Problem A. Sherlock and Parentheses
Category: Ad-Hoc, Math, Greedy
Let's define n = min(L,R). Then the claim is that the sequence ()()...() consisting of n pairs of "()" will give us the maximum answer n*(n+1) / 2
Proof:
Let's look at some sequence T = (S) where S is also a balanced parenthesis sequence of length m. Let's denote by X(S), the number of sub-strings of S that are also balanced and by X'(S) the number of balanced parenthesis sub-strings that include the first and last characters of S (In fact this will be just 1). Then X(T) = X(S) + X'(S).
It should be easy to prove that X'(S) <= X(S) (simply because X'(S) is a constrained version of X(S))
If we rearrange our sequence T to make another sequence T' = "S()", then X(T') >= X(S) + X'(S). The greater than equal to sign comes because we might have balanced sub-strings consisting of a suffix of S and the final pair of brackets.
Thus, we have that X(T') >= X(S) + X'(S) = X(T). Thus by rearranging the sequence in this manner, we will never perform worse. If we keep applying this operation on each string, we will finally end up with a sequence of the form "()()...()" possibly followed by non-zero number of "(" or ")"
Problem B. Sherlock and Watson Gym Secrets
Category: Math, DP
Let's look at all valid pairs (i,j) in which i is fixed. Then, if i^a mod k = r then j^b mod k = (k-r) mod k.
There are only k possible values of the remainder after division. Since k is small enough, if we can quickly determine how many numbers i exist such that i^a mod k = r and j^b mod k = (k-r) mod k then we can simply multiply these quantities and get our answer. Thus the reduced problem statement is —
Find the number of numbers i <= N such that i ^ a mod k = r for all r = 0 to k-1.
Let's denote by dp1[r] = number of i <= N such that i ^ a mod k = r. We will try to compute dp1 from r = 0 to k-1 in time O(K log a)
Let's make another observation —
i ^ a mod k = (i + k) ^ a mod k = (i + 2k) ^ a mod k = (i + x*k) ^ a mod k.
Proof:
(i + x*k) ^ a mod k = (i mod k + (x * k) mod k) ^ a mod k
= (i + 0) ^ a mod k = i ^ a mod k
Now i + x * k <= n. This implies that x <= (n-i) / k.
Thus if we know that for some i <= K, i ^ a mod k is r then there are x + 1 such numbers which also have the same remainder.
This gives us a quick way to compute dp1
for i = 1 to k:
r = pow(i, a) mod k
x = (n-i) / k
dp1[r] += x + 1
Similarly, we can compute dp2[r] which denotes the number j <= N such that j ^ b mod k = r
Now our final answer is the sum of dp1[i] * dp2[(k-i) mod k] for i = 0 to k.
ans = 0
for i = 0 to k-1:
ans += dp1[i] * dp2[(k-i)%k]
But we need to exclude the cases when i = j.
To handle this we can simply iterate one last time from i = 1 to k and find all the cases when (i ^ a + i ^ b) mod k = 0. and subtract (1 + (n-i)/k) from the answer for all such i
for i = 1 to k
if pow(i,a) + pow(i,b) mod k == 0:
ans -= ((n-i)/k + 1)
Problem C. Watson and Intervals
Category: Greedy, Line Sweep
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