Yaroslav has n points that lie on the $Ox$ axis. The coordinate of the first point is $x_1$, the coordinate of the second point is $x_2$, ..., the coordinate of the n-th point is — $x_n$. Now Yaroslav wants to execute $m$ queries, each of them is of one of the two following types:
In mathematical terms, the sequence $F_n$ of Fibonacci numbers is defined by the recurrence relation
$$F_1 = 1; F_2 = 1; F_n = F_{n - 1} + F_{n - 2} (n > 2)$$
DZY loves Fibonacci numbers very much. Today DZY gives you an array consisting of n integers: $a_1, a_2, \dots, a_n$. Moreover, there are $m$ queries, each query has one of the two types:
$$ \begin{bmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \end{bmatrix} $$
每增加一个维度，世界便会增加无限的美感。
坠痛苦的是，POJ 的辣鸡 G++ 编译器不支持 long long，害得我调试调了半天……
For every pair of triplets, $T_a = (I_a, J_a, K_a)$ and $T_b = (I_b, J_b, K_b)$, we define the difference value between $T_a$ and $T_b$ as follows:
$$ D(T_a, T_b) = \max (I_a − I_b, J_a − J_b, K_a − K_b) − \min (I_a − I_b, J_a − J_b, K_a − K_b) $$
Now you are given $N$ triplets, could you write a program to calculate the sum of the difference values between every unordered pair of triplets?
Topcoder Single Round Match 635 Div2 T3 LonglongestPathTree 题解
Topcoder Single Round Match 638 Div2 T3 CandleTimerEasy 题解每天被 XY 题困扰的我们怎么会去做 Topcoder 呢
今天的 XY 题居然是递推专题，五道题目全都是递推，30+个人 AK 了……
递推是按照一定的规律来计算序列中的每个项，通常是通过计算前面的一些项来得出序列中的指定项的值。其思想是把一个复杂的庞大的计算过程转化为简单过程的多次重复，该算法利用了计算机速度快和不知疲倦的机器特点。
Topcoder Single Round Match 640 DIv2 T3 TwoNumberGroupsEasy 题解
（TC SRM 从 640 到 616 持续施工～）
（倒着开车，最为致命）（国庆之前怕是完不成了）