Overview

This problem asks us to find the day by which all \(N\) seeds are planted, planting one seed per day while skipping rainy days. Since rain periods can overlap, we use interval merging followed by binary search.

Analysis

Rephrasing the Problem

“The number of seeds that can be planted by day \(D\)” is \(D\) minus “the number of rainy days from day \(1\) to day \(D\).” That is:

\[\text{sunny}(D) = D - \text{rainy}(D)\]

We need to find the smallest \(D\) such that \(\text{sunny}(D) \geq N\).

Issues with a Naive Approach

Simulating day by day starting from day 1 is infeasible because \(N\) can be up to \(10^9\) and the endpoints of rain periods can be up to \(10^{18}\).

Key Observation

\(\text{sunny}(D)\) is monotonically non-decreasing as \(D\) increases. Therefore, we can use binary search to find the smallest \(D\) satisfying \(\text{sunny}(D) \geq N\).

Additionally, since rain periods can overlap, we perform interval merging (combining overlapping or adjacent intervals into one) beforehand so that we can efficiently compute the exact number of rainy days.

Concrete Example

If \(N = 5\) and the rain periods are \([2, 4]\) and \([3, 6]\): - After merging: \([2, 6]\) (5 rainy days) - Day 1: sunny → plant 1 seed - Days 2–6: rainy → cannot plant - Days 7–10: sunny → plant 4 seeds - All 5 seeds are planted by day 10

Algorithm

1. Interval Merging: Sort the rain periods by start day and merge overlapping or adjacent intervals.
2. Early Decision: Check whether there are at least \(N\) sunny days up to the end of the last merged rain period.
   • If not enough, then after the last rain period every day is sunny, so simply add the shortfall to get the answer.
3. Binary Search: If there are at least \(N\) sunny days within the range, binary search over \([1, \text{last\_rain\_end}]\).
   • For each candidate day \(D\), scan the merged intervals to compute \(\text{rainy}(D)\).
   • If \(D - \text{rainy}(D) \geq N\), the answer is at most \(D\); otherwise, it is greater than \(D\).

Computing rainy_days_up_to(D)

For each merged interval \([L_i, R_i]\): - If \(L_i > D\), skip (all subsequent intervals are also after \(D\)) - Otherwise, add \(\min(R_i, D) - L_i + 1\)

Complexity

• Time complexity: \(O(M \log M + M \log V)\)
  • \(M \log M\): sorting the intervals
  • \(M \log V\): binary search (\(\log V\) iterations, where \(V\) is the search range up to about \(10^{18}\)), each step scanning the merged intervals (at most \(M\))
• Space complexity: \(O(M)\)

Implementation Notes

• Condition for interval merging: Using \(L \leq \text{prev\_R} + 1\) ensures that adjacent intervals (e.g., \([2,3]\) and \([4,5]\)) are correctly merged.

• Upper bound for binary search: After the last rain period ends, every day is sunny, so we can simply add the remaining number of needed seeds. Binary search is only needed when the answer falls within a rain period.

• Overflow prevention: \(R_i\) can be as large as \(10^{18}\), but since Python handles arbitrary-precision integers, this is not an issue.

  Source Code

import sys

def solve():
    input_data = sys.stdin.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    M = int(input_data[idx]); idx += 1
    
    intervals = []
    for i in range(M):
        L = int(input_data[idx]); idx += 1
        R = int(input_data[idx]); idx += 1
        intervals.append((L, R))
    
    if M == 0:
        print(N)
        return
    
    # Merge overlapping intervals
    intervals.sort()
    merged = []
    for L, R in intervals:
        if merged and L <= merged[-1][1] + 1:
            merged[-1] = (merged[-1][0], max(merged[-1][1], R))
        else:
            merged.append((L, R))
    
    # For each merged interval, we know rainy days in [L, R].
    # We need to find the day D such that the number of non-rainy days in [1, D] = N.
    
    # Number of rainy days in [1, D] can be computed using merged intervals.
    # non_rainy(D) = D - rainy_days(D)
    # We need non_rainy(D) = N, i.e., D - rainy_days(D) = N.
    
    # Binary search on D.
    def rainy_days_up_to(D):
        """Count total rainy days in [1, D] using merged intervals."""
        total = 0
        for L, R in merged:
            if L > D:
                break
            total += min(R, D) - L + 1
        return total
    
    # Lower bound: N (if no rain), Upper bound: N + total_rainy_days (but rain might extend beyond)
    # We need to be careful with upper bound.
    # The answer D satisfies D - rainy_days(D) = N.
    # D = N + rainy_days(D) >= N.
    # Also, total rainy days is at most sum of all interval lengths.
    # But intervals can go up to 10^18, so we need a smarter upper bound.
    
    # Actually, after the last merged interval ends, every day is sunny.
    # So the answer is at most max(merged[-1][1], 0) + N (very loose but safe... actually not if intervals are huge)
    # Better: after all rain periods, remaining seeds can be planted consecutively.
    # Let's compute how many sunny days exist up to the end of the last rain interval.
    
    last_rain_end = merged[-1][1]
    total_rain = sum(R - L + 1 for L, R in merged)
    sunny_before_end = last_rain_end - total_rain
    
    if sunny_before_end >= N:
        # Answer is within [1, last_rain_end], binary search
        lo, hi = 1, last_rain_end
    else:
        # Need more days after last rain period
        remaining = N - sunny_before_end
        print(last_rain_end + remaining)
        return
    
    while lo < hi:
        mid = (lo + hi) // 2
        sunny = mid - rainy_days_up_to(mid)
        if sunny >= N:
            hi = mid
        else:
            lo = mid + 1
    
    print(lo)

solve()

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