请问， 正则表达式能完成这样的功能么？
别人有个一字符串数组， 已知数组的每个元素匹配 ^[a-z]+$
，现在那个人把数组中的所有字符串串成一个，分隔符用空格。
比如数组 p = {“abc”, “cdd”, “xxx”,”abc”}， 我收到的结果是 “abc cdd xxx abc”.

我的问题是能不能写一个正则表达式，来判定他原来的数组里没有重复元素。
也就是假设这个正则表达式是 regex = ….
那么regex.IsMatch(“abc cdd xxx abc”) 返回 false.
Re

Jack has bought a rectangular table containing a grid of lamps. Each lamp is initially either "on" or "off". There is a switch underneath each column, and when the switch is flipped, all the lamps in that column reverse their states ("on" lamps become "off" and vice versa).

A row in the grid is considered lit if all the lamps in that row are "on". Jack must make exactly K flips. The K flips do n

SRM423d1

Two players play a simple game on a n x n board. The first player has a single white checker which is initially located at (rowWhite, colWhite). The second player has a single black checker which is initially located at (rowBlack, colBlack). All coordinates are 1-based. The two players alternate turns, and the first player moves first.

When it is the first p

所谓曼哈顿距离：
假设点A的坐标为x0,y0, 点B的坐标为x1,y1
那么A和B的曼哈顿距离为： abs(x0-x1)+abs(y0-y1).

现在有这样一个题目：
平面上有n个点，给出他们坐标（可能有多个点的坐标是一样的）。 一次操作可以将一个点上移、下移、左移或者右移一个单位。
现在需要平面上某个坐标上至少有k个点，问需要最好的操作次数？

solution:

Let's fix some K and try to find the best way to gather some K checkers in the same cell. To do this, we first consider a very slow approach: i

There are N trees numbered 0 to N-1, and you must plant them along a straight line. Tree i will be planted at coordinate X[i], where X is constructed using the following recursive definition:
X[0] = X0 MOD L
X[i] = (X[i-1]*A+B) MOD L (note that X[i-1]*A+B may overflow a 32-bit integer)
The price of planting tree i is the sum of the

过桥问题。 一堆人要过桥，只有一个手电筒，桥本身有最大载重。给出每个人的重量，给出每个人过桥需要的时间。一组人过桥，时间按速度最慢的算。过桥的时候必须使用手电筒。这个相对来说很常见了。现在题目又给出了任意两个人之间的信任关系，要求一组人过桥的时候，每个人至少信任组内另外一个人。当然也可以一个人单独过桥。
问所有人过桥的最短时间。

这个题目典型的宽搜，但是需要注意的是，从桥的对面回来（把手电筒带回来）时，可能最优情况需要不止一个人回来。原因是题目中加上了信任模型。
srm422 d1 level2.

经典著作《人工智能：现代方法》的作者之一 Peter Norvig 曾经写过一篇介绍如何写一个拼写检查/纠正器的文章（原文在这里，徐宥的翻译版在这里，这篇文章很深入浅出，强烈建议读一读），里面用到的就是贝叶斯方法--- by ponda.
也可以看看ponda老大写的这篇文章。