin>>N;
        int result[64];
        memset(result,0,sizeof(result));
        for(int j = 0; j < k; j++){
            if(N%2 == 0){
                result[j] = 0;
                N = N/2;
            }else{
                result[j] = 1;
                if(flag[j] == 1){
                    N = (N-1)/2;
                }else{
                    N = (N+1)/2;
                }
            }

        }
        if(N != 0){
            cout<<"Impossible"<<endl;
        }else{
            for(int i = k-1; i >= 0; i--){
                cout<<result[i];
            }
            cout<<endl;
        }

    }
}