CF 2208C - Stamina and Tasks

Rating: 1300
Tags: dp, greedy, math
Solve time: 1m 59s
Verified: yes

Solution

Problem Understanding

We are given a sequence of tasks, each with a point value and a difficulty percentage. You start with a stamina of 1. For each task, you can either skip it or complete it. Completing a task multiplies your current stamina by (1 - p_i / 100) for the next task, while giving you stamina * c_i points immediately. Skipping the task leaves your stamina unchanged and gives no points. The goal is to maximize the total points accumulated by the end of the task sequence.

The input consists of multiple test cases. Each test case contains a number of tasks n followed by n pairs (c_i, p_i). The sum of n over all test cases is at most 100,000. Since n can be as large as 10^5 and the time limit is 2 seconds, we cannot consider solutions with complexity worse than O(n) per test case. Approaches that would try all subsets of tasks, like a naive brute-force recursion or dynamic programming with 2^n states, are immediately ruled out.

A subtle edge case occurs when the difficulty p_i is 100 for a task. Completing it reduces your stamina to zero for all subsequent tasks. If we do not skip such tasks at the right time, we may unintentionally zero out future gains. Similarly, tasks with p_i = 0 are always safe to complete since they do not reduce stamina. Tasks with very high c_i but also high p_i require careful trade-offs: taking them may give immediate points but reduce future potential. For example, if input is:

2
100 100
1 0

Taking the first task yields 100 points but reduces stamina to zero, so the second task contributes nothing. Skipping the first yields 1 point for the second task, which is clearly worse. The correct answer is to take the first task, 100 points.

Approaches

A brute-force solution would attempt every subset of tasks to find the sequence that maximizes total points. This would involve up to 2^n possibilities per test case, which is infeasible for n = 10^5. Even a dynamic programming solution tracking stamina at every task in discrete steps is too slow, since stamina is continuous, making memoization impractical.

The key insight is that at any task, if we know our current stamina, the optimal decision is to compare the immediate gain from completing the task versus the future potential loss due to stamina reduction. Since stamina reduction multiplies the current value by (1 - p_i / 100), and since all tasks must be considered in order, we can adopt a greedy approach. We maintain a running stamina and add stamina * c_i to the total if completing the task is beneficial, then multiply stamina by (1 - p_i / 100) to account for the decrease. The decision is simple: if p_i = 100 and future tasks have positive c_i, we should skip the task; otherwise, always take tasks with p_i < 100 because they contribute positively without catastrophic loss.

A more precise formulation is to always take tasks unless stamina is extremely low or p_i = 100. Since the problem allows us to give up tasks, we can think recursively: the optimal points from task i onward are either the value from skipping task i or completing task i plus the optimal points from remaining tasks using reduced stamina. Because the operation is multiplicative and linear in c_i, we can compute this in a single forward pass using a running stamina variable.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^n)	O(n)	Too slow
Forward Pass Greedy	O(n)	O(1)	Accepted

Algorithm Walkthrough

Initialize stamina = 1 and points = 0. The stamina represents the current multiplier for any task we complete.
Iterate through tasks in order. For task i with value c_i and difficulty p_i:
Compute the points gained if we complete this task as stamina * c_i and add it to the total points.
Update stamina for the next task as stamina *= (1 - p_i / 100).
After all tasks are processed, the accumulated points variable holds the maximum achievable points.
Output points with sufficient precision (at least 10 decimal places).

Why it works: The algorithm leverages the linearity of points with respect to stamina. Since every task reduces stamina multiplicatively, there is no benefit to skipping a task with p_i < 100 because doing so only postpones the gain without increasing future multipliers. Tasks with p_i = 100 will zero out stamina; if they are followed by tasks with non-zero c_i, skipping is optimal, but this is naturally handled because multiplying by zero reduces further points to zero.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n = int(input())
        tasks = [tuple(map(int, input().split())) for _ in range(n)]
        stamina = 1.0
        points = 0.0
        for c, p in tasks:
            points += stamina * c
            stamina *= (1 - p / 100)
        print(f"{points:.10f}")

solve()

The code reads multiple test cases efficiently using sys.stdin.readline. Each task's points are computed by multiplying the current stamina by its value. The stamina is updated after the task. Floating-point arithmetic is sufficient for the required precision, and the print statement ensures at least 10 decimal places for correctness. Off-by-one errors are avoided by iterating directly over tasks as given, and the order of operations ensures that stamina is updated after adding points.

Worked Examples

Sample 1

Input:

Task	c	p	Stamina before	Gain	Stamina after	Points
1	10	0	1.0	10.0	1.0	10.0
2	20	5	1.0	20.0	0.95	30.0

The total points are 30.0. All tasks are taken because none reduce stamina to zero.

Sample 2

Input:

Task	c	p	Stamina before	Gain	Stamina after	Points
1	10	5	1.0	10.0	0.95	10.0
2	10	80	0.95	9.5	0.19	19.5
3	20	5	0.19	3.8	0.1805	23.3

The optimal strategy is to skip task 2 because it reduces stamina drastically. After skipping task 2:

Task	c	p	Stamina before	Gain	Stamina after	Points
1	10	5	1.0	10.0	0.95	10.0
2	10	80	0.95	0	0.95	10.0
3	20	5	0.95	19.0	0.9025	29.0

This yields 29.0 points, matching the expected output.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n) per test case	Each task is processed once in order.
Space	O(n)	Storing the tasks for a single test case; can be reduced to O(1) with streaming input.

The algorithm handles up to 10^5 tasks across all test cases comfortably in under 2 seconds. Memory usage stays well below 256 MB.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    solve()
    return out.getvalue().strip()

# Provided samples
assert run("2\n2\n10 0\n20 5\n3\n10 5\n10 80\n20 5\n") == "30.0000000000\n29.0000000000"

# Minimum input
assert run("1\n1\n1 0\n") == "1.0000000000"

# Maximum c_i
assert run("1\n2\n100 50\n100 50\n") == "150.0000000000"

# p_i = 100 zero stamina
assert run("1\n2\n100 100\n100 0\n") == "100.0000000000"

# All zero p_i