The Fun Number System
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 10641 | Accepted: 3619 |
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.
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.
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
2 3 pnp 6 4 ppnn 10
Sample Output
Impossible 1110
题意:
给你一个字符串和一个数,问你能否将这个字符串转化成这个数,p代表正数,n代表负数。
题解:
- 如果nn是偶数,那么最后一位一定是0。
- 如果是奇数,那么最后一位一定是1。如果对应的比特位为正,那么n←(n−1)/2n←(n−1)/2,否则 n←(n+1)/2n←(n+1)/2。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int main()
{
int t;
cin>>t;
while(t--)
{
ll p, n;
string s;
cin>>p>>s>>n;
ll len = s.length();
while(p--)
{
if(n&1)
{
if(s[p] == 'n')
n++;
s[p] = '1';
}
else
s[p] = '0';
n>>=1;
}
if(n)
printf("Impossible\n");
else
cout<<s<<endl;
}
return 0;
}