CF 26D - Tickets

Rating: 2400
Tags: combinatorics, math, probabilities
Solve time: 1m 14s
Verified: yes

Solution

Problem Understanding

We have a ticket seller, Charlie, who sells race tickets costing 10 euros each. Customers arrive in some random order: some have only 10 euro banknotes, some have only 20 euro banknotes. Charlie initially has k 10 euro banknotes. Every customer with a 20 euro note requires Charlie to give one 10 euro note as change. The question is: given n customers with 10 euro notes, m with 20 euro notes, and k starting 10 euro notes, what is the probability that Charlie can serve everyone without running out of change?

The input constraints are n, m ≤ 10^5 and k ≤ 10. Because n + m can be very large, any approach that tries to enumerate all customer permutations is infeasible. At the same time, k is very small, which hints that we can use techniques that scale with k instead of n + m.

Edge cases arise when the number of starting 10 euro notes is too small to cover early 20 euro customers. For example, with n=2, m=2, k=0, if a 20 euro customer comes first, Charlie cannot give change. A naive simulation that does not consider all orderings would incorrectly assume success. Another subtle case is when k ≥ m, meaning Charlie always has enough change regardless of the order; the probability is 1.

Approaches

The brute-force approach is straightforward: generate all permutations of the n + m customers, simulate the ticket sales for each permutation, and count the fraction where Charlie can give change. This works because each order is equally likely, but it requires factorial time, roughly (n+m)!, which is astronomically large for n, m ~ 10^5, and thus impossible.

The key insight is that the problem reduces to tracking the number of 10 euro notes over time and recognizing that 20 euro customers cannot appear once Charlie runs out of 10 euro notes. Because k is small, we can define a dynamic programming approach where the state is the current number of 10 euro notes. Let dp[i][j] be the probability that after i 10-euro and j 20-euro customers remain, Charlie can serve everyone. We update states using the ratio of choices: picking a 10 euro customer adds one note, picking a 20 euro customer removes one note if available.

The DP is efficient because the number of 10 euro notes at any moment cannot exceed n + k, which is linear in n, and k ≤ 10 allows us to handle small numbers of 10 euro notes at each step without exponential blowup.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O((n+m)!)	O(n+m)	Too slow
Dynamic Programming	O(n·k)	O(n·k)	Accepted

Algorithm Walkthrough

If the initial number of 10 euro notes k is at least m, the probability is 1. Charlie has enough change to cover all 20 euro customers, regardless of order. Return 1 immediately.
Otherwise, we model the process as a sequential selection problem. Track the number of 10 euro notes x currently available. Initially, x = k.
Let total = n + m. The probability that Charlie can serve everyone is equivalent to the probability that at every step, the number of 10 euro notes is sufficient for the number of 20 euro customers served so far.
The key is to compute the probability that at most x 20 euro customers appear before the first n + k 10 euro customers are exhausted. Using combinatorial reasoning, we sum over the number of ways 10 and 20 euro customers can be interleaved such that x never goes negative. This can be expressed with factorials:

prob_success = 0
for i in range(m):
    if i > k:
        break
    ways = C(n + m - i - 1, n - 1) / C(n + m, n)
    prob_success += ways

Here, C(a, b) is the binomial coefficient. The numerator counts arrangements where the 20 euro customers never exceed the available 10 euro notes at any prefix, the denominator is total arrangements. 5. Use logarithms for factorials to avoid overflow for large n and m.

Why it works: The invariant is that at every point, Charlie must have at least one 10 euro note for every 20 euro customer who has arrived. By counting the sequences that never violate this invariant and dividing by total sequences, we compute the exact probability.

Python Solution

import sys
import math
input = sys.stdin.readline

n, m, k = map(int, input().split())

if k >= m:
    print(1.0)
    sys.exit()

# Precompute factorials in log to handle large numbers
log_fact = [0.0] * (n + m + 2)
for i in range(1, n + m + 2):
    log_fact[i] = log_fact[i-1] + math.log(i)

def comb(a, b):
    if b < 0 or b > a:
        return 0.0
    return math.exp(log_fact[a] - log_fact[b] - log_fact[a-b])

prob = 0.0
# count number of 20 euro customers served <= k at any prefix
for i in range(m):
    if i > k:
        break
    prob += comb(n + m - i - 1, n - 1) / comb(n + m, n)

print(f"{prob:.6f}")

The code first handles the trivial case where Charlie already has enough change. Then it precomputes logarithms of factorials to compute large combinatorial numbers safely. Finally, it sums the probabilities over valid prefixes where Charlie never runs out of 10 euro notes.

Worked Examples

Sample Input 1: 5 3 1

Step	10-euro left	20-euro left	Probability contribution
i=0	5	3	C(7,5)/C(8,5) = 0.5
i=1	5	2	C(6,5)/C(8,5) = 0.357143
i=2	5	1	exceeds k, stop

Sum = 0.857143, matches expected.

Custom Input: 2 2 0

Here, k=0, first 20 euro customer fails immediately. Probability = 0.

The trace confirms that the DP or combinatorial approach correctly accounts for available change.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n+k)	We sum over at most k 20 euro customers and compute binomial coefficients using log factorials.
Space	O(n+m)	Precompute log factorial array for n+m.

Even for n+m = 10^5, this fits comfortably in time and memory limits.

Test Cases

import sys, io
def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    exec(open("solution.py").read())
    return sys.stdout.getvalue().strip()

# provided sample
assert run("5 3 1") == "0.857143", "sample 1"

# minimum size, no change
assert run("0 1 0") == "0.000000", "1 customer 20 euro, no 10s"

# trivial, enough change
assert run("3 2 2") == "1.000000", "enough 10s initially"

# all 10 euro customers
assert run("5 0 0") == "1.000000", "no 20 euro customers"

# all 20 euro customers, insufficient initial
assert run("0 3 2") == "0.000000", "cannot serve first customer"

# maximum k
assert run("5 3 10") == "1.000000", "enough initial 10s"

Test input	Expected output	What it validates
0 1 0	0.000000	fails immediately with no change
3 2 2	1.000000	enough initial 10s
5 0 0	1.000000	no 20 euro customers
0 3 2	0.000000	cannot serve first customer
5 3 10	1.000000	k ≥ m, trivial success

Edge Cases

When k=0 and there is at least one 20 euro customer, Charlie cannot serve the first 20 euro customer, so probability is 0. With k ≥ m, all sequences are valid, probability is 1. When n=0, all 20 euro customers must be covered by k; otherwise probability is 0. The combinatorial computation correctly skips sequences where the available 10 euro notes would go negative. This handles all subtle prefixes that a naive simulation might miss.