Overview

This problem involves multiple range addition queries on consecutive elements, where we need to find the value of each element after all operations. This can be solved efficiently using the “imos method” (cumulative sum technique).

Analysis

Problems with the Naive Approach

A straightforward approach would be, for each watering query \((L_j, R_j)\), to add \(K\) to all flower beds from \(L_j\) to \(R_j\).

for j in range(Q):
    for i in range(L[j], R[j]+1):
        C[i] += K

However, this method has a worst-case time complexity of \(O(N \times Q)\). When both \(N\) and \(Q\) are at most \(2 \times 10^5\), this requires \(4 \times 10^{10}\) operations, resulting in TLE (Time Limit Exceeded).

Key Insight

We realize that we only need to record the count of “how many times watering occurred” for each query, and multiply by \(K\) at the end. In other words, for each flower bed \(i\), we find the “number of times watered” and finally calculate \(C_i + (\text{count}) \times K\).

The imos method can be used as a technique to “efficiently process range additions.”

Algorithm

What is the Imos Method?

It’s a technique that records range \([L, R]\) additions in \(O(1)\) using a difference array, then restores the value at each position by taking the cumulative sum at the end.

Specifically: 1. Prepare a difference array diff (all initial values are 0) 2. When you want to add +1 to range \([L, R]\): - diff[L] += 1 (the +1 starts from this position) - diff[R+1] -= 1 (the +1 ends at this position) 3. Taking the cumulative sum of diff gives the number of additions at each position

Concrete Example

For \(N = 5\), watering ranges \([2, 4]\) and \([3, 5]\):

Initial:  diff = [0, 0, 0, 0, 0, 0]  (size N+1)

+1 to [2,4]:  diff[1] += 1, diff[4] -= 1
              diff = [0, 1, 0, 0, -1, 0]

+1 to [3,5]:  diff[2] += 1, diff[5] -= 1
              diff = [0, 1, 1, 0, -1, -1]

Cumulative sum:    count = [0, 1, 2, 2, 1, 0]
                   → Flower bed 1: 1 time, bed 2: 2 times, bed 3: 2 times, bed 4: 1 time, bed 5: 0 times

(※ For 0-indexed. The code uses L-1 and R)

Complexity

• Time Complexity: \(O(N + Q)\)
  • Processing each query: \(O(1) \times Q = O(Q)\)
  • Cumulative sum calculation and output: \(O(N)\)
• Space Complexity: \(O(N)\)
  • \(O(N)\) for the difference array diff and result array result

Implementation Notes

1. Index Conversion: The problem uses 1-indexed flower beds, but Python arrays are 0-indexed, so we use diff[L-1] and diff[R].

2. Difference Array Size: Since we access diff[R], the size needs to be \(N+1\) (to access diff[N] when \(R\) is \(N\)).

3. Overflow Prevention: The product of \(K\) and the watering count can become large, but Python has no integer overflow, so there’s no need to worry.

4. Fast Input: Since \(N\) and \(Q\) are large, reading all input at once with sys.stdin.read() is faster.

Source Code

import sys

def main():
    input_data = sys.stdin.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    K = int(input_data[idx]); idx += 1
    Q = int(input_data[idx]); idx += 1
    
    C = [int(input_data[idx + i]) for i in range(N)]
    idx += N
    
    # Difference array for imos method
    diff = [0] * (N + 1)
    
    for _ in range(Q):
        L = int(input_data[idx]); idx += 1
        R = int(input_data[idx]); idx += 1
        diff[L - 1] += 1
        diff[R] -= 1
    
    # Calculate cumulative sum to get watering count for each flower bed
    count = 0
    result = []
    for i in range(N):
        count += diff[i]
        result.append(C[i] + count * K)
    
    print(' '.join(map(str, result)))

if __name__ == '__main__':
    main()

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This editorial was generated by claude4.5opus.