# CodeForces 295E - Yaroslav and Points 题解：又是线段树！

## Description

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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# CodeForces 447E - DZY Loves Fibonacci Numbers 题解：线段树

## Description

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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# 矩阵乘法在图论中的简单应用

$$C[i,j]=\sum_{k=1}^{b} A[i,k]\ast B[k,j]$$

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# 矩阵乘法在动态规划中的应用

$$\begin{bmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \end{bmatrix}$$

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# POJ 3244 Difference between Triplets 题解：（线段树或树状数组）或（排序 + 前缀和）

## Description

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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# Topcoder SRM 635 Div2 T3 LonglongestPathTree 题解

Topcoder Single Round Match 635 Div2 T3 LonglongestPathTree 题解

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# Topcoder SRM 638 Div2 T3 CandleTimerEasy 题解

Topcoder Single Round Match 638 Div2 T3 CandleTimerEasy 题解

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# Topcoder SRM 640 Div2 T3 TwoNumberGroupsEasy 题解

Topcoder Single Round Match 640 DIv2 T3 TwoNumberGroupsEasy 题解

（TC SRM 从 640 到 616 持续施工～）

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