CF 102697031 - 2001: A Space Odyssey

CF 102697031 - 2001: A Space Odyssey

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Solve time: 1m 49s
Verified: yes

Solution

Problem Understanding

We are given one positive composite integer n. The task is to find the smallest divisor of n that is greater than 1.

For example, the divisors of 2001 begin with 1, 3, 23, ..., so the required answer is 3. The official statement gives this example and specifies that the input number is guaranteed to be composite.

The input contains only one integer, so there is no need for multiple test cases or any complicated input handling. The official problem has a 1 second time limit and 256 MB memory limit. The statement does not expose an explicit upper bound for n, so the natural approach is to avoid unnecessary work and stop as soon as a divisor is found.

The key mathematical fact is that if n is composite, it has a factor d with 2 <= d <= sqrt(n). If no integer in that range divides n, then n would be prime. Since the input guarantees that n is composite, searching up to sqrt(n) is sufficient.

One edge case is when the smallest divisor is 2. For input 8, the correct output is 2. An implementation that starts checking at 3 would incorrectly return 4 or might fail entirely.

Python8

The correct output is:

Python2

Another edge case is a perfect square. For input 49, the smallest divisor is 7, and 7 * 7 = 49. A loop that stops before checking sqrt(n) would miss the answer.

Python49

The correct output is:

Python7

A third useful boundary case is the smallest possible composite input, 4.

Python4

The answer is 2. The algorithm must include 2 in its search and must allow the divisor to equal sqrt(n).

Approaches

The direct brute-force approach is to try every integer from 2 through n - 1 and return the first one that divides n. It is correct because every possible answer is examined in increasing order, so the first successful divisor is necessarily the smallest one. The problem is that for a large composite number whose smallest divisor is close to n / 2, this performs roughly n / 2 modulo operations. Even worse, if the input bound were large, a linear scan would spend most of its time checking numbers that cannot possibly be the smallest factor.

The structure of factor pairs gives us a much better stopping point. If n = a * b and both factors were greater than sqrt(n), their product would be greater than n, which is impossible. Thus every composite number has at least one factor no larger than sqrt(n). We only need to search that range.

There is an even simpler observation specific to this problem. We are looking for the smallest divisor, so we can test candidates in increasing order starting from 2. The first successful candidate is the answer, and if the answer is not encountered earlier, we only need to continue through sqrt(n). For a composite input, the loop is guaranteed to terminate there at the latest.

The optimized method therefore reduces the worst-case number of divisibility checks from linear in n to linear in sqrt(n).

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n)	O(1)	Too slow for large n
Optimal	O(sqrt(n))	O(1)	Accepted

Algorithm Walkthrough

1. Read the composite integer n.
2. Start a candidate divisor at 2. This is the smallest possible positive divisor greater than 1, so it must be tested first.
3. Test candidates in increasing order while candidate * candidate <= n. If n % candidate == 0, print candidate and stop. Because candidates are examined in increasing order, the first divisor found is the smallest one.
4. The loop must always find a divisor for the promised composite input. If it did not, n would have no divisor from 2 through sqrt(n), which would imply that n is prime.

The condition candidate * candidate <= n is preferable to computing a floating-point square root. It avoids floating-point precision issues and directly expresses the mathematical boundary.

Why it works

The invariant is that before testing a candidate d, every integer from 2 through d - 1 has been checked and does not divide n. Consequently, when d divides n, no smaller valid divisor exists, so d is exactly the required answer. A composite number must have a divisor at most sqrt(n), so the search cannot pass the square-root boundary without finding an answer.

Python Solution

Pythonimport sysinput = sys.stdin.readline
def solve():    n = int(input())
    d = 2    while d * d <= n:        if n % d == 0:            print(d)            return        d += 1
if __name__ == "__main__":    solve()

The program reads the only input value and initializes d to 2, matching the smallest divisor we are allowed to output.

The loop uses d * d <= n instead of d <= int(sqrt(n)). The multiplication stays entirely in integer arithmetic, so there is no rounding problem at perfect squares such as 49.

When n % d == 0, the current candidate divides n. Since all smaller candidates have already failed, printing d immediately gives the correct answer and avoids unnecessary work.

The input is guaranteed to be composite, so there is no required output for the prime case. If the problem did allow prime inputs, the algorithm would need an additional decision after the loop. Here that case cannot occur.

Worked Examples

Example 1

For the official sample, n = 2001. The first candidate, 2, does not divide 2001, while 3 does.

Candidate d	d * d <= n	2001 % d	Action
2	Yes	1	Continue
3	Yes	0	Output 3

The algorithm stops immediately at 3. This demonstrates the main advantage of testing divisors in increasing order: as soon as the first factor is found, no further factorization is necessary. The official sample output is 3.

Example 2

Consider the composite square n = 49.

Candidate d	d * d <= n	49 % d	Action
2	Yes	1	Continue
3	Yes	1	Continue
4	Yes	1	Continue
5	Yes	4	Continue
6	Yes	1	Continue
7	Yes	0	Output 7

The condition uses <=, so d = 7 is actually tested. This is exactly the boundary that a < condition would incorrectly skip.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(sqrt(n))	At most all integers from 2 through sqrt(n) are tested
Space	O(1)	Only the input and one candidate divisor are stored

The 1 second limit makes a linear scan undesirable when n is large, while the square-root search performs dramatically fewer operations. The algorithm also uses constant extra memory, comfortably within the 256 MB limit specified by the problem.

Test Cases

Pythonimport sysimport io
def solve():    input = sys.stdin.readline    n = int(input())
    d = 2    while d * d <= n:        if n % d == 0:            print(d)            return        d += 1

def run(inp: str) -> str:    old_stdin = sys.stdin    old_stdout = sys.stdout
    sys.stdin = io.StringIO(inp)    sys.stdout = io.StringIO()
    solve()    result = sys.stdout.getvalue()
    sys.stdin = old_stdin    sys.stdout = old_stdout
    return result

# Provided sampleassert run("2001\n") == "3\n", "sample 1"
# Minimum-size compositeassert run("4\n") == "2\n", "minimum composite"
# All factors larger than 2assert run("49\n") == "7\n", "perfect square boundary"
# Small composite with smallest divisor 3assert run("9\n") == "3\n", "small square"
# Large power of two, catches unnecessary scanningassert run("1048576\n") == "2\n", "smallest divisor at the first candidate"

Test input	Expected output	What it validates
4	2	Minimum composite input
49	7	Perfect-square boundary and <= sqrt(n)
9	3	Smallest divisor greater than 2
1048576	2	Immediate termination when the first candidate works

Edge Cases

For n = 4, the algorithm begins with d = 2. The condition 2 * 2 <= 4 is true, and 4 % 2 == 0, so it immediately prints 2. This handles the smallest possible composite number without any special case.

For n = 8, the first candidate is again 2, and 8 % 2 == 0. The answer is printed before candidates such as 3 or 4 are considered. This is why the search must begin at 2, rather than at 3.

For n = 49, candidates 2 through 6 fail. When d reaches 7, the loop condition is still true because 7 * 7 == 49. The divisibility test succeeds and the program prints 7. Using d * d < n instead would silently produce the wrong result.

For n = 2001, 2 fails and 3 succeeds, so the program prints 3. The algorithm does not need to factor the remaining value 667, because the problem asks only for the smallest non-unit divisor.