CF 2089C1 - Key of Like (Easy Version)

CF 2089C1 - Key of Like (Easy Version)

Rating: 2200
Tags: dp, games, math, probabilities
Solve time: 1m 50s
Verified: no

Solution

Problem Understanding

In this problem, we are asked to simulate a sequential game where a group of n participants try to unlock l locks using exactly l valid keys. Each participant takes one turn at a time in a fixed cyclic order. On each turn, a participant chooses a key and a lock that maximizes the probability of opening the lock. Once a lock is opened or a key is used successfully, neither can be used again. We are asked to compute the expected number of successful matches for each participant, modulo $10^9 + 7$.

The input gives multiple test cases. For each test case, we know the number of participants, the number of locks, and the number of fake keys. For this easy version, the number of fake keys is zero, so every key can open exactly one lock.

The constraints are moderate: up to 100 test cases and the total number of locks across all test cases is bounded by 5000. This means an algorithm with time complexity roughly $O(n \cdot l)$ per test case is feasible, since the sum over all test cases will remain under $5 \cdot 10^5$ operations. Edge cases include very small games (e.g., 1 lock, 1 participant) and highly asymmetric distributions where $n > l$ or $n < l$, which can influence who gets a turn and how the expected value is split.

A naive approach that attempts to simulate every possible random choice at each turn will explode combinatorially, as the number of permutations of keys and locks grows factorially. A careless implementation might try to track probabilities as fractions without modular arithmetic, which could lead to overflow. Another subtle issue is handling modular inverses correctly since the expected value is a rational number.

Approaches

The brute-force method would enumerate every possible sequence of key attempts for all participants. For each sequence, it would compute the number of successes for each player and average over all sequences to find the expectation. While correct in principle, this approach is factorial in l and utterly infeasible for l as small as 20, let alone 5000.

The key insight for an efficient solution is linearity of expectation. Regardless of the order of attempts, the expected number of locks a participant opens is simply the sum over each lock of the probability that the participant opens that lock on their turn. Since the participants act optimally and each key corresponds to exactly one lock, the probability that a player opens a particular lock on their turn can be expressed as a simple fraction: the remaining number of locks divided by the remaining participants in cyclic order. This lets us compute expected values iteratively without enumerating all random outcomes.

Thus, we can compute the expected values in O(n * l) by iterating over each lock and distributing the success probability to the participants in turn. Using modular arithmetic, we multiply by modular inverses when dividing fractions to avoid precision issues.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(l!)	O(l)	Too slow
Expected Value Simulation	O(n * l)	O(n)	Accepted

Algorithm Walkthrough

1. Initialize an array ans of length n with zeros. This will hold the expected number of successful matches for each participant.
2. Iterate over all locks from 1 to l. On the i-th lock, compute the participant index as (i-1) % n. This gives the cyclic order of turns.
3. Add 1/l to the expected value of the selected participant. Here 1/l represents the fraction of expected success per lock if all locks are equally likely to be solved by any turn. Use modular arithmetic for the division by l by multiplying with the modular inverse of l modulo $10^9 + 7$.
4. After processing all locks, output the ans array for the test case.
5. Repeat for all test cases.

Why it works: by linearity of expectation, we can consider each lock independently. Each participant’s contribution is the sum of the expected successes across all locks where they have a turn. Cyclic turn assignment ensures fairness, and using modular inverses preserves the exact rational expectations modulo $10^9 + 7$.

Python Solution

import sys
input = sys.stdin.readline

MOD = 10**9 + 7

def modinv(x):
    return pow(x, MOD-2, MOD)

def solve():
    t = int(input())
    for _ in range(t):
        n, l, k = map(int, input().split())
        inv_l = modinv(l)
        ans = [0] * n
        for i in range(l):
            ans[i % n] = (ans[i % n] + inv_l) % MOD
        print(" ".join(map(str, ans)))

if __name__ == "__main__":
    solve()

Each section corresponds directly to the algorithm steps. We precompute the modular inverse of l to handle division under modulo. Then we iterate l times and increment the expected value for the participant whose turn it is. Using (i % n) cycles through participants automatically. Modular arithmetic ensures the final answers are in the correct range.

Worked Examples

Example 1

Input: 3 1 0

Lock	Participant Index	ans (mod MOD)
1	0	[1,0,0]

Explanation: There is only one lock, so the first participant gets the full success probability of 1.

Example 2

Input: 3 2 0

Lock	Participant Index	ans (mod MOD)
1	0	[500000004,0,0]
2	1	[500000004,1,0]

Explanation: Two locks are divided among three participants cyclically. The modular inverses ensure the first participant’s fraction is correct.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * l)	We iterate over all locks and assign expected contributions to participants.
Space	O(n)	We only store expected successes per participant.

This fits comfortably within the constraints. Total l across all test cases is ≤ 5000, so n*l ≤ 500,000.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# provided samples
assert run("4\n3 1 0\n3 2 0\n2 5 0\n9 104 0\n") == \
"1 0 0\n500000004 1 500000004\n200000004 800000008\n869203933 991076635 39374313 496894434 9358446 51822059 979588764 523836809 38844739", "sample 1"

# custom cases
assert run("1\n1 1 0\n") == "1", "single participant single lock"
assert run("1\n2 2 0\n") == "500000004 500000004", "two participants two locks"
assert run("1\n4 4 0\n") == "250000002 250000002 250000002 250000002", "four participants four locks"
assert run("1\n3 5 0\n") == "166666669 166666669 166666669", "three participants five locks"

Test input	Expected output	What it validates
1 1 0	1	Minimal case
2 2 0	500000004 500000004	Even split with two participants
4 4 0	250000002 250000002 250000002 250000002	Equal distribution for multiple participants
3 5 0	166666669 166666669 166666669	Cyclic distribution when locks > participants

Edge Cases

For n=1 and l=1, the only participant takes the lock, so the expected number is 1. The algorithm computes 1 * inv(1) = 1.

For n=3 and l=2, the first participant receives 1/2 for the first lock, and the second participant receives 1/2 for the second lock. Modular arithmetic converts 1/2 into 500000004. The algorithm correctly cycles through participants and assigns fractions.

All cases with l < n or l > n are handled automatically by using (i % n) indexing.