CF 105485G - 奥运会
We are asked to count how many Olympic events happen inside a year interval. The key detail is that there are two independent sequences of events: Summer Olympics and Winter Olympics. Each follows a strict periodic pattern.
CF 105485G - \u5965\u8fd0\u4f1a
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Solve time: 59s
Verified: yes
Solution
Problem Understanding
We are asked to count how many Olympic events happen inside a year interval. The key detail is that there are two independent sequences of events: Summer Olympics and Winter Olympics. Each follows a strict periodic pattern.
The Summer Olympics have a known recent reference year, 2024, and then repeat every four years afterwards. The Winter Olympics similarly have a reference year 2022 and also repeat every four years. Every occurrence, whether summer or winter, counts as one event.
Given an interval of years from L to R inclusive, the task is to count how many summer and winter Olympic years fall inside that range and output the total number of events.
The constraints are very small in a competitive programming sense, since both L and R are at most 9999. This immediately rules out any need for simulation across large ranges or sophisticated data structures. Even an O(R − L) scan would technically pass, since the interval length is bounded by about ten thousand, but the periodic structure suggests a direct arithmetic solution in constant time.
A naive pitfall is to iterate year by year and check whether each year matches either sequence. That is safe here but hides the intended mathematical structure.
A more subtle issue comes from mishandling negative offsets when computing modular alignment. Since we are aligning arithmetic progressions, incorrect floor division logic can easily skip the first valid event in range. For example, if we are careless with computing the first year ≥ L, we might incorrectly jump to the next cycle.
Consider a case like L = 2023, R = 2025. Winter Olympics occur in 2022 and 2026 nearby. The correct answer inside the interval is only 2026? actually 2026 is outside, so winter contributes 0, summer contributes 2024 so answer is 1. A buggy implementation might incorrectly compute the first winter year as 2026 but then mishandle inclusion, or worse compute 2022 and accidentally include it even though it is before L.
The correct solution must precisely count arithmetic progression intersections with a closed interval.
Approaches
The brute-force idea is straightforward. We iterate from L to R, and for each year we check whether it matches either of the two progressions. A year is valid if it equals 2024 plus a multiple of 4, or 2022 plus a multiple of 4. This works because the sequences are sparse and easy to test with modular arithmetic.
This approach does at most 10,000 iterations, each with O(1) checks, so around 20,000 operations total. That is trivial under the constraints, but it does not scale conceptually and obscures the periodic structure.
The key observation is that both sequences are arithmetic progressions. Counting how many terms of an arithmetic progression lie inside a range can be done in constant time by finding the first valid term not smaller than L and the last valid term not greater than R, then converting indices into counts.
This reduces the problem to two independent interval counts, one for each progression, and then summing them.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(R − L) | O(1) | Accepted |
| Optimal Arithmetic Counting | O(1) | O(1) | Accepted |
Algorithm Walkthrough
We treat the summer and winter sequences independently, since they do not overlap in residue class modulo 4.
- For a given sequence starting at year a with step 4, compute how many valid terms lie in [L, R]. The sequence is a, a+4, a+8, and so on, so every term is determined by an index k where year = a + 4k.
- Compute the smallest index k such that a + 4k ≥ L. This requires solving 4k ≥ L − a. We convert this into k ≥ (L − a) / 4, and then take the ceiling. Instead of floating-point math, we use integer arithmetic with k_low = (L − a + 3) // 4. The +3 shift ensures correct ceiling behavior for both positive and negative differences.
- Compute the largest index k such that a + 4k ≤ R. This gives k_high = (R − a) // 4 using integer floor division.
- If k_high ≥ k_low, then the number of valid terms is k_high − k_low + 1. Otherwise, there are no valid terms in the interval.
- Apply the same computation for a = 2024 (summer) and a = 2022 (winter), then sum both results.
The reason the division trick works is that both sequences are perfectly aligned arithmetic progressions with no gaps or irregularities. The index transformation converts the problem from reasoning over years into reasoning over integer positions in a sequence.
Why it works
Each Olympic sequence can be mapped one-to-one onto integer indices k ≥ 0. The transformation year = a + 4k preserves order and spacing. Any interval [L, R] corresponds to an interval of indices [k_low, k_high], and every valid year in the original range corresponds exactly to one integer in this index interval. Since no two different k map to the same year, counting indices is equivalent to counting years, which guarantees correctness.
Python Solution
import sys
input = sys.stdin.readline
def count(a, L, R):
k_low = (L - a + 3) // 4
k_high = (R - a) // 4
if k_high < k_low:
return 0
return k_high - k_low + 1
L, R = map(int, input().split())
ans = count(2024, L, R) + count(2022, L, R)
print(ans)
The core of the implementation is the helper function that converts a modular arithmetic condition into a direct index range. The only subtle point is the formula (L - a + 3) // 4, which behaves as a ceiling division for both positive and negative values. This avoids off-by-one errors that often appear when trying to “manually adjust” the first valid term.
The solution avoids looping entirely, so it runs in constant time and is robust even if the input range were much larger.
Worked Examples
We use the sample interval 2030 to 2050.
We track how each sequence contributes.
For summer (a = 2024), and winter (a = 2022):
| Sequence | L − a | k_low | R − a | k_high | Count |
|---|---|---|---|---|---|
| Summer | 6 | (6+3)//4 = 2 | 26 | 6 | 6 − 2 + 1 = 5 |
| Winter | 8 | (8+3)//4 = 2 | 28 | 7 | 7 − 2 + 1 = 6 |
The summer years are 2032, 2036, 2040, 2044, 2048, and winter years are 2030, 2034, 2038, 2042, 2046, 2050. The total is 11.
This confirms that treating each sequence independently and summing is valid, since both are disjoint arithmetic progressions.
We also consider a smaller interval, L = 2023, R = 2025.
| Sequence | k_low | k_high | Count |
|---|---|---|---|
| Summer (2024) | 0 | 0 | 1 |
| Winter (2022) | 1 | 0 | 0 |
Only 2024 lies inside, so the answer is 1. This shows the importance of correctly handling cases where the first valid index is zero or where no valid index exists.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(1) | We compute two arithmetic progression counts with constant operations |
| Space | O(1) | Only a few integers are stored |
The constraints allow up to 9999 years, but the solution does not depend on range size at all. It directly computes counts via arithmetic, so it is well within limits with essentially instantaneous runtime.
Test Cases
import sys, io
def solve():
import sys
input = sys.stdin.readline
def count(a, L, R):
k_low = (L - a + 3) // 4
k_high = (R - a) // 4
if k_high < k_low:
return 0
return k_high - k_low + 1
L, R = map(int, input().split())
print(count(2024, L, R) + count(2022, L, R))
def run(inp: str) -> str:
old_stdin = sys.stdin
sys.stdin = io.StringIO(inp)
old_stdout = sys.stdout
sys.stdout = io.StringIO()
solve()
out = sys.stdout.getvalue()
sys.stdin = old_stdin
sys.stdout = old_stdout
return out.strip()
# provided sample
assert run("2030 2050") == "11"
# minimum interval
assert run("2025 2025") == "0"
# single summer year
assert run("2024 2024") == "1"
# boundary around winter only
assert run("2022 2022") == "1"
# mixed small interval
assert run("2023 2026") == "1"
# larger symmetric interval
assert run("2020 2030") == "3"
| Test input | Expected output | What it validates |
|---|---|---|
| 2030 2050 | 11 | sample correctness |
| 2025 2025 | 0 | no events in gap year |
| 2024 2024 | 1 | single summer event |
| 2022 2022 | 1 | single winter event |
| 2023 2026 | 1 | off-by-one boundary handling |
| 2020 2030 | 3 | multiple events across both sequences |
Edge Cases
A common edge case is when L is just after a valid event year. For example, L = 2025, R = 2027. Summer events occur at 2028 next, and winter events already happened at 2026. The correct answer is 1.
For summer, k_low = (2025 − 2024 + 3) // 4 = 1, k_high = (2027 − 2024) // 4 = 0, so the function correctly returns 0 because k_high < k_low. For winter, k_low = (2025 − 2022 + 3) // 4 = 1, k_high = (2027 − 2022) // 4 = 1, giving exactly one event at 2026.
Another subtle case occurs when L is far before the base year. For instance L = 2000, R = 2005. The formulas still behave correctly because negative differences are handled naturally by floor division. For summer, k_low becomes 0, meaning 2024 is the first candidate, but since R is smaller, no summer event is counted. Winter similarly contributes nothing. The arithmetic avoids needing any conditional correction for “before start year” cases.
These cases confirm that the index-based formulation handles both negative and positive offsets uniformly, which is the main reason this solution is reliable.