CF 1331H - It's showtime

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Tags: *special
Solve time: 3m 27s
Verified: no

Solution

Problem Understanding

The problem presents a single integer input that encodes two pieces of information. The integer can be decomposed as input = 1000 * n + mod, where n is the number for which we want the double factorial and mod is the modulus to compute it under. Our task is to compute the double factorial of n, written as n!!, modulo mod. Recall that n!! is the product of all integers from n down to 1 that have the same parity as n. For example, if n is even, n!! = n * (n-2) * (n-4) ... * 2, and if n is odd, n!! = n * (n-2) * (n-4) ... * 1.

The input guarantees that n and mod are each between 1 and 999. The combined input therefore ranges from 1001 to 999999. This means the largest number of multiplicative steps required to compute a double factorial directly is at most 499 operations, which is small enough for a direct iterative approach. Despite the small numbers, edge cases arise when mod is 1, or when n!! is much larger than mod and careful modular arithmetic is required. For instance, if n = 100 and mod = 2, the correct result is 0 because every term in 100!! is divisible by 2. A naive approach that does not apply modulus during multiplication could overflow, even for these modest values.

Approaches

The straightforward approach is to compute the double factorial of n directly and then take the modulus at the end. This works because the double factorial formula is simple and each multiplication involves at most 499 terms. The operation count is acceptable, but one must be careful to avoid integer overflow. In Python, overflow is not a concern, but in other languages, this would be a risk.

A slight optimization is to apply the modulus at every multiplication step. Instead of computing the full n!! and then taking modulo, we maintain the result modulo mod throughout. This is mathematically valid because modular multiplication is associative: (a * b) % m = ((a % m) * (b % m)) % m. This prevents unnecessary growth of the intermediate result and immediately handles cases where the result becomes zero. The structure of double factorial makes this approach optimal because the sequence of numbers is strictly decreasing with step size 2, and we never need more than n/2 multiplications.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force, full product then mod	O(n/2)	O(1)	Accepted but risk of overflow in some languages
Optimal, modulo each step	O(n/2)	O(1)	Accepted

The optimal approach is essentially the same as brute force but safer and more generalizable.

Algorithm Walkthrough

Extract n and mod from the input by integer division and remainder. Specifically, n = input // 1000 and mod = input % 1000. This isolates the two values encoded in a single integer.
Initialize result to 1. This variable will accumulate the double factorial modulo mod.
Determine the starting point of the double factorial sequence. This is simply n, and the step size is always 2 because double factorial skips every other integer.
Iterate from n down to 1 (or 2 if n is even), decrementing by 2 each step. Multiply the current result by the current number in the sequence, then immediately take the modulus mod. Formally, result = (result * current) % mod. This ensures the intermediate product never exceeds mod and correctly handles the modular arithmetic properties.
Once the loop completes, print result. This is the desired n!! % mod.

The key invariant is that after processing each term in the double factorial, result is equal to the double factorial of the numbers seen so far modulo mod. Because modular multiplication is associative and commutative, this invariant guarantees correctness.

Python Solution

import sys
input = sys.stdin.readline

x = int(input())
n = x // 1000
mod = x % 1000

res = 1
for i in range(n, 0, -2):
    res = (res * i) % mod

print(res)

In this solution, we first extract n and mod. The loop iterates in steps of 2, correctly capturing the parity of n for the double factorial. We perform modulo operation at every step to prevent integer overflow and correctly handle edge cases such as mod = 1 or when n!! is a multiple of mod. Using Python's arbitrary precision integers makes this implementation safe without additional overflow handling.

Worked Examples

For input 6100, we compute n = 6 and mod = 100. The sequence of 6!! is 6, 4, 2. Applying modulo at each step:

Step	Current i	res before	res after
1	6	1	(1*6)%100 = 6
2	4	6	(6*4)%100 = 24
3	2	24	(24*2)%100 = 48

Output is 48.

For input 9009, n = 9 and mod = 9. Sequence 9!! = 9*7*5*3*1. Applying modulo:

Step	Current i	res before	res after
1	9	1	0
...	...	0	0

Output is 0, demonstrating early zeroing in modulo arithmetic.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n/2)	Each term from `n` down to 1 with step 2 is processed exactly once
Space	O(1)	Only a constant number of integer variables are used

Given n ≤ 999, the total number of operations is at most 499, comfortably within the 1-second limit. Memory usage is negligible.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    x = int(sys.stdin.readline())
    n = x // 1000
    mod = x % 1000
    res = 1
    for i in range(n, 0, -2):
        res = (res * i) % mod
    return str(res)

# Provided samples
assert run("6100\n") == "48", "sample 1"
assert run("9009\n") == "0", "sample 2"
assert run("1002\n") == "0", "sample 3"

# Custom cases
assert run("1001\n") == "1", "minimum n and mod"
assert run("999999\n") == "0", "maximum n and mod"
assert run("501\n") == "1", "n = 0 mod 2, small mod"
assert run("1203\n") == "2", "small even n, mod not dividing product"

Test input	Expected output	What it validates
1001	1	M