Poj3239 Solution to the n Queens Puzzle (n queen problem)

Source: Internet
Author: User

Poj3239 Solution to the n Queens Puzzle (n queen problem)

Solution to NQueens Puzzle
Time Limit:1000 MS Memory Limit:131072 K
Total Submissions:3494 Accepted:1285 Special Judge

Description

The eight queens puzzle is the problem of putting eight chess queens on an 8 × 8 chessboard such that none of them is able to capture any other. The puzzle has been generalized to arbitraryN×NBoards. GivenN, You are to find a solution toNQueens puzzle.

Input

The input contains multiple test cases. Each test case consists of a single integerNBetween 8 and 300 (inclusive). A zero indicates the end of input.

Output

For each test case, output your solution on one line. The solution is a permutation of {1, 2 ,...,N}. The number inITh place meansITh-column queen in placed in the row with that number.

Sample Input

80

Sample Output

5 3 1 6 8 2 4 7

Solution:

1. When n mod 6! = 2 or n mod 6! = 3:


[,..., N], [,..., n-1] (n is an even number)


[,..., N-1], [,..., n] (n is an odd number)

2. When n mod 6 = 2 or n mod 6 = 3


(When n is an even number, k = n/2; when n is an odd number, k = (n-1)/2)


[K, k + 2, k + 4 ,..., n], [2, 4 ,..., k-2], [k + 3, k + 5 ,..., n-1], [1, 3, 5 ,..., k + 1] (k is an even number, n is an even number)

[K, k + 2, k + 4 ,..., n-1], [2, 4 ,..., k-2], [k + 3, k + 5 ,..., n-2], [, 5 ,..., k + 1], [n] (k is an even number, n is an odd number)

[K, k + 2, k + 4 ,..., n-1], [1, 3, 5 ,..., k-2], [k + 3 ,..., n], [2, 4 ,..., k + 1] (k is an odd number, n is an even number)

[K, k + 2, k + 4 ,..., n-2], [, 5 ,..., k-2], [k + 3 ,..., n-1], [2, 4 ,..., k + 1], [n] (k is an odd number, n is an odd number)

(There are six sequences above. One sequence in a row, and the brackets are added to help you identify the subsequence. The subsequence and the subsequence are in a continuous relationship. Ignore the brackets. The number of I is ai, indicating that there is a queen in the ai column of row I; In the sequence omitted by..., the numbers of adjacent two increase by 2 .)

Reference code:

# Include
 
  
# Include
  
   
Using namespace std; int main (int I) {int n; // Number of Queens while (cin> n) {if (! N) break; if (n % 6! = 2 & n % 6! = 3) {if (n % 2 = 0) // n is an even number {for (I = 2; I <= n; I + = 2) cout <
   
    


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