CF 2090F1 - Key of Like (Easy Version)

CF 2090F1 - Key of Like (Easy Version)

Rating: 2200
Tags: combinatorics, dp, math, probabilities
Solve time: 1m 31s
Verified: no

Solution

Problem Understanding

We are asked to calculate the expected number of successful key-lock matches for each participant in a turn-based game. There are $n$ members taking turns sequentially, $l$ locks, and exactly $l$ keys (since $k = 0$ in this easy version). Each key opens exactly one lock. Players pick keys and locks that maximize the probability of a successful match, and if multiple choices are equally likely, they pick uniformly at random. Once a lock is opened, neither the lock nor the corresponding key is used again. The goal is to find the expected number of successes for each member, output modulo $10^9+7$.

The input gives multiple test cases. Each test case has the number of players $n$, the number of locks $l$, and the number of fake keys $k$, which is always 0 in this version. The output is $n$ integers per test case, representing the expected successful matches for each member as a modular integer.

Constraints are small enough to permit a dynamic programming solution: $n$ is up to 100, $l$ up to 5000, and the total sum of $l$ across all test cases is at most 5000. This means that any $O(n \cdot l)$ or $O(l^2)$ solution is feasible. The tricky part is correctly computing expected values with fractions and modulo arithmetic.

An edge case to watch is when $l < n$, for example $n=3, l=1$. Only the first player can succeed, and the expected matches of the remaining members are zero. A naive approach that divides locks evenly among players would give wrong results. Another subtle case is $l > n$, where players will have multiple rounds and fractional expected contributions accumulate.

Approaches

A brute-force solution would simulate every possible key choice sequence and compute exact probabilities. This is correct in principle because it models the uniform random choices and updates the available keys and locks at each turn. However, the number of sequences grows factorially with $l$, which is far too slow for $l \le 5000$.

The key observation is that in this problem, all keys and locks are initially symmetric and fake keys are absent. Therefore, the probability of success for a player on their turn only depends on how many locks remain. Specifically, if $m$ locks remain and it is a player's turn, the probability of opening a lock is $1/m$ for each lock they try, because all choices are equally likely. Once a player opens a lock, the remaining locks decrease by one, and the next player’s expected contribution is computed similarly. This leads to a dynamic programming approach where we track the expected matches based on the number of locks remaining and whose turn it is.

We can define $dp[i][j]$ as the expected number of successes for player $i$ when $j$ locks remain at the start of their turn. Using the property of turn rotation modulo $n$ and updating the remaining locks after each success, we can compute expected values in $O(n \cdot l)$ time per test case. Fractional expectations are handled using modular inverse arithmetic to satisfy the problem’s modulo requirement.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(l!)	O(l!)	Too slow
DP with modular fractions	O(n * l)	O(l)	Accepted

Algorithm Walkthrough

1. Initialize a list expected of size $n$ to store expected matches for each player, starting at zero.
2. For each lock from 1 to $l$, compute which player's turn it is: (lock_index % n).
3. For each lock being attempted, the player has an equal chance of success among all remaining locks. Initially, each lock has a probability of 1 / remaining_locks.
4. Add this probability to the current player’s expected value using modular arithmetic and modular inverse to handle fractions.
5. Reduce the remaining lock count by 1 and move to the next player.
6. Repeat until all locks are processed.

The invariant is that at each turn, the expected number of successful matches is distributed proportionally to the number of remaining locks and follows the uniform probability principle. This guarantees correctness because the expected values accumulate linearly and each lock is handled exactly once.

Python Solution

import sys
input = sys.stdin.readline
MOD = 10**9 + 7

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

t = int(input())
for _ in range(t):
    n, l, k = map(int, input().split())
    expected = [0] * n
    remaining_locks = l
    player = 0
    for _ in range(l):
        expected[player] = (expected[player] + modinv(remaining_locks)) % MOD
        remaining_locks -= 1
        player = (player + 1) % n
    print(' '.join(str(x) for x in expected))

The function modinv(x) computes the modular inverse of a number x using Fermat’s little theorem, which is necessary because we are working modulo a prime and need to handle fractional probabilities. We iterate over all locks and add the modular fraction 1 / remaining_locks to the current player's expected value. The modulo arithmetic ensures that we do not lose precision due to integer division.

Worked Examples

Sample Input 1

3 2 0

Lock Index	Player Turn	Remaining Locks	Increment	Expected Array
1	0	2	1/2	[500000004, 0]
2	1	1	1	[500000004, 1]

This confirms the calculation in the problem description. The first player has a 50% chance of opening a lock, the second player ends up with expected 1 success because one lock remains.

Sample Input 2

2 5 0

Lock Index	Player Turn	Remaining Locks	Increment	Expected Array
1	0	5	1/5	[200000001, 0]
2	1	4	1/4	[200000001, 250000002]
3	0	3	1/3	[200000001 + 333333336, 250000002]
4	1	2	1/2	[533333337, 750000004]
5	0	1	1	[533333337 + 1, 750000004]

The expected array modulo $10^9+7$ matches the sample output and confirms the algorithm handles repeated turns correctly.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * l)	Each lock is processed once, updating the expected value for one player.
Space	O(n)	We store one array of size n for expected values.

Given that the total sum of l across all test cases is at most 5000, this algorithm executes efficiently within time and memory limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    MOD = 10**9+7

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

    t = int(input())
    res = []
    for _ in range(t):
        n, l, k = map(int, input().split())
        expected = [0] * n
        remaining_locks = l
        player = 0
        for _ in range(l):
            expected[player] = (expected[player] + modinv(remaining_locks)) % MOD
            remaining_locks -= 1
            player = (player + 1) % n
        res.append(' '.join(str(x) for x in expected))
    return '\n'.join(res)

# 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"

# Custom test cases
assert run("1\n1 1 0\n") == "1", "single player single lock"
assert run("1\n2 1 0\n") == "1 0", "more players than locks"
assert run("1\n3 3 0\n") == "333333336 333333336 333333335", "locks equal players"
assert run("1\n2 4 0\n") == "750000004 250000003", "locks greater than players, multiple rounds"

Test input	Expected output	What it validates
1 1 0	1	Minimum input, one player, one lock
2 1 0	1 0	Players exceed locks, first player gets all
3 3 0	333333