0935HardArrayMathSegment TreeNumber Theory

Matched Pairs After Each Rewrite

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Two positions i < j form a matched pair when nums[i] * nums[j] divides evenly by p.

Each entry of queries is [ind, val] and rewrites nums[ind] to val, a change that stays for every later query. Count the matched pairs after each rewrite, and return the total of those counts modulo 10^9 + 7.

Examples

Example 1

Input: nums = [6, 10, 15, 4, 9], p = 12, queries = [[0, 8], [2, 6], [4, 3]]
Output: 14

Each rewrite is counted after it lands and the changes carry forward, so the three counts are added together.

Example 2

Input: nums = [1, 1], p = 1, queries = [[0, 1]]
Output: 1

Every product divides evenly by one, so the single pair always matches and the one rewrite reports one.

Example 3

Input: nums = [3, 5, 7, 11], p = 2, queries = [[0, 2], [1, 4]]
Output: 8

The values start out all odd, so no product is even. The first rewrite puts a 2 in place and still leaves no pair, since one even value is not enough on its own; the second puts a 4 in, and that pair matches.

Constraints

2 <= nums.length <= 5 * 10^4
1 <= nums[i] <= 5 * 10^4
1 <= p <= 5 * 10^4
1 <= queries.length <= 5 * 10^4
queries[i].length == 2
0 <= queries[i][0] <= nums.length - 1
1 <= queries[i][1] <= 5 * 10^4

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_good_subseq(nums: list[int], p: int, queries: list[list[int]]) -> int:

Java

public int countGoodSubseq(int[] nums, int p, int[][] queries)

The intended approach

Time O(V log V + q * D) for values up to V, q rewrites and D divisors of p.Space O(D) for the tally and the class table.

A value matters only through what it shares with `p`, because whether `p` divides a product depends, prime by prime, on how much of `p`'s share of that prime each side supplies, and anything beyond `p`'s own share is never needed. So replace each value by its greatest common divisor with `p`. Those divisors are few, at most the number of divisors of `p`, so the whole register collapses into a tally over a short list of classes. Two classes match when their product divides evenly by `p`, which is worked out once for every pair of classes. Then the pair count follows from the tally, and a rewrite changes only one entry: take the old class out, counting off the pairs it was part of, and put the new class in, counting on the pairs it joins. That is one pass over the classes per rewrite rather than over the register.

Pythonentry point: count_good_subseq

def count_good_subseq(nums, p, queries):
    from math import gcd
    MOD = 1000000007
    classes = []
    index_of = {}
    for d in range(1, p + 1):
        if p % d == 0:
            index_of[d] = len(classes)
            classes.append(d)
    size = len(classes)
    matches = [[False] * size for _ in range(size)]
    for a in range(size):
        for b in range(size):
            if classes[a] * classes[b] % p == 0:
                matches[a][b] = True
    tally = [0] * size
    holding = []
    for v in nums:
        at = index_of[gcd(v, p)]
        holding.append(at)
        tally[at] += 1
    pairs = 0
    for a in range(size):
        for b in range(a, size):
            if not matches[a][b]:
                continue
            if a == b:
                pairs += tally[a] * (tally[a] - 1) // 2
            else:
                pairs += tally[a] * tally[b]
    total = 0
    for ind, val in queries:
        old = holding[ind]
        tally[old] -= 1
        for other in range(size):
            if matches[old][other]:
                pairs -= tally[other]
        new = index_of[gcd(val, p)]
        for other in range(size):
            if matches[new][other]:
                pairs += tally[other]
        tally[new] += 1
        holding[ind] = new
        total = (total + pairs) % MOD
    return total

The obvious version

Apply each rewrite to the register and then count the matched pairs by testing every pair of positions outright. Quadratic in the register size per rewrite and it never groups anything.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_good_subseq(nums, p, queries):
    MOD = 1000000007
    values = list(nums)
    n = len(values)
    total = 0
    for ind, val in queries:
        values[ind] = val
        pairs = 0
        for i in range(n):
            for j in range(i + 1, n):
                if values[i] * values[j] % p == 0:
                    pairs += 1
        total = (total + pairs) % MOD
    return total

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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