sum += a[lo]
==

The query takes less than <tex>k + n/k + k = 2k + n/k</tex> time. 2k + n/k is minimized when k ~ sqrt(n). For k = sqrt(n) the query takes O(sqrt(n)) time.
This trick also works for other associative operations, like: min, gcd, product etc.
h2. Nearest neighbour

Given a set S of n points and a query point p, find the point in S closest to p.

bq. Given a set S of n points and a query point p, find the point in S closest to p.

For uniformly distributed points, a good strategy is to represent the space as a grid and maintain a list of inner points for each cell. For a given query point, we can check the cell the point falls into and its neighbouring cells. For a sqrt(n) * sqrt(n) grid we’ll have one point per cell, on average. So, on average, finding the point in S closest to p, requires traversing a constant number of cells.
h2. Longest common subsequence solution

Given two strings A (n characters) and B (m characters), find their longest common subsequence. (eg. The longest common sub sequence for abcabc and abcbcca is abcbc.)

bq. Given two strings A (n characters) and B (m characters), find their longest common subsequence. (eg. The longest common sub sequence for abcabc and abcbcca is abcbc.)

There is a standard dynamic programming solution which uses an array best[i][j] to mean the longest common sub sequence for A[0:i] and B[0:j], computed as below:

h2. Additional problems

1. (Josephus problem) n people numbered from 1 to n sit in a circle and play a game. Starting from the first person and every kth person is eliminated. Write an algorithm that prints out the order in which people are eliminated.
2. (Level ancestor) You are given an tree of size n. ancestor(node, levelsUp) finds the node’s ancestor that is levelsUp steps up. For example, ancestor(node, 1) returns the father and ancestor(node, 2) returns the grandfather. Implement ancestor(node, levelsUp) efficiently. (O(sqrt(n)) per query)
3. (Range median) You are given an array of size n. Implement a data structure to perform update operations a[i] = k and range median operations efficiently. The range median query, median(l, r)  returns the median element of the sorted subsequence a[l..r]. ($O(\log n)$ per update and O(sqrt n log n) per query)

* (Josephus problem) n people numbered from 1 to n sit in a circle and play a game. Starting from the first person and every kth person is eliminated. Write an algorithm that prints out the order in which people are eliminated.
*  (Level ancestor) You are given an tree of size n. ancestor(node, levelsUp) finds the node’s ancestor that is levelsUp steps up. For example, ancestor(node, 1) returns the father and ancestor(node, 2) returns the grandfather. Implement ancestor(node, levelsUp) efficiently. (O(sqrt(n)) per query)
* (Range median) You are given an array of size n. Implement a data structure to perform update operations a[i] = k and range median operations efficiently. The range median query, median(l, r)  returns the median element of the sorted subsequence a[l..r]. ($O(\log n)$ per update and O(sqrt n log n) per query)

You can discuss the problems in the comments section.