CF 105530B - Modular MEX
We are looking at all remainders produced by fixing a number $n$ and dividing it by every integer $i$ from $1$ to $n$. This gives a set of values of the form $n bmod i$.
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Solve time: 52s
Verified: yes
Solution
Problem Understanding
We are looking at all remainders produced by fixing a number $n$ and dividing it by every integer $i$ from $1$ to $n$. This gives a set of values of the form $n \bmod i$. Each choice of divisor produces a remainder between $0$ and $i - 1$, so across all divisors we collect a structured but not arbitrary set of integers.
The task is to determine the smallest non-negative integer that never appears among these remainders. In other words, we are computing the MEX of the set ${ n \bmod i \mid 1 \le i \le n }$.
The constraint structure is extremely tight. Since the input is a single integer, any solution up to $O(n)$ or even $O(\sqrt{n})$ is trivially fast enough, and the real challenge is not computation but recognizing the pattern in the set of remainders.
A naive approach would explicitly compute all values $n \bmod i$, insert them into a set, and then scan upward from zero until finding the first missing integer. This works mechanically, but it hides the structure and can mislead intuition about what values are actually reachable.
A subtle pitfall appears when reasoning about whether large values can occur as remainders. For example, one might incorrectly assume that values close to $n$ can appear because the modulus changes with $i$, but in fact every remainder is strictly less than its divisor, which constrains the maximum possible value in a global sense.
For instance, when $n = 5$, the set is:
$5 \bmod 1 = 0$,
$5 \bmod 2 = 1$,
$5 \bmod 3 = 2$,
$5 \bmod 4 = 1$,
$5 \bmod 5 = 0$,
so the MEX is $3$.
The correct answer turns out to depend only on how many small integers are guaranteed to appear, not on the exact distribution of all remainders.
Approaches
A direct brute-force strategy computes every remainder $n \bmod i$ for $i$ from $1$ to $n$. This produces $n$ values, and then we insert them into a boolean array or set and linearly search for the smallest missing integer. This is correct because it exhaustively constructs the target set.
The cost of this approach is linear in $n$, both for generating remainders and for scanning. That is at most about $2n$ operations, which is fine for typical constraints but gives no structural insight.
The key observation is that the set of remainders has a sharp threshold around $\lceil n/2 \rceil$. Every integer smaller than this threshold can be shown to appear as a remainder for some divisor, while the threshold itself is never guaranteed to be present. This creates a contiguous prefix of achievable values starting from zero, and the MEX is exactly the first value outside this prefix.
The brute-force works because it directly enumerates all residues, but it becomes unnecessary once we recognize that the reachable values form a continuous interval. The problem reduces to identifying the length of this interval rather than constructing it explicitly.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | $O(n)$ | $O(n)$ | Accepted |
| Optimal | $O(1)$ | $O(1)$ | Accepted |
Algorithm Walkthrough
The optimal solution is based on recognizing the structure of reachable remainders.
- Compute $k = \lceil n/2 \rceil$. This value acts as a boundary separating guaranteed remainders from impossible ones.
- Return $k$ as the answer. This is the first integer that cannot be represented as $n \bmod i$ for any $i \le n$.
The reason this step is valid comes from how modular reduction behaves: small values can be formed by choosing divisors just slightly larger than the remainder we want, while values at or above the threshold cannot survive the constraint $n \bmod i < i$ for all $i$.
Why it works
For any integer $m < \lceil n/2 \rceil$, we can construct a divisor $i = n - m$. This choice ensures that $n \bmod i = m$, since $n = 1 \cdot (n - m) + m$. The condition $i \le n$ holds automatically because $m > 0$ implies $i < n$, and for $m = 0$, we can take $i = n$.
This shows that every value below $\lceil n/2 \rceil$ is achievable. On the other hand, any value $m \ge \lceil n/2 \rceil$ would require a divisor $i \le n - m \le n/2$, but such construction no longer guarantees that the remainder equals $m$, and in fact the arithmetic structure prevents covering all larger values consecutively. The reachable set forms exactly the prefix $[0, \lceil n/2 \rceil - 1]$, so the MEX is $\lceil n/2 \rceil$.
Python Solution
import sys
input = sys.stdin.readline
n = int(input())
print((n + 1) // 2)
The entire solution reduces to computing the ceiling of half of $n$. The expression $(n + 1) // 2$ handles both even and odd cases cleanly. No additional data structures or loops are required because the reachable structure is fully determined by arithmetic reasoning rather than enumeration.
A common mistake here is attempting to simulate remainders or build a set, which is unnecessary and risks confusion about bounds. The solution depends only on integer division behavior.
Worked Examples
Consider $n = 5$.
We compute remainders:
| i | n mod i |
|---|---|
| 1 | 0 |
| 2 | 1 |
| 3 | 2 |
| 4 | 1 |
| 5 | 0 |
The set is ${0, 1, 2}$. The first missing integer is $3$, which matches $\lceil 5/2 \rceil = 3$.
This confirms that all values below the boundary appear, and the boundary itself does not.
Now consider $n = 8$.
| i | n mod i |
|---|---|
| 1 | 0 |
| 2 | 0 |
| 3 | 2 |
| 4 | 0 |
| 5 | 3 |
| 6 | 2 |
| 7 | 1 |
| 8 | 0 |
The set becomes ${0, 1, 2, 3}$. The MEX is $4$, matching $\lceil 8/2 \rceil = 4$.
These examples show that the reachable prefix grows exactly up to half of $n$, and nothing beyond it fills the gap immediately after.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | $O(1)$ | Only a single arithmetic computation is required |
| Space | $O(1)$ | No auxiliary storage is used |
The solution runs in constant time regardless of input size, which is far below the limits of any competitive programming environment.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
n = int(sys.stdin.readline())
print((n + 1) // 2)
return sys.stdout.getvalue().strip()
# simple checks
assert run("1\n") == "1", "n=1"
assert run("2\n") == "1", "n=2"
assert run("5\n") == "3", "example"
assert run("8\n") == "4", "even case"
assert run("1000000000\n") == "500000000", "large case"
| Test input | Expected output | What it validates |
|---|---|---|
| 1 | 1 | minimum boundary |
| 2 | 1 | smallest non-trivial case |
| 5 | 3 | standard odd case |
| 8 | 4 | standard even case |
| 1000000000 | 500000000 | large input stability |
Edge Cases
For $n = 1$, the only remainder is $0$, so the MEX is $1$. The formula $(n + 1) // 2$ gives $1$, matching directly.
For $n = 2$, remainders are $0$ and $0$, so the MEX is $1$. The formula gives $1$, consistent with the set structure.
For large values of $n$, no iteration is performed at all, so there is no risk of overflow or performance degradation. The computation remains a single arithmetic operation regardless of magnitude.