poj 1147 Binary codes

Binary codes

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5647 Accepted: 2201

Description

Consider a binary string (b₁…b_N) with N binary digits. Given such a string, the matrix of Figure 1 is formed from the rotated versions of the string.

b₁b₂…b_N−1b_Nb₂b₃…b_Nb₁…b_N−1b_N…b_N−3b_N−2b_Nb₁…b_N−2b_N−1

Figure 1. The rotated matrix

Then rows of the matrix are sorted in alphabetical order, where ‘0’ is before ‘1’. You are to write a program which, given the last column of the sorted matrix, finds the first row of the sorted matrix.

As an example, consider the string (00110). The sorted matrix is

0001100110011001000111000

and the corresponding last column is (1 0 0 1 0). Given this last column your program should determine the first row, which is (0 0 0 1 1).

Input

The first line contains one integer N ≤ 3000, the number of binary digits in the binary string. The second line contains N integers, the binary digits in the last column from top to bottom.

Output

The first line contains N integers: the binary digits in the first row from left to right.

Sample Input

51 0 0 1 0

Sample Output

0 0 0 1 1

对由0,1组成的n个数，照题中的旋转，最后根据每行的字典序排序，组成n*n的矩阵，给出矩阵的最后1列，求矩阵

的第首行。

给出最后一列可以求出第0列，因为是按字典序排的，所以第0列肯定0在前，1在后，而第0列为0的相对位置在最后

1列不变，因为第0列都为0，又是按字典序排的。第0列为1也一样。根据第0列和最后一列就可以将对应关系求出，

也就是next数组。

例如样例的

0 0 0 1 1

0 0 1 1 0

0 1 1 0 0

1 0 0 0 1

1 1 0 0 0

next[ ]={1,2,4,0,3}

第0行第0列为0，第0行的第1列下次旋转后为第0列的第0行，所以第0行第1列为第0列的第1行为0，第1列的第0行

为第2列的第1行，为第0列的第2行，所以第0行的第2列为第0列的第2行为0，通过推理发现为next数组中元素递推

代码：

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int maxn=5000+100;int last[maxn];int first[maxn];int next[maxn];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)        {           scanf("%d",&last[i]);           first[i]=last[i];        }        sort(first,first+n);        int cur=0;        int i;        for(i=0;i<n;i++)        {            if(first[i])            break;            while(last[cur]&&cur<n)            cur++;            next[i]=cur++;        }        cur=0;        for(i=i;i<n;i++)        {            while(last[cur]==0&&cur<n)            cur++;            next[i]=cur++;        }        int  k=0;        for(int i=0;i<n-1;i++)        {           printf("%d ",first[k]);           k=next[k];        }        printf("%d\n",first[k]);    }    return 0;}