多查询任务：最多K次操作下数组A可化简为1的最大元素数

Hey there, let's work through this problem together. I've dealt with similar factorization-based optimization questions before, so here's a clear, efficient approach to solve it.

First, let's clarify what each operation does: when you divide an element by its smallest factor greater than 1, you're essentially stripping off one prime factor (the smallest one) each time. For example:

• 18 → 9 (divide by 2, 1 operation) → 3 (divide by 3, 2 operations) → 1 (divide by 3, 3 operations)
• The total number of operations needed to reduce a number to 1 is exactly the total count of prime factors (including multiplicities) — this is known as the big omega function, Ω(n).

So the core of the problem reduces to:

1. Calculate how many operations each element needs to become 1.
2. Use these counts to answer each query: given K operations, what's the maximum number of elements we can fully reduce to 1?

1. Precompute Operation Counts for Each Element #

To efficiently calculate the operation count for every element, we'll use a smallest prime factor (SPF) array precomputed via sieve of Eratosthenes. This lets us factorize any number in O(log n) time.

Here's how it works:

• Precompute the SPF array up to the maximum value in the input array. The SPF array stores the smallest prime factor for every number up to that max value.
• For each element in the array, use the SPF array to decompose it into primes, then sum the exponents (this sum is the number of operations needed).

2. Sort Counts & Compute Prefix Sum #

To maximize the number of elements we can reduce, we should always prioritize elements that need fewer operations first. So:

• Sort the list of operation counts in ascending order.
• Compute a prefix sum array where prefix[i] is the total number of operations needed to reduce the first i elements (after sorting) to 1.

For example, if sorted counts are [1,1,2,3,4], the prefix sum would be [0,1,2,4,7,11] (we add a leading 0 to handle edge cases easily).

3. Answer Queries with Binary Search #

For each query K, we need to find the largest index m such that prefix[m] ≤ K. This m is our answer (the maximum number of elements we can reduce). Binary search is perfect here because the prefix sum array is strictly increasing, letting us find the answer in O(log N) per query.

def compute_smallest_prime_factors(max_num):
    # Initialize SPF array where spf[i] is the smallest prime factor of i
    spf = list(range(max_num + 1))
    for i in range(2, int(max_num ** 0.5) + 1):
        if spf[i] == i:  # i is a prime number
            for j in range(i * i, max_num + 1, i):
                if spf[j] == j:
                    spf[j] = i
    return spf

def calculate_operations(x, spf):
    operations = 0
    while x > 1:
        prime = spf[x]
        # Count how many times this prime divides x
        while x % prime == 0:
            operations += 1
            x = x // prime
    return operations

# Read input
n, q = map(int, input().split())
array = list(map(int, input().split()))

# Precompute SPF array if array isn't empty
max_val = max(array) if array else 0
spf_array = compute_smallest_prime_factors(max_val)

# Get operation counts for each element
operation_counts = [calculate_operations(num, spf_array) for num in array]

# Sort counts and compute prefix sums
operation_counts.sort()
prefix_sums = [0] * (n + 1)
for i in range(n):
    prefix_sums[i + 1] = prefix_sums[i] + operation_counts[i]

# Process each query
for _ in range(q):
    k = int(input())
    left, right = 0, n
    result = 0
    while left <= right:
        mid = (left + right) // 2
        if prefix_sums[mid] <= k:
            result = mid
            left = mid + 1
        else:
            right = mid - 1
    print(result)

• Elements equal to 1: These need 0 operations, so they'll be at the start of the sorted list and automatically included in the count as long as K ≥ 0.
• K=0: Only elements already equal to 1 can be counted.
• Large input sizes: The sieve precomputation is O(M log log M) (M is max array value), sorting is O(N log N), and each query is O(log N) — this is efficient even for large N and Q.

内容的提问来源于stack exchange，提问作者Rohit Sroch