Overview

Given warehouses lined up in a row, you carry luggage from them, stopping when the total weight exceeds the limit \(K\) (or when you reach the end). The problem asks you to compute the sum of stopping positions \(f(s)\) over an interval for each starting position \(s\).

Analysis

1. Properties of \(f(s)\)

Takahashi starts from warehouse \(s\) and proceeds through warehouses \(s, s+1, \dots, t\). The stopping conditions are as follows: - The smallest \(t\) such that \(A_s + A_{s+1} + \dots + A_t > K\) - If no such \(t\) exists, then \(t = N\)

The key observation here is the monotonicity: as the starting position \(s\) moves to the right, the stopping position \(f(s)\) also moves to the right (or stays at the same position). For example, if starting from warehouse \(1\) the weight is exceeded at warehouse \(5\), then starting from warehouse \(2\), the weight cannot be exceeded before warehouse \(5\) (since all weights \(A_i\) are positive).

2. Naive Approach and Its Issues

For each query \((L_j, R_j)\), if we compute \(f(L_j), f(L_j+1), \dots, f(R_j)\) each time, and for each \(f(s)\) we scan through the warehouses, the worst-case time complexity is \(O(Q \times N^2)\), which will not fit within the time limit.

To solve this problem, we apply the following two-step speedup: 1. Use the two-pointer technique to compute \(f(s)\) for all \(s\) together in \(O(N)\). 2. Use prefix sums to answer each query in \(O(1)\).

Algorithm

Step 1: Enumerating \(f(s)\) with the Two-Pointer Technique

We perform the operation “extend the right endpoint \(t\) until the total weight exceeds \(K\)” while moving the left endpoint \(s\) from \(1\) to \(N\). - When advancing \(s\) by one, we can resume from the previous position of \(t\), so \(t\) never moves backward. - To efficiently compute the sum of an interval, we precompute the prefix sums of the weights \(A_i\).

Step 2: Answering Queries with Prefix Sums

Once \(f(s)\) is determined for all \(s\), we create a prefix sum array \(P\) of \(f(s)\) itself. - \(P[i] = f(1) + f(2) + \dots + f(i)\) With this prepared, the total sum of \(f(s)\) over any interval \([L, R]\) can be obtained as \(P[R] - P[L-1]\) in just a single computation.

Complexity

• Time complexity: \(O(N + Q)\)
  • \(O(N)\) for computing the prefix sums of the weights
  • \(O(N)\) for computing \(f(s)\) using the two-pointer technique (since \(s\) and \(t\) each move at most \(N\) times)
  • \(O(N)\) for computing the prefix sums of \(f(s)\)
  • \(O(Q)\) for answering \(Q\) queries
• Space complexity: \(O(N)\)
  • Memory proportional to \(N\) is used for storing prefix sums and the array \(f(s)\).

Implementation Notes

• Two-pointer condition: Advance \(t\) as long as the total weight is at most \(K\). When the loop exits, \(t\) is either “the first position where \(K\) is exceeded” or “the end.”

• Fast I/O: Since \(N\) and \(Q\) can be as large as \(2 \times 10^5\), in Python you can reduce execution time by using sys.stdin.read().split() and sys.stdout.write().

• Index management: When using prefix sums, making the array size \(N+1\) and placing \(0\) at the beginning makes range sum calculations easier to handle.

  Source Code

import sys

# Increase recursion depth just in case, though the solution is iterative
sys.setrecursionlimit(200005)

def solve():
    # Read all input at once for faster processing
    try:
        input_data = sys.stdin.read().split()
    except EOFError:
        return
    
    if not input_data:
        return
    
    # N: number of warehouses, K: weight limit, Q: number of queries
    N = int(input_data[0])
    K = int(input_data[1])
    Q = int(input_data[2])
    
    # A: weights of the warehouses 1 to N
    A = list(map(int, input_data[3:3+N]))
    
    # S: prefix sums of A to calculate range sums in O(1)
    # S[i] = A_1 + A_2 + ... + A_i
    S = [0] * (N + 1)
    for i in range(N):
        S[i+1] = S[i] + A[i]
        
    # f[s]: the warehouse where Takahashi stops if he starts at warehouse s.
    # Takahashi visits s, s+1, ..., t until:
    # 1. The total weight A[s] + ... + A[t] > K, OR
    # 2. He reaches the last warehouse (t = N)
    f = [0] * (N + 1)
    t = 1
    for s in range(1, N + 1):
        # Ensure t is at least the current warehouse s
        if t < s:
            t = s
        
        # Use a two-pointer approach: as s increases, the stopping point t also increases.
        # We move t forward as long as we haven't reached the end and the cumulative weight is within K.
        while t <= N and S[t] - S[s-1] <= K:
            t += 1
        
        # If the loop ended because t > N, the sum never exceeded K, so he stopped at warehouse N.
        if t > N:
            f[s] = N
        else:
            # If the loop ended because the weight exceeded K, he stopped at warehouse t.
            f[s] = t
            
    # P: prefix sums of f to answer range sum queries for f in O(1).
    # P[i] = f[1] + f[2] + ... + f[i]
    P = [0] * (N + 1)
    for i in range(1, N + 1):
        P[i] = P[i-1] + f[i]
        
    # Process each of the Q queries.
    # Convert query range strings to integers efficiently.
    queries = list(map(int, input_data[3+N:]))
    results = []
    for j in range(Q):
        L = queries[2*j]
        R = queries[2*j+1]
        # The sum f(L) + f(L+1) + ... + f(R) is calculated as P[R] - P[L-1].
        results.append(str(P[R] - P[L-1]))
        
    # Output all query results joined by newlines for efficiency.
    if results:
        sys.stdout.write('\n'.join(results) + '\n')

if __name__ == '__main__':
    solve()

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This editorial was generated by gemini-3-flash-thinking.