POJ_1023 The Fun Number System解题报告
The Fun Number System
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 8173 | Accepted: 2640 |
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
Source
Tehran 2002, First Iran Nationwide Internet Programming Contest
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
#include
#include
#include
__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