CF 102556D - Riana and Distribution of Pie

CF 102556D - Riana and Distribution of Pie

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Solve time: 2m 30s
Verified: yes

Solution

Problem Understanding

Let the chosen percentage of the i-th person be P_i, written as a fraction between 0 and 1.

When person i takes a turn, two things happen. They take P_i of the untouched pie, then they also take P_i of every slice already owned by previous people. After every person has played, the untouched part of the pie must be zero, and we want the final slice sizes to be as equal as possible.

The input contains only the number of people. The output is the percentage chosen by each person, in order.

The only constraint is N ≤ 1000, so even an O(N^2) simulation would easily fit. The challenge is not efficiency, it is discovering the mathematical structure of the process.

The first non-obvious edge case is N = 1. The only valid answer is 100%, because otherwise some pie remains uneaten.

The second subtle case is that making the last person choose 100% is not required. For example, with N = 2, choosing 100%, 50% leaves no untouched pie because the first player already consumed all of it. A greedy strategy that always forces the last player to take everything misses the optimal distribution.

The final trap is confusing the percentage a player chooses with the percentage they finally own. Later players repeatedly steal from earlier ones, so the chosen percentages and final shares are different quantities.

Approaches

A direct simulation keeps every player's current slice. Whenever a new player arrives, every previous slice is multiplied by (1 - P_i), the new player receives everything removed from those slices plus P_i of the untouched pie, and the untouched pie is also multiplied by (1 - P_i). This faithfully models the game, but it does not tell us how to choose the percentages.

The key observation is that a player's gain from untouched pie and from stealing always adds up to the same value.

Suppose the untouched fraction before person i is R. The total amount already owned by previous players is 1 - R. The new player receives

P_i · R + P_i · (1 - R) = P_i.

So immediately after their own turn, player i always owns exactly P_i of the original pie.

Afterward, every later player simply scales that slice by (1 - P_j). Hence the final share is

$$F_i=P_i\prod_{j=i+1}^{N}(1-P_j).$$

Now define

$$S_i=\prod_{j=i}^{N}(1-P_j),$$

the untouched fraction just before person i.

Then

$$F_i=S_{i+1}-S_i.$$

The leftover pie is S_1, and the requirement of no leftovers means S_1=0. Also S_{N+1}=1.

Since the final shares sum to 1, the smallest possible difference between the largest and smallest share is achieved when every final share is exactly 1/N.

Setting every F_i=1/N gives

$$S_i=\frac{i-1}{N}.$$

Recovering the percentages,

$$P_i=\frac{S_{i+1}-S_i}{S_{i+1}}.$$

For i=1,

$$P_1=1.$$

For every i≥2,

$$P_i=\frac{1/N}{i/N}=\frac1i.$$

So the answer is remarkably simple.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force simulation and search	Exponential	O(N)	Too slow
Closed-form formula	O(N)	O(1)	Accepted

Algorithm Walkthrough

1. Read N.
2. Print 100.0 for the first person. This guarantees the untouched pie immediately becomes zero, satisfying the no-leftovers condition.
3. For every person i from 2 to N, print 100 / i. This is exactly the value obtained from the derived formula P_i = 1/i.

The derivation guarantees that every player's final share becomes exactly 1/N, which is the best possible because all shares sum to one.

Why it works

The invariant is that player i owns exactly P_i immediately after finishing their own turn. Every later turn simply multiplies every existing slice by the same factor. This gives the formula

$$F_i=P_i\prod_{j>i}(1-P_j).$$

Writing the suffix products as S_i converts the expression into

$$F_i=S_{i+1}-S_i.$$

Equal final shares uniquely determine every S_i, and those suffix products uniquely determine every percentage. Since equal shares make the maximum and minimum identical, no better distribution exists.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())

    print(f"{100.0:.10f}")
    for i in range(2, n + 1):
        print(f"{100.0 / i:.10f}")

if __name__ == "__main__":
    solve()

The program directly prints the closed-form answer. The first value is always 100%. Every later value is simply 100/i. No simulation is required because the mathematical derivation already characterizes the unique optimal solution.

Floating point precision is more than sufficient here. The required error is 1e-4, while printing ten decimal places keeps the accumulated error many orders of magnitude smaller.

Worked Examples

Example 1

Input:

1

Person	Printed percentage
1	100.0000000000

The only player takes the whole pie, so the final distribution is already perfectly equal.

Example 2

Input:

4

Person	Percentage printed
1	100.0000000000
2	50.0000000000
3	33.3333333333
4	25.0000000000

The final shares become

Person	Final share
1	25%
2	25%
3	25%
4	25%

This confirms that every player finishes with exactly one quarter of the pie.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(N)	One value is printed for each person.
Space	O(1)	Only a loop variable is stored.

With at most 1000 people, this runs comfortably within the limits.

Test Cases

import io
import sys

def solve():
    input = sys.stdin.readline
    n = int(input())
    print(f"{100.0:.10f}")
    for i in range(2, n + 1):
        print(f"{100.0 / i:.10f}")

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

assert run("1\n") == "100.0000000000\n"

assert run("2\n") == (
    "100.0000000000\n"
    "50.0000000000\n"
)

assert run("3\n") == (
    "100.0000000000\n"
    "50.0000000000\n"
    "33.3333333333\n"
)

out = run("1000\n").splitlines()
assert len(out) == 1000
assert abs(float(out[-1]) - 0.1) < 1e-9

Test input	Expected output	What it validates
1	100	Minimum input size
2	100, 50	First non-trivial case
3	100, 50, 33.333...	General formula
1000	Last value is 0.1	Largest input size

Edge Cases

When N = 1, the algorithm prints only 100. The untouched pie immediately becomes zero, and the only player owns the entire pie. Both requirements are satisfied.

When N = 2, the algorithm prints 100 and 50. The first player initially owns the whole pie. The second player steals half of it, leaving both players with exactly 50%. This demonstrates that the first player, not the last one, is the person who must choose 100%.

For larger values such as N = 3, the algorithm prints 100, 50, and 33.333.... The resulting final shares are all exactly 1/3, confirming that the construction equalizes every player's final ownership while still leaving no untouched pie.