m 元 n 次式的每一项（同类项）都是以下形式： product(x_i ^ k_i, i=1 to m), where x_i 's are the variables and k_i 's are the powers with k_i >=0 and sum(k_i, i=1 to m) <= n . Easy to see that a unique feasible (k1, k2, ... k_m) vector corresponds to a unique term. Therefore # of feasible vectors (k1, k2, ..., k_m) subject to k_i >=0, sum(k_i) <=n is exactly the # of terms requested. www.ddhw.com Furthermore, let (w1, w2, ..., w_m) be a vector of increasing positive numbers chosen from {1, 2, 3,..., m+n}, subject to: 0 The mapping: k_i := w_i - w_(i-1) - 1 is apparently 1-1, and it satisfies: k_i>=0 and sum(k_i, i=1 to m) = sum( w_i - w_(i-1) -1, i=1 to m) = w_m - w0 -m = w_m - m <= m+n-m=n. Therefore, the k_i 's will form a feasible vector. And for every feasible vector (k_i), the reverse mapping is obvious. www.ddhw.com Therefore the # of terms = the # of choices of m out from m+n numbers = C(m, m+n). |
哈哈哈哈 O(∩_∩)O哈哈~ 好久不来了，不小心看了一眼自己主页，发现自动关联了自己以往发过的帖，点进这个看了一眼，笑个半死。。。敢情我当初还发过这么搞笑的东西啊，我自己完全不记得了 
雷达卡
发表于 2007-11-19 09:39:05
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千斤顶
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发表于 2007-11-20 00:31:19
HF兄的数学功底和逻辑思维厉害啊）。但当 m 或 n 较大时，恐怕难以“扳”出个正确的结果来。例如，对该院士学术报告中采用的 m=100 和 n=3 ，用递推公式来计算 f(m,n)，实在是一条漫长的、容易出错的路。朋友们能否给出 f(m,n) 的一个显式 (explicit expression)，并用那显式算出 f(100,3) 来？
。（2）呢？扳个100元3次的试试。