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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:

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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:

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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……

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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?

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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。

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