The Fun Number System

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.

Sample Input

23pnp64ppnn10

Sample Output

Impossible1110

#include<iostream>#include<string>using namespace std;void init(string str,int fuhao[]){int i=0;for(;i<str.size();i++){if(str[str.size()-i-1]=='p')fuhao[i]=1;if(str[str.size()-i-1]=='n')fuhao[i]=-1;}}long long int cheng2(int n){long long i=1;while(n--){i=i*2;}return i;}void print(int num[],int l){for(int i=l-1;i>=0;i--)cout<<num[i];cout<<endl;}int main(){int thetimes;cin>>thetimes;while(thetimes--){int l;string str;long long int res;cin>>l>>str>>res;int result[100]={0};int fuhao[100]={0};init(str,fuhao);//long long all=0;for(int i=0;i<l;i++){if(res%2==0){result[i]=0;res=res/2;}else{result[i]=1;res=(res-fuhao[i])/2;}}if(res==0){print(result,l);}elsecout<<"Impossible"<<endl;}}

思考要更加严密，每个位置都可能是0，最好是由后向前进行判断，若为奇数，则为1，若为偶数，应为0