Overview

This problem asks us to calculate the sum of products for all ways of selecting \(K\) consecutive elements from a sequence. We solve it efficiently using a sliding window and modular inverse.

Analysis

Issues with the Naive Approach

A straightforward approach would calculate the product of \(K\) elements for each of the \(N - K + 1\) possible selections. However, with this method: - Number of selections: \(O(N - K + 1)\) - Calculation of each product: \(O(K)\)

The total is \(O((N - K + 1) \times K)\), which becomes \(O(N^2)\) in the worst case (\(K \approx N/2\)) and results in TLE.

Key Insight

When calculating products of consecutive intervals in order, adjacent intervals share \(K - 1\) elements.

For example, when \(N = 5, K = 3\): - Interval 1: \(A_1 \times A_2 \times A_3\) - Interval 2: \(A_2 \times A_3 \times A_4\) - Interval 3: \(A_3 \times A_4 \times A_5\)

When moving from interval 1 to interval 2, only \(A_1\) is removed and \(A_4\) is added. In other words: $\(\text{Product of interval 2} = \frac{\text{Product of interval 1}}{A_1} \times A_4\)$

The Division Problem

When working with remainders modulo \(10^9 + 7\), we cannot use ordinary division. Instead, we use the modular inverse.

Dividing by \(a\) is equivalent to multiplying by \(a^{-1}\) (the modular inverse of \(a\)). By Fermat’s Little Theorem, when \(p\) is prime: $\(a^{-1} \equiv a^{p-2} \pmod{p}\)$

Algorithm

1. Initialization: Calculate the product of the first \(K\) elements (\(A_1, A_2, \ldots, A_K\)) and store it as product. Add product to total.

2. Sliding Window: Repeat the following for \(i = 0, 1, \ldots, N - K - 1\):

   • Calculate the modular inverse of \(A_{i+1}\) (which is \(A[i]\) in 0-indexed)
   • “Divide” product by \(A[i]\) and multiply by \(A[i+K]\)
   • Add the new product to total

3. Output Result: Print total

[A0, A1, A2, A3, A4]  (Example with N=5, K=3)
 └──────┘            First product: A0×A1×A2
     └──────┘        Next product: (A0×A1×A2) ÷ A0 × A3 = A1×A2×A3
         └──────┘    Next product: (A1×A2×A3) ÷ A1 × A4 = A2×A3×A4

Complexity

• Time Complexity: \(O(N \log MOD)\)

  • Initial product calculation: \(O(K)\)
  • Sliding window: \(O(N - K)\) loop iterations, each with modular inverse calculation (using fast exponentiation) in \(O(\log MOD)\)
  • Total: \(O(K + (N - K) \log MOD) = O(N \log MOD)\)

• Space Complexity: \(O(N)\)

  • Storing array \(A\): \(O(N)\)

Implementation Notes

1. Modular Inverse Calculation: In Python, pow(a, MOD - 2, MOD) efficiently computes \(a^{MOD-2} \mod MOD\).

2. Overflow Prevention: By taking % MOD after each calculation, we prevent values from becoming too large.

3. Be Careful with 0-indexed vs 1-indexed: The code uses 0-indexed while the problem statement uses 1-indexed, so be careful not to confuse the correspondence.

Source Code

def solve():
    MOD = 10**9 + 7
    
    N, K = map(int, input().split())
    A = list(map(int, input().split()))
    
    # Calculate the product of the first K elements
    product = 1
    for i in range(K):
        product = (product * A[i]) % MOD
    
    total = product
    
    # Calculate the remaining products using sliding window
    # Remove A[i] and add A[i+K]
    for i in range(N - K):
        # Divide by A[i] and multiply by A[i+K]
        # Division uses modular inverse
        inv_ai = pow(A[i], MOD - 2, MOD)
        product = (product * inv_ai) % MOD
        product = (product * A[i + K]) % MOD
        total = (total + product) % MOD
    
    print(total)

solve()

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