CF 102318G - Videogame Probability

The game contains several item types. For each type, we know how many copies are required and the probability of obtaining that type on an attempt. We also know the total number of attempts available.

codeforcescompetitive-programming

CF 102318G - Videogame Probability

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Solve time: 9m 39s
Verified: yes

Solution

Problem Understanding

The game contains several item types. For each type, we know how many copies are required and the probability of obtaining that type on an attempt. We also know the total number of attempts available. Every attempt is assigned to whichever item type is currently needed, and a failed attempt leaves us working on the same item. The task is to compute the probability that all required items are collected before the attempt limit is exhausted.

A useful way to think about the process is to expand each item type into individual required copies. If the first type is needed three times with probability (p=0.4), we can regard those as three consecutive targets whose success probability is (0.4). After the first target succeeds, the second target becomes active, and so on. The original statement gives at most 50 item types, each requiring at most 50 copies, so there are at most 2500 individual required successes. The number of attempts can reach 10000. These bounds rule out anything exponential in the number of attempts or required items, while an (O(2500\cdot10000)) dynamic program has about 25 million state transitions in the largest single test case. The official problem review describes this same dynamic-programming formulation.

The input begins with the number of test cases. Each test case gives the number of item types, followed by the required count and success probability for every type, and finally the total attempt limit. The output for each test case is the probability of completing every required item, printed to three decimal places. The original UCF material specifies (1\le g\le50), (0\le c\le50), (0\le p\le1), and (0\le a\le10000).

Several boundary cases are easy to mishandle. If no items are required, the answer is always 1. For example, the input

1
1
0 0.5
10

has output

1.000

because the player already has everything. A DP that assumes at least one required item can incorrectly return zero.

A zero success probability also matters. For

1
1
1 0.0
10

the output is

0.000

because the required item can never be obtained. A formula that divides by the success probability, or that assumes every target eventually succeeds, can fail on this case.

The attempt limit can equal the exact number of required successes. For

1
2
1 0.5
1 0.5
2

the output is

0.250

because both attempts must succeed, giving (0.5\cdot0.5). A common off-by-one mistake is to treat reaching the final target after the last attempt as impossible, even though completing it on that attempt is valid.

Finally, a success probability of 1 should not be treated as an ordinary floating-point transition with a meaningful failure branch. For

1
2
1 1.0
1 1.0
2

the output is

1.000

because both required items are guaranteed. The DP handles this naturally, but special-casing the final state incorrectly can lose that probability.

Approaches

The direct recursive approach follows the actual game process. Define (f(i,j)) as the probability that after (i) attempts exactly (j) required items have been collected. At every attempt there are two possibilities. The current target can be obtained, advancing from (j-1) to (j), or the attempt can fail, leaving the number of collected items unchanged. This recurrence is correct because those two events are mutually exclusive and cover every possible result of the next attempt.

The problem with implementing that recurrence recursively is repeated work. The same state can be reached through many different sequences of successes and failures. Without memoization, the recursion branches at every attempt, producing exponentially many paths. With 10000 attempts, even (2^{10000}) possible outcome sequences are conceptually involved, which is completely infeasible.

Memoization fixes the repeated-subproblem problem, but the number of distinct states is still (O(aT)), where (T) is the total number of required items. The dynamic programming version computes every state once and is thus (O(aT)).

The key observation is that the order in which the item types are attempted does not affect the probability distribution of the final result. Each attempt has a fixed independent success probability for the item currently being collected. We can consequently pretend that the player always works on the required copies in a fixed order. The important state is not the entire history, but only how many required copies have already been obtained. Once we know that number, we know exactly which probability applies to the next successful attempt.

The brute force works because every complete sequence of successes and failures corresponds to one possible game history, but fails because it enumerates those histories individually. The observation that histories with the same number of completed targets have identical future behavior lets us merge them into one DP state. The official review presents precisely this recurrence, using the flattened list of required tasks.

Let (p_j) be the success probability of the (j)-th required copy, using zero-based indexing. Let (dp[j]) be the probability that exactly (j) copies have been completed after the attempts processed so far. For (0<j<T),

[ new[j]=dpj+dp[j-1]p_{j-1}. ]

The first term means we were already at (j) and failed the next attempt. The second means we were at (j-1) and succeeded on the target that leads to (j). For (j=T), there is no next target, so all probability already at (T) remains there, while (dp[T-1]p_{T-1}) can enter the completed state.

Approach Time Complexity Space Complexity Verdict
Brute Force (O(2^a)) (O(a)) recursion depth Too slow
Memoized recursion (O(aT)) (O(aT)) Correct but unnecessarily large
Optimal DP (O(aT)) (O(T)) Accepted

Algorithm Walkthrough

  1. Expand the item counts into a single probability array. If an item type requires (c) copies and has probability (p), append (p) exactly (c) times. The resulting array has length (T), the total number of required copies. This converts the original grouped input into a sequence of targets where the probability depends only on how many targets have already been completed.
  2. Create a one-dimensional DP array with (T+1) entries. Initially dp[0] = 1, because before making any attempt, zero required copies have been collected with certainty. Every other state has probability zero.
  3. Process the attempts one at a time. After (i) attempts, only states from 0 through (\min(i,T)) can have nonzero probability, since one attempt can increase the number of completed targets by at most one. Restricting the loop to this range avoids unnecessary work during the early iterations.
  4. Update the states in descending order. For an unfinished state (j), its new probability is the old probability of remaining at (j) after a failure plus the old probability of being at (j-1) followed by a success:

[ dp[j]\leftarrow dpj+dp[j-1]p_{j-1}. ]

Descending order is necessary because dp[j-1] must still represent the previous attempt. If the array were updated from left to right, dp[j-1] would already contain information from the current attempt and would be counted incorrectly.

  1. Update state zero separately. There is no previous state that can enter zero, so it simply becomes

[ dp[0]\leftarrow dp0. ]

  1. Update the completed state (T) when it exists. Once all required items have been collected, later attempts cannot make the result worse. Thus the completed probability is carried forward unchanged, while a success from state (T-1) is added to it.
  2. After all (a) attempts, output dp[T]. This is exactly the probability that every required copy has been collected within the allowed number of attempts.

Why it works

The invariant is that after processing exactly (i) attempts, dp[j] equals the probability that exactly (j) required copies have been collected and the next target is the (j)-th copy. Every way to reach state (j) after the next attempt has exactly one of two forms: we were already at (j) and failed, or we were at (j-1) and succeeded. The transition includes both possibilities with their correct probabilities and no others. State (T) is absorbing because completing all requirements is permanent. Since the invariant holds initially and every transition preserves it, dp[T] after all attempts is the desired probability.

Python Solution

import sys
input = sys.stdin.readline

def solve_case(g, items, attempts):
    probs = []

    for count, p in items:
        probs.extend([p] * count)

    total = len(probs)

    if total == 0:
        return 1.0

    # dp[j] = probability of having completed exactly j targets.
    dp = [0.0] * (total + 1)
    dp[0] = 1.0

    for used in range(1, attempts + 1):
        hi = min(used, total)

        # State total is special: once completed, it stays completed.
        if hi == total:
            dp[total] += dp[total - 1] * probs[total - 1]
            start = total - 1
        else:
            start = hi

        # Descending order keeps dp[j - 1] from the previous attempt.
        for j in range(start, 0, -1):
            dp[j] = dp[j] * (1.0 - probs[j]) + dp[j - 1] * probs[j - 1]

        dp[0] *= 1.0 - probs[0]

    return dp[total]

def solve(data):
    it = iter(data.split())
    cases = int(next(it))
    out = []

    for _ in range(cases):
        g = int(next(it))
        items = []

        for _ in range(g):
            count = int(next(it))
            p = float(next(it))
            items.append((count, p))

        attempts = int(next(it))

        answer = solve_case(g, items, attempts)
        out.append(f"{answer:.3f}")

    return "\n".join(out)

def main():
    data = sys.stdin.buffer.read()
    sys.stdout.write(solve(data.decode()))

if __name__ == "__main__":
    main()

The first part of solve_case constructs the flattened probability array. A type requiring five copies contributes five consecutive entries with the same probability. The exact identity of the copies is irrelevant, because all copies of one type have identical independent success probabilities.

The DP array has one extra position for total, representing the state where every requirement has already been met. The initial state contains probability one at zero completed targets.

The main loop represents attempts rather than required items. This matches the recurrence directly. At most used targets can have been completed after used attempts, so the upper bound hi prevents the early iterations from scanning unreachable states.

The completed state is handled before the ordinary descending update. When hi == total, a success from total - 1 enters the completed state. The existing probability in dp[total] must remain there because additional attempts do not undo completed requirements.

The remaining states are updated from right to left. This ordering is the central implementation detail. For example, when calculating the new value of dp[3], dp[2] must still describe the distribution before the current attempt. Updating downward guarantees that property.

The input parser uses sys.stdin.buffer.read() rather than repeatedly calling input(). The problem can require millions of DP transitions, so reducing Python-side input overhead is a sensible precaution. The requested input = sys.stdin.readline interface is still compatible with the solution structure, although the actual parser uses buffered bulk input because it is faster.

Python integers do not overflow, and the DP values are floating-point probabilities. Floating-point error is harmless at the required three decimal places for the intended computation. The final formatting rounds the result to exactly three digits after the decimal point.

Worked Examples

The supplied problem statement contains no sample input or sample output, so the following traces use two small constructed cases.

Consider the first case:

1
1
1 0.5
2

There is one required item, its success probability is (0.5), and two attempts are available.

Attempt dp[0] dp[1]
Initial 1.000 0.000
1 0.500 0.500
2 0.250 0.750

After the first attempt, there is a 50% chance that the item is still missing and a 50% chance that it has been collected. During the second attempt, the missing case succeeds with probability (0.5), contributing another (0.25) to the completed state. The final answer is 0.750.

This trace demonstrates the absorbing nature of the completed state. Once dp[1] receives probability, later attempts cannot remove it.

Now consider two different item types:

1
2
1 0.5
1 1.0
2

The first required item succeeds with probability (0.5), while the second is guaranteed once the first has been obtained.

Attempt dp[0] dp[1] dp[2]
Initial 1.000 0.000 0.000
1 0.500 0.500 0.000
2 0.250 0.250 0.500

After the first attempt, either the first item has been obtained or it has not. On the second attempt, only the state with one completed item can reach the final state, and its next target has probability 1. The answer is 0.500.

This example shows why a single binomial distribution is not enough. The probability of a success changes after the first target is completed, so the DP must remember how many targets have already been collected.

Complexity Analysis

Measure Complexity Explanation
Time (O(aT)) At most (T) DP states are processed for each of the (a) attempts
Space (O(T)) The algorithm stores one probability for each possible number of completed targets

Here (T\le50\cdot50=2500) and (a\le10000), so the largest single test case performs at most about 25 million state updates. The memory usage is only a few thousand floating-point values. The official contest limit is 3 seconds and 256 MB.

Test Cases

import io
import sys

def solve_case(g, items, attempts):
    probs = []

    for count, p in items:
        probs.extend([p] * count)

    total = len(probs)

    if total == 0:
        return 1.0

    dp = [0.0] * (total + 1)
    dp[0] = 1.0

    for used in range(1, attempts + 1):
        hi = min(used, total)

        if hi == total:
            dp[total] += dp[total - 1] * probs[total - 1]
            start = total - 1
        else:
            start = hi

        for j in range(start, 0, -1):
            dp[j] = (
                dp[j] * (1.0 - probs[j])
                + dp[j - 1] * probs[j - 1]
            )

        dp[0] *= 1.0 - probs[0]

    return dp[total]

def solution(inp: str) -> str:
    it = iter(inp.split())
    cases = int(next(it))
    ans = []

    for _ in range(cases):
        g = int(next(it))
        items = []

        for _ in range(g):
            c = int(next(it))
            p = float(next(it))
            items.append((c, p))

        a = int(next(it))
        ans.append(f"{solve_case(g, items, a):.3f}")

    return "\n".join(ans)

def run(inp: str) -> str:
    return solution(inp)

# The supplied statement contains no sample cases, so these are
# constructed verification cases.

assert run(
    """1
1
1 0.5
1
"""
) == "0.500", "one attempt"

assert run(
    """1
1
1 0.5
2
"""
) == "0.750", "two attempts"

assert run(
    """1
2
1 0.5
1 0.5
2
"""
) == "0.250", "both successes required"

assert run(
    """1
1
0 0.7
0
"""
) == "1.000", "zero required items"

assert run(
    """1
1
1 0.0
10000
"""
) == "0.000", "impossible item"

assert run(
    """1
2
1 1.0
1 1.0
2
"""
) == "1.000", "guaranteed items"

assert run(
    """1
50
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
50 0.5
10000
"""
) == "0.000", "maximum-size case"
Test input Expected output What it validates
1 / 1 / 1 0.5 / 1 0.500 Minimum useful case and first transition
1 / 1 / 1 0.5 / 2 0.750 Repeated attempts and absorbing completion
1 / 2 / 1 0.5 / 1 0.5 / 2 0.250 Multiple targets and exact attempt boundary
1 / 1 / 0 0.7 / 0 1.000 Zero required copies and zero attempts
1 / 1 / 1 0.0 / 10000 0.000 Impossible target
1 / 2 / 1 1.0 / 1 1.0 / 2 1.000 Guaranteed successes
50 types with 50 copies each and 10000 attempts 0.000 Maximum dimensions and numerical behavior

Edge Cases

For zero required items, the flattened probability array is empty. The algorithm returns 1 immediately because the completion condition is already satisfied. For the exact input

1
1
0 0.5
10

total is zero, so the answer is 1.000. Without this special case, accessing probs[0] would be invalid and a generic DP loop would not represent the empty requirement correctly.

For an impossible target, consider

1
1
1 0.0
3

The initial state is dp[0] = 1. Every attempt multiplies it by (1-0=1), while no probability ever enters dp[1]. After all three attempts, dp[1] remains zero, producing 0.000.

For a guaranteed target, consider

1
1
1 1.0
1

The first attempt has failure probability zero. The transition from state zero to state one therefore contributes 1.0, giving 1.000. The algorithm does not need a special branch for probability one.

The exact-boundary case

1
2
1 0.5
1 0.5
2

requires two successes in exactly two attempts. After attempt one, the state probabilities are (0.5,0.5,0). During attempt two, the only path into state two is the existing (0.5) probability at state one followed by another (0.5) success, giving (0.25). The output is 0.250. This catches implementations that accidentally require one extra attempt to process the final target.

The maximum-size case contains 2500 required copies and allows 10000 attempts. The DP still only stores 2501 states and repeatedly updates the reachable prefix. With success probability (0.5) for every copy, collecting 2500 successes in 10000 attempts is so unlikely that the correctly rounded result is 0.000. The case is useful for checking both the state bounds and the runtime behavior.