CF 1101A - Minimum Integer
Each query describes a forbidden interval on the number line and a fixed step size. We are asked to find the smallest positive integer that is a multiple of a given number $d$, but lies strictly outside the interval $[l, r]$.
Rating: 1000
Tags: math
Solve time: 5m 4s
Verified: yes
Solution
Problem Understanding
Each query describes a forbidden interval on the number line and a fixed step size. We are asked to find the smallest positive integer that is a multiple of a given number $d$, but lies strictly outside the interval $[l, r]$.
A useful way to think about it is that all valid candidates are points $d, 2d, 3d, \dots$, and among these we want the first one that does not fall inside the blocked segment. The blocked segment may cut out one or more multiples of $d$, but the answer must be the earliest multiple that survives this exclusion.
The constraints are small in terms of number of queries, but the values inside each query can be large up to $10^9$. This immediately rules out any approach that tries to iterate over all multiples up to $r$ for every query. Even if each query only costs $O(r/d)$, in the worst case this becomes $10^9$ operations per query, which is impossible under a 1 second limit.
The key edge cases come from how multiples interact with the interval boundaries. First, if $d > r$, then every multiple except $d$ itself might already be outside the segment, so the answer is trivially $d$. Second, if $d < l$, then $d$ is already valid and is the answer. The only interesting case is when a multiple of $d$ falls inside the interval and we must "jump" past it.
A subtle failure case for naive reasoning is assuming we only need to check $r$. For example, if $l = 5$, $r = 10$, $d = 4$, the multiples are $4, 8, 12, \dots$. The number 8 lies inside the interval, so skipping only by checking $r$ would incorrectly accept 8 if we only checked $r$-based bounds without verifying membership in $[l, r]$.
Approaches
A brute-force approach generates multiples of $d$: start from $d$, keep adding $d$, and stop when we find a value outside $[l, r]$. This is correct because it explicitly checks the definition of the answer. However, in the worst case where $d = 1$ and $l = 1, r = 10^9$, we would scan up to $10^9$ values, which is far beyond the allowed operations.
The key observation is that we only ever need to reason about a single candidate relative to the interval. Either the first multiple $d$ is already outside the segment, or it is inside. If it is inside, then all smaller multiples are irrelevant and we only need to "jump" to the first multiple strictly greater than $r$. That value is exactly the next multiple of $d$ after $r$, which can be computed directly using arithmetic instead of iteration.
This reduces the problem to a constant-time computation per query: decide whether $d$ is inside the interval, and if so compute the next multiple above $r$.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(r/d) per query | O(1) | Too slow |
| Optimal | O(1) per query | O(1) | Accepted |
Algorithm Walkthrough
- Read integers $l, r, d$. The goal is to locate the first multiple of $d$ that avoids $[l, r]$.
- Check whether $d$ lies inside the interval $[l, r]$. If $d < l$, then $d$ is already valid and is the smallest possible answer because it is the smallest positive multiple of $d$.
- If $d \ge l$, we check whether $d \le r$. If this is false, then again $d$ is outside the interval and is the answer.
- If $l \le d \le r$, then $d$ is forbidden, so we must skip all multiples that could still fall inside or overlap the interval.
- The next candidate multiple after $r$ is computed as $\left\lfloor \frac{r}{d} \right\rfloor d + d$. This gives the smallest multiple strictly greater than $r$.
- Output that computed value.
The crucial step is recognizing that once a single multiple lands inside the forbidden segment, the optimal answer cannot be any smaller multiple, because all smaller multiples are already checked and rejected. The structure of multiples guarantees that the next valid candidate is the first multiple beyond the right endpoint.
Why it works
All valid numbers form a strictly increasing sequence of multiples of $d$. The forbidden region removes a contiguous segment of the number line. If the first multiple $d$ is not removed, it must be optimal because it is minimal. Otherwise, once a multiple enters the forbidden segment, every subsequent multiple up to some point remains inside or above it in order. The next multiple after $r$ is the first point guaranteed to lie outside $[l, r]$, and no smaller multiple beyond $d$ can exist without violating divisibility or the interval constraint.
Python Solution
import sys
input = sys.stdin.readline
q = int(input())
for _ in range(q):
l, r, d = map(int, input().split())
if d < l or d > r:
print(d)
else:
print((r // d + 1) * d)
The code directly implements the logic derived in the walkthrough. The condition d < l or d > r captures the case where the smallest multiple already avoids the interval, so no adjustment is needed. Otherwise, we compute the first multiple strictly greater than r using integer division.
The expression (r // d + 1) * d is a standard way to jump to the next multiple. It avoids loops entirely and relies on floor division to find the last multiple not exceeding r.
Worked Examples
Example 1
Input:
l = 2, r = 4, d = 2
| Step | d in [l, r]? | Action | Result |
|---|---|---|---|
| 1 | Yes | compute next multiple after r | 6 |
The first multiple 2 is inside the interval, 4 is also inside, so we jump to 6.
This confirms the invariant that once the smallest valid multiple is blocked, we must skip the entire segment and move to the next block of multiples.
Example 2
Input:
l = 5, r = 10, d = 4
| Step | d in [l, r]? | Action | Result |
|---|---|---|---|
| 1 | No | return d directly | 4 |
Here the first multiple is already valid, so no further reasoning is needed. This demonstrates the shortcut case where no interval adjustment is required.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(q) | Each query is handled with a constant number of arithmetic operations |
| Space | O(1) | No additional structures beyond input variables |
The solution easily fits within limits since $q \le 500$ and each query is constant time, resulting in at most a few thousand operations overall.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
import sys
input = sys.stdin.readline
q = int(input())
out = []
for _ in range(q):
l, r, d = map(int, input().split())
if d < l or d > r:
out.append(str(d))
else:
out.append(str((r // d + 1) * d))
return "\n".join(out)
# provided samples
assert run("5\n2 4 2\n5 10 4\n3 10 1\n1 2 3\n4 6 5\n") == "6\n4\n1\n3\n10"
# custom cases
assert run("1\n1 100 101\n") == "101", "d just above range"
assert run("1\n10 20 3\n") == "3", "small multiple already valid"
assert run("1\n6 6 2\n") == "8", "tight interval containing multiple"
assert run("1\n1 1 1\n") == "2", "minimal boundary case"
| Test input | Expected output | What it validates |
|---|---|---|
| 1 100 101 | 101 | d greater than interval |
| 10 20 3 | 3 | d already optimal |
| 6 6 2 | 8 | single-point interval blocking multiples |
| 1 1 1 | 2 | smallest edge interval case |
Edge Cases
When $d > r$, the interval does not affect the smallest multiple at all. For example, with input $l=1, r=5, d=10$, the algorithm immediately returns 10 because it is already outside the segment.
When $d \in [l, r]$, the first multiple is blocked and we must jump. For $l=2, r=10, d=3$, the multiples are 3, 6, 9, 12. The first three lie in or near the interval, but the correct answer is 12, which is exactly the first multiple above $r$. The computation (r // d + 1) * d directly produces 12, matching the intended skip behavior without iteration.