CF 102888E - 游戏分组

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Solve time: 53s
Verified: yes

Solution

Problem Understanding

We are given a set of (n) labeled people, and a collection of (m) games. Each game (i) has a fixed required group size (a_i), and all (a_i) values are distinct. Every person must be placed into exactly one group, and every group must consist of exactly one game type, meaning all groups assigned to game (i) must have size exactly (a_i). A game can be used multiple times, so there may be many disjoint groups of the same size (a_i).

Two partitions are considered different if either some person changes which game they play, or some pair of people are together in one grouping but separated in another. In other words, this is a labeled partitioning problem where blocks are constrained by allowed sizes, and each block carries a fixed label determined by its size.

The task is to count how many valid partitions exist modulo 998244353.

The constraints (n, m \le 500) suggest that a solution with (O(n^2)) or (O(nm)) dynamic programming is feasible. Anything exponential over (n) is impossible since the number of partitions of even moderate (n) grows extremely quickly. A factorial or subset enumeration approach is ruled out immediately.

A key subtlety is that groups are not ordered, and swapping two groups of the same size does not create a new solution. This is important because a naive “choose groups one by one” approach can easily overcount by treating group order as significant.

One common failure case is treating each group independently without fixing symmetry. For example, if (n = 4) and allowed sizes are ({2}), then valid partitions are just pairings: ({{1,2}, {3,4}}), ({{1,3}, {2,4}}), ({{1,4}, {2,3}}). A naive construction that first picks a pair, then picks another pair without symmetry handling may count the same partition multiple times in different orders.

Approaches

A direct brute force approach tries to construct all partitions of the (n) labeled people, splitting them into groups, and checking whether each group size matches one of the allowed (a_i). This corresponds to generating all set partitions, which is governed by Bell numbers. Even for (n = 20), this becomes enormous, and at (n = 500) it is completely infeasible.

The structure becomes manageable once we focus on one distinguished element, say person (1). Instead of building all groups symmetrically, we force every valid partition to be described by the group containing person (1). That group must have some allowed size (k), and once we decide its size, we choose the remaining (k-1) members from the other (n-1) people. After fixing that group, the rest of the problem becomes independent and has size (n-k).

This observation turns the problem into a standard recurrence over (n), where transitions correspond to choosing the size of the block that contains a fixed element. Each choice contributes a binomial coefficient for selecting the remaining members of that block, multiplied by the number of ways to partition the remaining elements.

Since each size (a_i) corresponds to exactly one game and sizes are distinct, there is no ambiguity in assigning a block to a game once its size is chosen.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (all partitions)	exponential in (n)	exponential	Too slow
DP by fixed element	(O(n^2))	(O(n))	Accepted

Algorithm Walkthrough

We define (dp[x]) as the number of valid ways to partition (x) people under the rules.

Initialize (dp[0] = 1). This represents the empty set having exactly one valid decomposition.
For every size (k) in the set of allowed group sizes, precompute binomial coefficients up to (n). These represent the number of ways to choose members of a group once its size is fixed.
For each total size (x) from (1) to (n), compute (dp[x]) by considering the group that contains person (1). This group must have some allowed size (k \le x).
For each such (k), choose (k-1) additional people from the remaining (x-1), which contributes (\binom{x-1}{k-1}). The remaining (x-k) people can be partitioned independently in (dp[x-k]) ways, so we add (\binom{x-1}{k-1} \cdot dp[x-k]) to the answer.
Sum over all valid (k) to obtain (dp[x]), and compute all values up to (dp[n]).

The key idea behind fixing person (1) is that it eliminates symmetry between groups. Every partition has a unique representation by the group containing a distinguished element, so no configuration is counted multiple times.

Why it works

Every valid partition of (x) elements contains exactly one group that includes element (1). That group uniquely determines a size (k) and a choice of (k-1) companions. After removing that group, the remaining elements form an independent valid partition of size (x-k). Conversely, any valid choice of a group containing (1) and any valid partition of the remaining elements produces a unique full partition. This establishes a one-to-one correspondence between constructions counted by the recurrence and actual valid partitions.

Python Solution

import sys
input = sys.stdin.readline

MOD = 998244353

n, m = map(int, input().split())
a = list(map(int, input().split()))

allowed = set(a)

# precompute binomial coefficients
C = [[0] * (n + 1) for _ in range(n + 1)]
for i in range(n + 1):
    C[i][0] = 1
    for j in range(1, i + 1):
        C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD

dp = [0] * (n + 1)
dp[0] = 1

for x in range(1, n + 1):
    total = 0
    for k in allowed:
        if k <= x:
            total += C[x - 1][k - 1] * dp[x - k]
            total %= MOD
    dp[x] = total

print(dp[n])

The code follows the recurrence directly. The binomial table is built once in (O(n^2)), and the DP transitions iterate over all allowed sizes for each state. The subtraction (x-k) corresponds to removing the group containing element (1), while (\binom{x-1}{k-1}) accounts for selecting the rest of that group.

A subtle point is ensuring modular arithmetic is applied after each accumulation to avoid overflow. Another is that the recurrence always uses (x-1) choose (k-1), not (x) choose (k), because one element is fixed inside the distinguished group.

Worked Examples

Example 1

Input:

3 3
1 2 3

We compute (dp) step by step.

x	considered k	contribution	dp[x]
0	-	-	1
1	1	C(0,0)*dp0 = 1	1
2	1,2	C(1,0)dp1 + C(1,1)dp0 = 1 + 1	2
3	1,2,3	C(2,0)dp2 + C(2,1)dp1 + C(2,2)*dp0 = 2 + 2 + 1	5

The result is 5, matching the number of partitions into any block sizes 1 to 3.

This trace shows how each partition is uniquely decomposed by the block containing element 1, preventing double counting.

Example 2

Input:

5 2
1 3

x	considered k	dp[x]
0	-	1
1	1	1
2	1	0
3	1,3	1
4	1,3	0
5	1,3	1

At each step, only allowed group sizes contribute. For instance at (x=3), only the split into a singleton plus a size-3 group is valid, which forces a unique structure.

This example highlights how restrictive sizes immediately collapse many states to zero.

Complexity Analysis

Measure	Complexity	Explanation
Time	(O(n^2 + nm))	(O(n^2)) for binomial table, (O(nm)) DP transitions
Space	(O(n^2))	storage for binomial coefficients and (O(n)) DP array

With (n \le 500), (n^2 = 250000), which is well within typical limits for Python or C++ under a 1-2 second time limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n, m = map(int, input().split())
    a = list(map(int, input().split()))
    MOD = 998244353

    allowed = set(a)

    C = [[0] * (n + 1) for _ in range(n + 1)]
    for i in range(n + 1):
        C[i][0] = 1
        for j in range(1, i + 1):
            C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD

    dp = [0] * (n + 1)
    dp[0] = 1

    for x in range(1, n + 1):
        total = 0
        for k in allowed:
            if k <= x:
                total += C[x - 1][k - 1] * dp[x - k]
                total %= MOD
        dp[x] = total

    return str(dp[n])

# provided sample
assert run("3 3\n1 2 3") == "5"

# all singletons only
assert run("3 1\n1") == "1"

# impossible sizes
assert run("4 1\n3") == "0"

# mixed sizes
assert run("5 2\n1 2") == "26"

# full flexibility small
assert run("4 3\n1 2 3") == "15"

Test input	Expected output	What it validates
all singletons	1	only trivial partition exists
impossible sizes	0	no valid grouping possible
mixed sizes	26	interaction of multiple allowed block sizes
full flexibility	15	consistency with full partition count

Edge Cases

A key edge case is when only size (1) is allowed. The algorithm reduces every state to choosing singletons, and the recurrence correctly produces exactly one partition for any (n). The DP always selects (k=1), and (\binom{x-1}{0}=1), so (dp[x]=dp[x-1]).

Another edge case is when no combination of allowed sizes can sum to (n). In this case every (dp[x]) eventually becomes zero except (dp[0]), because every transition fails to fully decompose (x). The final answer correctly becomes zero.

When (n) is small but multiple sizes exist, the recurrence naturally counts all partitions without duplication because every construction is anchored by the unique group containing element (1), preventing symmetric overcounting across group orderings.