# Switch lights(HDU-3404)

## 题面

lxhgww is playing a game with his computer Deep Blue.
The game is played on a matrix containing lights. At first, some lights are on, while others are off. lxhgww and Deep Blue take turns to switch the lights. For each step, the player should choose a rectangle in the matrix: (x1 , y1) , (x1 , y2) , (x2 , y1) , (x2 , y2) , (x1<=x2,y1<=y2, the light at (x2, y2) should be on) and change the lights’ status on the four vertex of the rectangle, namely on to off, and off to on. The player turns all the lights off wins the game. Notice the rectangle is possibly degenerated to line or even a single cell so that the player may also switch two or one besides four lights in a move.
Deep Blue’s strategy is perfect, if it has a chance to win, never will it lose. Does lxhgww have a chance to win if he takes the first step?

## 输入

The first line is an integer T(T<=100) indicating the case number.
Each case has one integers n (n<= 1000 ), the number of on-lights at the beginning of the game.
Then come n lines, each line has two integers, xi , yi, (1<=xi<=10000, 1<=yi<=10000) , so light at (xi, yi) is on at first. (No two lights at the same position)

## 输出

If lxhgww still has a chance to win, output “Have a try, lxhgww.”, otherwise tell lxhgww “Don’t waste your time.”

## 思路

Nim积定义

$x \otimes y = sg(x, y) = mex\{(a \otimes y) \oplus (x \otimes b) \oplus (a \otimes b), 0 \le a \lt x, 0 \le b \lt y\}$

01234
000000
101234
202318
3031212
4048126

• 一个 Fermat 2-power 与任意小于它的数的 Nim 积为一般意义下乘法的积，即$x \otimes 2^{2^{k}} = x * 2^{2^{k}}$

• 一个 Fermat 2-power 与自己的 Nim 积为自己的 $\frac{3}{2}$ 倍，即 $2^{2^{k}} \otimes 2^{2^{k}} = \frac{3}{2} * 2^{2^{k}}$

• $x \otimes y < 2^{2^{k}}$

## 代码

_/_/_/_/_/ EOF _/_/_/_/_/