Inspector's Dilemma、The Bells are Ringing、 Photographic Tour
这三题好像当时没写summary。。。所以我们假设比较简单～～～

You are around me ...
首先旋转坐标，变成平行与xy轴的椭圆，然后坐标伸缩。。。变成圆。。。最近点对。。。贴模板。。。
ZJU2107 Quoit Design 一道测最近点对的题。

Infinite Matrix
显然，对于固定的j, Ri, j是一个关于i的多项式。
注意到数列Ri, j的差分序列R(i + 1, j) - R(i, j)是可以求出来的（利用Mj, k <= 10的条件，可以在O(10*n^2)的时间内算出）
然后有了差分序列求通项就是O(n^2)的事情。
然后记S(p, j)(n) = Sigma(i^p * R(i, j)) i <= n
继续利用差分序列之类的方法求这个，最后再求一个Sum_S(p, j)
预处理复杂度大致是O(10*n^3)的。
后面的就好办了，对每个询问，把(i + 1)^p二项式展开，最多10项，然后利用公式直接计算。
处理询问复杂度O(p*q*n)
POJ3529 Matrix Analysis，类似的思想，更简单～

Magnetic Train Tracks
给定n个点，求可以构成多少个锐角三角形。
n <= 1200
话说求锐角三角形不太好算是吧。。。补集转换，我们来求钝角/直角三角形 <=> 求钝角/直角个数。。。
后面的事情就简单了，是对每个点，将其他点按照极角排序 + 扫描。
Dhaka 2005 Counting Triangles 也是一道补集转换的题～（转化成求三点共线的个数）
Shanghai 2004 Amphiphilic Carbon Molecules 也是一道极角排序+扫描的题。

Who wins?

Chomp belongs to the category of impartial 2-player perfect information games.

It turns out that for any rectangular starting position bigger than 1 × 1 the 1st player can win. This can be shown using a strategy-stealing argument: assume that the 2nd player has a winning strategy against any initial 1st player move. Suppose then, that the 1st player takes only the bottom right hand square. By our assumption, the 2nd player has a response to this which will force victory. But if such a winning response exists, the 1st player could have played it as his first move and thus forced victory. The 2nd player therefore cannot have a winning strategy.

Computers can easily calculate winning moves for this game on two-dimensional boards of reasonable size.