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Exponentiating by squaring

ZelluX — Sat, 17 Nov 2007 16:56:00 GMT

http://www.wiki.cn/wiki/Exponentiation_by_squaring

Exponentiating by squaring is an algorithm used for the fast computation of large integer powers of a number. It is also known as the square-and-multiply algorithm or binary exponentiation. In additive groups the appropriate name is double-and-add algorithm. It implicitly uses the binary expansion of the exponent. It is of quite general use, for example in modular arithmetic.

ZelluX 2007-11-18 00:56 发表评论

拿糖果的博弈问题

ZelluX — Tue, 16 Oct 2007 03:30:00 GMT

更详�l�的分析google Nim Game
http://www.math.ucla.edu/~tom/Game_Theory/comb.pdf

发信�? flyskyf (flysky), 信区: Algorithm
�? �? 拿糖果问�?br /> 发信�? 水木�C�֌� (Mon Oct 15 19:07:51 2007), 站内

现有4堆糖�?分别�?,2,4,8
甲乙两�h分别从中拿糖�?br />
规则:
1 每�h可以从某一堆中拿�Q意多�?br /> 2 甲乙两�h交替�?br /> 3 谁拿到最后一个糖果或最后几个糖果算�?

请问谁有必胜把握?怎样实现?

发信�? meeme (�c�鸣), 信区: Algorithm
�? �? Re: 拿糖果问�?br /> 发信�? 水木�C�֌� (Mon Oct 15 19:26:32 2007), 站内

转成二进�?br />
1 =0001
2 =0010
4 =0100
8-1 =0111 +
-----------
0222
�q�样每个位上都有两个1�?br /> 比如个位上，1�?在个位上都有一�?
�Ҏ��不可能同时把�q�两�?拿走。所以对�Ҏ��拿不完的�?br /> �Ҏ��拿完之后�Q�自己再拿若�q�个调整成这�U�状态�?br />
中间应该有不��证�?..

ZelluX 2007-10-16 11:30 发表评论

Ferrers囑փ�

ZelluX — Tue, 16 Oct 2007 03:19:00 GMT

http://www.ekany.com/wdg98/zhsx/2/2_6.htm

Ferrers囑փ�

一个从上而下的n层格子，m_i 为第i层的格子敎ͼ�当m_i>=m_i+1(i=1,2,^...,n-1) �Q�即上层的格子数不少于下层的格子数时�Q�称之�ؓFerrers囑փ��Q�如�?2-6-2)�C��?nbsp;

�?nbsp; (2-6-2)

Ferrers囑փ��h��如下性质�Q?nbsp;

1.每一层至��有一个格子�?nbsp;

2.�W�一行与�W�一列互换，�W�二行于�W�二列互换，…�Q�即�?2-6-3)�l�虚�U��u旋�{所得的图仍然是Ferrers囑փ�。两个Ferrers 囑փ��U�Cؓ一对共轭的Ferrers囑փ��?nbsp;

利用Ferrers囑փ�可得关于整数拆分的十分有��的�l�果�?nbsp;

(a)整数n拆分成k个数的和的拆分数�Q�和数n拆分成个数的和的拆分数相�{��?nbsp;

因整数n拆分成k个数的和的拆分可用一k行的囑փ�表示。所得的Ferrers囑փ�的共轭图像最上面一行有k个格子。例如：

�?nbsp; (2-6-3)

(b)整数n拆分成最多不��过m个数的和的拆分数�Q�和n拆分成最大不��过m的拆分数相等�? 理由�?a)相类伹{�?nbsp;

因此�Q�拆分成最多不��过m个数的和的拆分数的母函数�?nbsp;

拆分成最多不��过m-1个数的和的拆分数的母函数�?nbsp;

所以正好拆分成m个数的和的拆分数的母函数�?nbsp;

(c)整数n拆分成互不相同的若干奇数的和的的拆分�?和n拆分成自��p��的Ferrers囑փ�的拆分数相等. �?nbsp;

其中n₁>n₂>^...>n_k

构造一个Ferrers囑փ��Q�其�W�一行，�W�一列都是n₁+1��|��对应�?n₁+1�Q�第二行�Q�第二列各n₂+1��|��对应�?n₂+1。以此类推。由此得到的Ferres囑փ�是共轭的。反�q�来也一栗��?nbsp;

例如 17=9+5+3 对应为Ferrers囑փ��?

�?nbsp; (2-6-4)

费勒斯（Ferrers�Q�图�?o:p>

假定n拆分为n=n1+n2+n3+……+nk�Q�且n1>=n2>=n3>=……>=nk

我们��它排列成阶梯�Ş�Q�左边看齐，我们可以得到一个类似倒阶梯图像，�q�种囑փ�我们�U�C��为Ferrers囑փ��Q�如对于20=10+5+4+1�Q�我们有囑փ��Q?/p>

对于Ferrers囑փ��Q�我们很�Ҏ��知道以下两条性质�Q?/p>

�Q?�Q?nbsp; 每层臛_��一个格�?o:p>

�Q?�Q?nbsp; 行列互换�Q�所对应的图像仍为Ferrers囑փ��Q�他应该��囑փ�的共轭图�?o:p>

��L��的Ferrers囑փ�对应一个整数的拆分�Q�而可用Ferrers囑փ�方便地证明：

�Q?�Q?nbsp; n拆分为k个整数的拆分敎ͼ�与n拆分成最大数为k的拆分数相等

�Q?�Q?nbsp; n拆分为最多不��过k个数的拆分数�Q�与n拆分成最大数不超�q�k的拆分数相等

�Q?�Q?nbsp; n拆分��Z��不相同的若干奇数的拆分数�Q�与n拆分成图像自��p��的拆分的拆分数相�{?o:p>

ZelluX 2007-10-16 11:19 发表评论

ZelluX — Mon, 17 Sep 2007 11:55:00 GMT

n个不同的物体按固定次序入栈，随时可以出栈�Q�求最后可能的出栈序列的��L��?br /> 只想到这个问题等价于把n个push和n个pop操作排列�Q�要求�Q意前几个操作中push数都不能��于pop敎ͼ�至于�q�个排列数怎么求就不知道了。请教了peter大牛后，原来�q�就是一个Catalan数的应用�?br />

Wikipedia上的Catalan numbers:

In combinatorial mathematics, the Catalan numbers form a sequence of natural numbers that occur in various counting problems, often involving recursively defined objects. They are named for the Belgian mathematician Eugène Charles Catalan (1814–1894).

The n^th Catalan number is given directly in terms of binomial coefficients by

The first Catalan numbers (sequence A000108 in OEIS) for n = 0, 1, 2, 3, … are

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, …

Properties

An alternative expression for C_n is

This shows that C_n is a natural number, which is not a priori obvious from the first formula given. This expression forms the basis for André's proof of the correctness of the formula (see below under second proof).

The Catalan numbers satisfy the recurrence relation

They also satisfy:

which can be a more efficient way to calculate them.

Asymptotically, the Catalan numbers grow as

in the sense that the quotient of the n^th Catalan number and the expression on the right tends towards 1 for n → ∞. (This can be proved by using for n!.)

[edit]

Applications in combinatorics

There are many counting problems in combinatorics whose solution is given by the Catalan numbers. The book Enumerative Combinatorics: Volume 2 by combinatorialist Richard P. Stanley contains a set of exercises which describe 66 different interpretations of the Catalan numbers. Following are some examples, with illustrations of the case C₃ = 5.

C_n is the number of Dyck words of length 2n. A Dyck word is a string consisting of n X's and n Y's such that no initial segment of the string has more Y's than X's (see also Dyck language). For example, the following are the Dyck words of length 6:

XXXYYY XYXXYY XYXYXY XXYYXY XXYXYY.

Re-interpreting the symbol X as an open parenthesis and Y as a close parenthesis, C_n counts the number of expressions containing n pairs of parentheses which are correctly matched:

((())) ()(()) ()()() (())() (()())

C_n is the number of different ways n + 1 factors can be completely parenthesized (or the number of ways of associating n applications of a binary operator). For n = 3 for example, we have the following five different parenthesizations of four factors:

Successive applications of a binary operator can be represented in terms of a binary tree. It follows that C_n is the number of rooted ordered binary trees with n + 1 leaves:

If the leaves are labelled, we have the quadruple factorial numbers.

C_n is the number of non-isomorphic full binary trees with n vertices that have children, usually called internal vertices or branches. (A rooted binary tree is full if every vertex has either two children or no children.)

C_n is the number of monotonic paths along the edges of a grid with n × n square cells, which do not cross the diagonal. A monotonic path is one which starts in the lower left corner, finishes in the upper right corner, and consists entirely of edges pointing rightwards or upwards. Counting such paths is equivalent to counting Dyck words: X stands for "move right" and Y stands for "move up". The following diagrams show the case n = 4:

Image:Catalan number 4x4 grid example.svg

C_n is the number of different ways a convex polygon with n + 2 sides can be cut into triangles by connecting vertices with straight lines. The following hexagons illustrate the case n = 4:

Error creating thumbnail:

C_n is the number of stack-sortable permutations of {1, ..., n}. A permutation w is called stack-sortable if S(w) = (1, ..., n), where S(w) is defined recursively as follows: write w = unv where n is the largest element in w and u and v are shorter sequences, and set S(w) = S(u)S(v)n, with S being the identity for one-element sequences.

C_n is the number of noncrossing partitions of the set { 1, ..., n }. A fortiori, C_n never exceeds the nth Bell number. C_n is also the number of noncrossing partitions of the set { 1, ..., 2n } in which every block is of size 2. The conjunction of these two facts may be used in a proof by mathematical induction that all of the free cumulants of degree more than 2 of the Wigner semicircle law are zero. This law is important in free probability theory and the theory of random matrices.

[edit]

Proof of the formula

There are several ways of explaining why the formula

solves the combinatorial problems listed above. The first proof below uses a generating function. The second and third proofs are examples of bijective proofs; they involve literally counting a collection of some kind of object to arrive at the correct formula.

[edit]

First proof

We start with the observation that several of the combinatorial problems listed above can easily be seen to satisfy the recurrence relation

For example, every Dyck word w of length ≥ 2 can be written in a unique way in the form

w = Xw₁Yw₂

with (possibly empty) Dyck words w₁ and w₂.

The generating function for the Catalan numbers is defined by

As explained in the article titled Cauchy product, the sum on the right side of the above recurrence relation is the coefficient of xⁿ in the product

Therefore

Multiplying both sides by x, we get

So we have

and hence

The square root term can be expanded as a power series using the identity

which can be proved, for example, by the binomial theorem, (or else directly by considering repeated derivatives of ) together with judicious juggling of factorials. Substituting this into the above expression for c(x) produces, after further manipulation,

Equating coefficients yields the desired formula for C_n.

[edit]

Second proof

This proof depends on a trick due to D. André, which is now more generally known as the reflection principle (not to be confused with the Schwarz reflection theorem in complex analysis). It is most easily expressed in terms of the "monotonic paths which do not cross the diagonal" problem (see above).

Figure 1. The green portion of the path is being flipped.

Suppose we are given a monotonic path in an n × n grid that does cross the diagonal. Find the first edge in the path that lies above the diagonal, and flip the portion of the path occurring after that edge, along a line parallel to the diagonal. (In terms of Dyck words, we are starting with a sequence of n X's and n Y's which is not a Dyck word, and exchanging all X's with Y's after the first Y that violates the Dyck condition.) The resulting path is a monotonic path in an (n − 1) × (n + 1) grid. Figure 1 illustrates this procedure; the green portion of the path is the portion being flipped.

Since every monotonic path in the (n − 1) × (n + 1) grid must cross the diagonal at some point, every such path can be obtained in this fashion in precisely one way. The number of these paths is equal to

Therefore, to calculate the number of monotonic n × n paths which do not cross the diagonal, we need to subtract this from the total number of monotonic n × n paths, so we finally obtain

which is the nth Catalan number C_n.

[edit]

Third proof

The following bijective proof, while being more involved than the previous one, provides a more natural explanation for the term n + 1 appearing in the denominator of the formula for C_n.

Figure 2. A path with exceedance 5.

Suppose we are given a monotonic path, which may happen to cross the diagonal. The exceedance of the path is defined to be the number of pairs of edges which lie above the diagonal. For example, in Figure 2, the edges lying above the diagonal are marked in red, so the exceedance of the path is 5.

Now, if we are given a monotonic path whose exceedance is not zero, then we may apply the following algorithm to construct a new path whose exceedance is one less than the one we started with.

Starting from the bottom left, follow the path until it first travels above the diagonal.
Continue to follow the path until it touches the diagonal again. Denote by X the first such edge that is reached.
Swap the portion of the path occurring before X with the portion occurring after X.

The following example should make this clearer. In Figure 3, the black circle indicates the point where the path first crosses the diagonal. The black edge is X, and we swap the red portion with the green portion to make a new path, shown in the second diagram.

Figure 3. The green and red portions are being exchanged.

Notice that the exceedance has dropped from three to two. In fact, the algorithm will cause the exceedance to decrease by one, for any path that we feed it.

Figure 4. All monotonic paths in a 3×3 grid, illustrating the exceedance-decreasing algorithm.

It is also not difficult to see that this process is reversible: given any path P whose exceedance is less than n, there is exactly one path which yields P when the algorithm is applied to it.

This implies that the number of paths of exceedance n is equal to the number of paths of exceedance n − 1, which is equal to the number of paths of exceedance n − 2, and so on, down to zero. In other words, we have split up the set of all monotonic paths into n + 1 equally sized classes, corresponding to the possible exceedances between 0 and n. Since there are

monotonic paths, we obtain the desired formula

Figure 4 illustrates the situation for n = 3. Each of the 20 possible monotonic paths appears somewhere in the table. The first column shows all paths of exceedance three, which lie entirely above the diagonal. The columns to the right show the result of successive applications of the algorithm, with the exceedance decreasing one unit at a time. Since there are five rows, C₃ = 5.

[edit]

Hankel matrix

The n×n Hankel matrix whose (i, j) entry is the Catalan number C_i+j has determinant 1, regardless of the value of n. For example, for n = 4 we have

Note that if the entries are ``shifted", namely the Catalan numbers C_i+j+1, the determinant is still 1, regardless of the size of n. For example, for n = 4 we have

The Catalan numbers is the unique sequence with this property.

[edit]

Quadruple factorial

The quadruple factorial is given by , or . This is the solution to labelled variants of the above combinatorics problems. It is entirely distinct from the multifactorials.

[edit]

History

The Catalan sequence was first described in the 18th century by Leonhard Euler, who was interested in the number of different ways of dividing a polygon into triangles. The sequence is named after Eugène Charles Catalan, who discovered the connection to parenthesized expressions. The counting trick for Dyck words was found by D. André in 1887.

ZelluX 2007-09-17 19:55 发表评论

Functional Programming For The Rest of Us

ZelluX — Fri, 24 Aug 2007 14:54:00 GMT

摘要: http://www.defmacro.org/ramblings/fp.html�q�是攑֜�Mathematics�c�d�� Functional Programming For The Rest of Us Monday, June 19, 2006 Introduction Programmers are procrastinators. Get in, get some coffee, ... 阅读全文

ZelluX 2007-08-24 22:54 发表评论

��L��学习�W�记 - ��?(1)

ZelluX — Tue, 07 Aug 2007 14:31:00 GMT

1. 一个有限非�I�集合M到自�w�的一个双��称�?span style="color: red; font-weight: bold;">�|�换�?br>若M的元素个��Cؓn�Q�记M的所有置换的集合��C��S_n�Q�则不难得出S_n的元素个��Cؓn个元素的排列敎ͼ��?br>| S_n | = n!

2. Sn的一个把i₁变到i₂�Q�i₂变到i₃�Q?#8230;…�Q�i_k-1变到i_k�Q�i_k变到i₁�Q�而其余的元（如果�q�有�Q�不变的�|�换�U�Cؓk�?span style="color: red; font-weight: bold;">循环�|�换
�?br>(1 2 3 4 5 6) = (1) = (2) = ... = (6)�Q�称为恒�{�置�?br> 1 2 3 4 5 6

3.几个定理�Q?br>1) 设f�Q�g��Z��个不�怺�的��@环置换，则fg = gf
2) ��L��|�换均可唯一地分解成不相交��@环置换的乘积
�q�个定理可由构造法证得�Q�该证法同时也给��Z��分解为��@环置换的乘积的方�?br>3) ��L��|�换均可分解为对换的乘积�Q�不唯一�Q�，例如
(12)(345) = (12)(35)(34) = (12)(14)(41)(35)(34)

4. �|�换的奇偶�?br>1) 设f ∈Sn�Q�规定f的符号�ؓ
Sgn f = ∏[ f(i) - f(j) ] / (i - j)
貌似��是逆序�Ҏ��的奇偶性，奇�ؓ-1�Q�偶�?
2) Sgn(fg) = (Sgn f)(Sgn g)
3) n > 1�Ӟ��Sn中奇�|�换与偶�|�换的个数相�{�，均�ؓn! / 2
可通过分离一�l�对换积证得

ZelluX 2007-08-07 22:31 发表评论

ZelluX — Sat, 28 Jul 2007 11:20:00 GMT

1. 设A, B��Z��个集合，若存在从A到B的双��函敎ͼ�则称A与B�?strong style="COLOR: red">�{�势的，��CؓA≈B
N*N �?N的一�U�证明：构造双��函�?n = 2^a * (2b - 1)�?br>

2. 设A, B, C��Z�Q意的集合�Q�则
(1) A≈A
(2) 若A≈B�Q�则B≈A
(3) 若A≈B且B≈C�Q�则A≈C

3. Cantor定理
(1) N不与R�{�势
(2) 设A��Z�Q意的集合�Q�则A不与P(A)�{�势

4. 若一个集合A与某个自然数n�{�势�Q�则�U�A�?strong style="COLOR: red">有穷集合�Q�否则称A�?span style="COLOR: red">无穷集合�?/p>

ZelluX 2007-07-28 19:20 发表评论

��L��学习�W�记 - 自然�?(2)

ZelluX — Sat, 28 Jul 2007 08:37:00 GMT

1. 数学归纳法证明自然数的性质P�Q?br>�W�一�Q�构�?S = { n | n ∈N �?P(n) }
�W�二�Q�证明S是归�U�集

2. 设A��Z��个集合，如果A中�Q何元素的元素也是A的元素，则称A�?strong style="COLOR: red">传递集。每个自然数都是传递集�?br>以下命题�{��h�Q?br>(1) A是传递集
(2) ∪A 包含�?A
(3) 对于��L��的y∈A�Q�y包含于A
(4) A包含于P(A)
(5) P(A)��Z��递集

3. 设A��Z��个集合，�U�C��A*A到A的函��CؓA上的二元�q�算�?br>�?: N*N -> N�Q�且对于��L��的m, n ∈N�Q?() = A_m(n), ��C��m + n�Q�称+为N上的加法�q�算
�?#183;: N*N -> N�Q�且对于��L��的m, n ∈N�Q?#183;() = M_m(n)�Q�记作m·n�Q�称·为N上的乘法�q�算�?br>

ZelluX 2007-07-28 16:37 发表评论

��L��学习�W�记 - 自然�?(1)

ZelluX — Fri, 27 Jul 2007 11:36:00 GMT

1. Peano �pȝ��

Peano�pȝ��是满��以下公讄��有序三元�l?lt;M, F, e>�Q�其中M��Z��个集合，F为M到M的函敎ͼ�e为首元素�?条公设�ؓ�Q?/p>

(1) e ∈M
(2) M在F下是��闭�?br>(3) e �K?ranF �Q�暂时只扑ֈ��q�个�W�号表示“不属�?#8221; 囧）
(4) F是单��的
(5) 如果M的子集A满�� (e属于A) �?(A在F下是��闭�?�Q�则A=M
(5)�U�Cؓ极小性公�?br>

2. 设A��Z��个集合，�U?A∪{A} 为A�?strong style="COLOR: red">后��Q�记作A⁺, �q�称求集合的后��?span style="COLOR: red">后��q�算�?br>

3. 设A��Z��个集合，若A满��Q?br>(1) Ø ∈A�Q?br>(2) 若对于一�?a ∈A�Q�则 a⁺ ∈A�Q?br>则称A�?strong style="COLOR: red">归纳�?/strong>�?br>

4. 从归�U�集的定义可以看出，Ø�Q?#216;⁺, Ø⁺⁺,... 是所有归�U�集的元素，于是可以��它们定义成自然�?/strong>�?br>自然数是属于每个归纳集的集合�Q�将Ø�Q?#216;⁺, Ø⁺⁺,...分别��Cؓ0, 1, 2, ...
设D={v | v是归�U�集|�Q�称∩D为全体自然数集合�Q�记为N.
设N��然数集合�Q?#963;: N -> N�Q�且σ(n) = n⁺, �?lt;N, σ, Ø>是Peano�pȝ��?br>�q�个Peano�pȝ��的第(5)条公设提��Z��证明自然数性质的一�U�方法，�?span style="COLOR: red">数学归纳�?/strong>�Q�称此公设�ؓ数学归纳法原�?/strong>�?/p>

ZelluX 2007-07-27 19:36 发表评论

ZelluX — Wed, 25 Jul 2007 04:26:00 GMT
各符��L��定义与上文中最�q�处的定义相同�?br>1. 设R属于A*A�Q�若R是自反的、反对称的和传递的�Q�则�U�R是A上的偏序关系�?partial order
2. �U�C��个非�I�集合A及其A上的一个偏序关�p?lt;=�l�成的有序二元组(a, <=)��Z��个偏序集。partially ordered set, or simply poset.
3. �?A, <=)��Z��个偏序集�Q�若对于一切x,y属于A�Q�如果x<=y或者y<=x成立�Q�则�U�x与y是可比的�?br>4. 若x与y是可比的�Q�且x5. 哈斯图作法： Hasse diagram
(1) 省去关系图中每个��点处的环�?br>(2) 若y覆盖x�Q�则��代表y的顶�Ҏ��在代表x的顶点之上，�q�连�U�，省去有向边的��头�?br>6. 若对于一切x,y属于A�Q�x与y均可比，则称<=为A上的全序关系�Q�或�U�性关�p�R��linear order
7. 若R是反自反的和传递的�Q�则�U�R为A上的拟序关系�Q�常��R��C��<�?br>8. 最大元最��元极大�?极小�?上界下界最��上�?最大下�?br>其实�q�些词的区别和高��C��的很相近�Q?#8220;最”针对自��n集合的所有元素，“�?#8221;针对自��n集合的部分元素，“�?#8221;指有一个更大的包含自��n集合的参照系下的情况�?br>9. �U�性关�p�M��׃��M��两个元素均可比，因此哈斯图中��可以表�C�Zؓ一条直�U�，从而容易理解线性的由来。又�U�Cؓ链，元素个数�U�Cؓ铄��长度�?br>10. 良序关系�Q�拟全序�?A, <)中�Q何非�I�子集均有最��元�?br>

ZelluX 2007-07-25 12:26 发表评论

��L��学习�W�记 - 二元关系

ZelluX — Wed, 25 Jul 2007 03:03:00 GMT
1.闭包 Closure
(1) 自反 reflexive
对称 symmetric
传�?transitive
(2) 其中�Q�设R属于A*A�Q�A为非�I�集合）�Q�则r(R) = R与A上恒�{�关�pȝ��qӞ��s(R) = R与R的逆的�qӞ��
t(R) = R �q?R^2 �q?R^3 �q?..�q?R^l�?br>(3) rs(R) = sr(R)
rt(R) = tr(R)
st(R) 属于 ts(R)

2. �{��h关系和划�?br>(1) 设R属于A*A�Q�若R是自反的、对�U�的和传递的�Q�则�U�R为A上的�{��h关系�?br>(2) 令[x]_R为x的关于R的等��L��Q�在不引��h؜乱时可简��Cؓ[x]�?br>(3) 以关于R的全体不同的�{��h�c�Mؓ元素的集合称为A关于R的商集，��C��A/R�?br>(4) 设A为非�I�集合，若存在A的一个子集族S满��
a. S中不包含�I�集元素
b. 对于一切x,y属于S�Q�且x,y不相�{�，则x与y不相交的(disjoint)
c. S中所有集合的�q��ؓA
则称S为A的一个划分，S中元素称为划分块�?br>(5) 非空集合A上的�{��h关系与A的划分是一一对应的�?br>(6) �W�二�c�Stirling敎ͼ�表示��n个不同的球放入r个相同的盒子中的�Ҏ��敎ͼ�可以�׃��列递归式计��：
f(n, r) = r * f(n - 1, r) + f(n - 1, r - 1)
很容易理解的一个递归式，其中初始状态�ؓ
f(n, 0) = 0, f(n, 1) = 1, f(n, 2) = 2^(n-1) - 1, f(n, n - 1) = C(n, 2), f(n, n) = 1
(7) A上等价关�pȝ��数量可以通过Stiring数求出，以A={a,b,c,d}��Z��
f(4,1) + f(4,2) + f(4,3) + f(4,4) = 15

ZelluX 2007-07-25 11:03 发表评论

一些数独技�?Very Easy - Hard)

ZelluX — Fri, 20 Jul 2007 03:59:00 GMT

from www.BrainBashers.com
1. Intersection 横断�Q�游戏一开始就使用的常见技巧�?/p>

2. Forced Moves 排除所有其他可能性后唯一的答�?br>

3. Pinned Squares
Intersection 的加强版�Q�根据更多的情况��定某一个数字在该区域的唯一可能位置�?br>

4. Locked Sets

如图一R5C1和R6C1只能�?�?�Q�由此可排除与他们有关的域中的其他格�?�?的可能性，从而R6C2只能�?�?/p>

ZelluX 2007-07-20 11:59 发表评论

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Exponentiating by squaring

拿糖果的博弈问题

Ferrers囑փ�

Properties

Applications in combinatorics

Proof of the formula

First proof

Second proof

Third proof

Hankel matrix

Quadruple factorial

History

Functional Programming For The Rest of Us

���L��学习�W�记 - ��?(1)

���L��学习�W�记 - 自然�?(2)

���L��学习�W�记 - 自然�?(1)

���L��学习�W�记 - 二元关系

一些数独技�?Very Easy - Hard)

在线播放国产精品 ,夜色福利资源站www国产在线视频夜色资源站国产www在线视频 ,精品国产乱码久久久久久88av

��L��学习�W�记 - ��?(1)

��L��学习�W�记 - 自然�?(2)

��L��学习�W�记 - 自然�?(1)

��L��学习�W�记 - 二元关系