print N % 20. (With a string you can use up to the last 3 digits.)

The pattern is n//2*3 — n%2. In general second row bishops want 2 moves and first row bishops want 1 move, except when there is only one bishop on the "/" diagonal, or the (2,2) bishop.

Binary search for the answer. Standard dp lets you query the sum of a radius d box centered around (r, c), that also needs A[r][c] >= 1 for the mentor to stay there.

https://www.codechef.com/viewsolution/94358156

Precalc lpf (least prime factor) of all numbers <= 1e6. For each A[i], find all the primes that divide it and store in a map p -> A[i]. Now for each query, find all primes that divide it to get the lowest candidate. Store A in a multiset and remove that candidate (or the lowest).

https://www.codechef.com/viewsolution/94359936

We can represent the answer in terms of one sided half-blasts like : [0, 0, 1, 2, 3, 4, 0, 0, 0, 0].

Line sweep. The events are of the form "started" or "ended length=4" (following the example). Keep track of how many blasts are active (active += 1), what the running sum of the actives is (s += active), and what the running sums of s is (t += s). When you get an "ended length=r", then active -= 1, s -= r, t -= r*(r+1)/2. The answer at each step is ans[i] = t.

https://www.codechef.com/viewsolution/94359537

Let F(u) = sum_{v child of u} d(u, v). Represent binary numbers as an array<int, 32>. We can add binary numbers without carrying (eg. (1, 0, 1) + (1, 1, 0) = (2, 1, 1)) and evaluate them into base 10 later. This makes circular shifting and xor-ing easy, in particular calculating [(X xor shift(d1)) + (X xor shift(d2)) + ...] from [d1 + d2 + ...]. Now a standard dp gets you F(0).

Rerooting gets you the rest of the F's. When rerooting u to v, handwaving a bit but basically F(u) -= F(v), size[u] -= size[v], size[v] += size[u], F(v) += F(u) happens.

https://www.codechef.com/viewsolution/94360578