poj3239

题意：给出正方形棋盘边长n（最大300），要求输出一种摆放n皇后不冲突的方案。

分析：

数据范围较大，只能用构造的方法，不能用搜索。

下面用一个数列表示一种方案，第i个数表示棋盘第i行上的皇后所在的列号

n皇后问题构造法：

一、当n mod 6 != 2 且 n mod 6 != 3时，有一个解为：
2,4,6,8,...,n,1,3,5,7,...,n-1        (n为偶数)
2,4,6,8,...,n-1,1,3,5,7,...,n        (n为奇数)
(上面序列第i个数为ai,表示在第i行ai列放一个皇后;...省略的序列中,相邻两数以2递增。下同)
二、当n mod 6 == 2 或 n mod 6 == 3时，
(当n为偶数,k=n/2；当n为奇数,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为偶数,n为偶数)
k,k+2,k+4,...,n-1,2,4,...,k-2,k+3,k+5,...,n-2,1,3,5,...,k+1,n    (k为偶数,n为奇数)
k,k+2,k+4,...,n-1,1,3,5,...,k-2,k+3,...,n,2,4,...,k+1            (k为奇数,n为偶数)
k,k+2,k+4,...,n-2,1,3,5,...,k-2,k+3,...,n-1,2,4,...,k+1,n        (k为奇数,n为奇数)

第二种情况可以认为是，当n为奇数时用最后一个棋子占据最后一行的最后一个位置，然后用n-1个棋子去填充n-1的棋盘，这样就转化为了相同类型且n为偶数的问题。

若k为奇数，则数列的前半部分均为奇数，否则前半部分均为偶数。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
usingnamespace std;

int n;

int main()
{
    //freopen("t.txt", "r", stdin);
while (scanf("%d", &n), n)
    {
        if (n %6!=2&& n %6!=3)
        {
            printf("2");
            for (int i =4; i <= n; i +=2)
                printf(" %d", i);
            for (int i =1; i <= n; i +=2)
                printf(" %d", i);
            putchar('\n');
            continue;
        }
        int k = n /2;
        printf("%d", k);
        if (!(k &1) &&!(n &1))
        {
            for (int i = k +2; i <= n; i +=2)
                printf(" %d", i);
            for (int i =2; i <= k -2; i +=2)
                printf(" %d", i);
            for (int i = k +3; i <= n -1; i +=2)
                printf(" %d", i);
            for (int i =1; i <= k +1; i +=2)
                printf(" %d", i);
        }
        elseif (!(k &1) && (n &1))
        {
            for (int i = k +2; i <= n -1; i +=2)
                printf(" %d", i);
            for (int i =2; i <= k -2; i +=2)
                printf(" %d", i);
            for (int i = k +3; i <= n -2; i +=2)
                printf(" %d", i);
            for (int i =1; i <= k +1; i +=2)
                printf(" %d", i);
            printf(" %d", n);
        }
        elseif ((k &1) &&!(n &1))
        {
            for (int i = k +2; i <= n -1; i +=2)
                printf(" %d", i);
            for (int i =1; i <= k -2; i +=2)
                printf(" %d", i);
            for (int i = k +3; i <= n; i +=2)
                printf(" %d", i);
            for (int i =2; i <= k +1; i +=2)
                printf(" %d", i);
        }
        elseif ((k &1) && (n &1))
        {
            for (int i = k +2; i <= n -2; i +=2)
                printf(" %d", i);
            for (int i =1; i <= k -2; i +=2)
                printf(" %d", i);
            for (int i = k +3; i <= n -1; i +=2)
                printf(" %d", i);
            for (int i =2; i <= k +1; i +=2)
                printf(" %d", i);
            printf(" %d", n);
        }
        putchar('\n');
    }
    return0;
}