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强悍,赞!!!!!!In Reply To:Re:大牛们是如何想出构造的??偶至今还没有想明白呀!!!!! Posted by:nuciedh at 2007-06-18 23:48:28 > 一、当n mod 6 != 2 或 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) Followed by: Post your reply here: |
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