N皇后问题公式解!!!

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)

 

 

下面是POJ 3239的代码:

 

#include<iostream>
using namespace std;
int main()
{
    int n,k,i;
    while(1)
    {
       cin>>n;
       if(n==0)break;

       if(n%6!=2 && n%6!=3)
       {
           if(n%2==0)
           {
              for(i=2;i<=n;i+=2)cout<<i<<' ';
              for(i=1;i<n-1;i+=2)cout<<i<<' ';
              cout<<n-1<<endl;
           }else
           {
              for(i=2;i<n;i+=2)cout<<i<<' ';
              for(i=1;i<n;i+=2)cout<<i<<' ';
              cout<<n<<endl;
           }
       }else
       {
           k=n/2;
           if(k%2==0 && n%2==0)
           {
              for(i=k;i<=n;i+=2)cout<<i<<' ';
              for(i=2;i<=k-2;i+=2)cout<<i<<' ';
              for(i=k+3;i<=n-1;i+=2)cout<<i<<' ';
              for(i=1;i<k+1;i+=2)cout<<i<<' ';
              cout<<k+1<<endl;
           }
           else if(k%2==0 && n%2==1)
           {
              for(i=k;i<=n-1;i+=2)cout<<i<<' ';
              for(i=2;i<=k-2;i+=2)cout<<i<<' ';
              for(i=k+3;i<=n-2;i+=2)cout<<i<<' ';
              for(i=1;i<=k+1;i+=2)cout<<i<<' ';
              cout<<n<<endl;
           }
           else if(k%2==1 && n%2==0)
           {
              for(i=k;i<=n-1;i+=2)cout<<i<<' ';
              for(i=1;i<=k-2;i+=2)cout<<i<<' ';
              for(i=k+3;i<=n;i+=2)cout<<i<<' ';
              for(i=2;i<k+1;i+=2)cout<<i<<' ';
              cout<<k+1<<endl;
           }
           else
           {
              for(i=k;i<=n-2;i+=2)cout<<i<<' ';
              for(i=1;i<=k-2;i+=2)cout<<i<<' ';
              for(i=k+3;i<=n-1;i+=2)cout<<i<<' ';
              for(i=2;i<=k+1;i+=2)cout<<i<<' ';
              cout<<n<<endl;
           }
       }
    }
    return 0;
}