Square Root Decomposition

You should already have done this problem in Point Update Range Sum, but here we'll present two more approaches. Both run in $\mathcal{O}((N+Q)\sqrt N)$ time.

Blocking

See the first example in CPH.

int n,q;
vi x;
ll block[450];
int BLOCK;

ll sum(int a) { // sum of first a elements of array in O(sqrt(n)) time
	ll ans = 0;
	F0R(i,a/BLOCK) ans += block[i];
	FOR(i,a/BLOCK*BLOCK,a) ans += x[i];
	return ans;

Batching

See the CPH section on "batch processing."

Maintain a "buffer" of the latest updates (up to $\sqrt N$). The answer for each sum query can be calculated with prefix sums and by examining each update within the buffer. When the buffer gets too large ($\ge \sqrt N$), clear it and recalculate prefix sums.

int n,q;
vi x;
vl cum;

void build() {
	cum = {0};
	trav(t,x) cum.pb(cum.bk+t);
}

int main() {

Mo's Algorithm

See CPH 27.3 and the CF link above.

Additional Notes

Low constraints (ex. $n=5\cdot 10^4$) and/or high time limits (greater than 2s) can be signs that sqrt decomp is intended.

CPH 262:

In practice, it is not necessary to use the exact value of $\sqrt n$ as a parameter, and instead we may use parameters $k$ and $n/k$ where $k$ is different from $\sqrt n$. The optimal parameter depends on the problem and input. For example, if an algorithm often goes through the blocks but rarely inspects single elements inside the blocks, it may be a good idea to divide the array into $k<\sqrt n$ blocks, each of which contains $n/k > \sqrt n$ elements.

As another example, if an update takes time proportional to the size of one block ($\mathcal{O}(n/k)$) while a query takes time proportional to the number of blocks times $\log n$ ($\mathcal{O}(k\log n)$) then we can set $k\approx \sqrt{\frac{n}{\log n}}$ to make both updates and queries take time $\mathcal{O}(\sqrt{n\log n})$.

Solutions with worse complexities are not necessarily slower (at least for problems with reasonable input sizes, ex. $n\le 5\cdot 10^5$). I recall an instance where a fast $\mathcal{O}(n\sqrt n\log n)$ solution passed (where $\log n$ came from a BIT) while an $\mathcal{O}(n\sqrt n)$ solution did not ... So constant factors are important!

On Trees

The techniques mentioned in the blogs below are extremely rare but worth a mention.

Some more discussion about how sqrt decomp can be used:

Problems

A

Problems where the best solution I know of involves sqrt decomp.

B

Problems that can be solved without it. But you might as well try to use it!!