poj1023 二进制-十进制

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如题：http://poj.org/problem?id=1023

The Fun Number System

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 10018 Accepted: 3381

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

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest

思路：一开始我的思路是从高位来搜索，判断这一位是0或1，然后判断后面低位的数能否组成余下的数，由于负数的存在和N (-2^63 ≤ N < 2^63）的范围，在余下的数的时候会超界。

其实一个十进制的数肯定由唯一的二进制数来组成，即便是这几位正负给出，也是唯一的，因此每次判断最低位的数，如果N是奇数，最低位肯定是1，如果这位要求为正，

剩下的数，也就是这一位前面得数是(N-1)/2. 如果这一位要求为负，N=（N+1）/2.如果N偶数，这一位是0，N/=2;

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define ll _int64

char s[70];
int k;
ll n;
int res[70];

int main()
{
// freopen("C:\\1.txt","r",stdin);
int t;
cin>>t;
int i;
while(t--)
{
  memset(res,0,sizeof(res));
  cin>>k;
  cin>>s;
  scanf("%I64d",&n);
  for(i=k-1;i>=0;i--)
  {
   if(n%2)
   {
    res[i]=1;
    if(s[i]=='p')
     n=(n-1)/2;
    else
     n=(n+1)/2;
   }
   else
   {
    res[i]=0;
    n/=2;
   }
  }
  if(n)
   printf("Impossible\n");
  else
  {
   for(i=0;i<=k-1;i++)
    printf("%d",res[i]);
   printf("\n");
  }
}
return 0;
}

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