G - The Fun Number System解题报告（贺奕凯）

G - The Fun Number System

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 1023

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

2
3
pnp
6
4
ppnn
10

Sample Output

Impossible
1110

题目链接：http://poj.org/problem?id=1023
算法类型：数论
解题思路：这道题的思路是从最后一位开始判断，如果N是奇数，那么说明最后一位肯定是1。因为只有最后一位才是0次方，才有可能得到唯一的奇数1。这时如果最后那位是+时，就把N-1再除以2，得到前面的k-1位数；如果最后那位是-时，就把N+1再除以2，得到前面的k-1位数。如果N是偶数，可以断定最后一位是0，那么直接除以2得到前面k-1位得到的数。依次循环至剩下0位即可。如果循环完毕N没有等于0，可以判断为impossible，因为可以存在负数，上述考虑都是排除负数的情况，有负数也没关系，判断N%2==-1就行了。
算法实现：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
__int64 N;    
char P[64];
int Er[64];
int d;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&d);
		memset(P,0,sizeof(P));
		memset(Er,0,sizeof(Er));
		getchar();
		gets(P);
		scanf("%I64d",&N);
		int i=d-1;
		for(i;i>=0;i--)
		{
			if(N%2!=0)
			{
				Er[i]=1;
			    if(P[i]=='n')
				{N=N+1;
				}
				else
				{N=N-1;}
			    N=N/2;
			}
			else
			{
				Er[i]=0;
				N=N/2;
			}
		}
		if(N==0)
		{
			for(int j=0;j<d;j++)
				printf("%d",Er[j]);
			printf("\n");
		}
		else
		{
			printf("Impossible\n");
		}
	}
	return 0;
}