CF 105494A - Problem Statement

Rating: -
Tags: -
Solve time: 52s
Verified: yes

Solution

Problem Understanding

We are given a very simple process that starts from a single object, called a part. Every time we press Enter, the number of parts increases by exactly one. After performing some number of presses, the system ends with exactly $n$ parts.

The question is essentially asking how many times Enter was pressed to reach from the initial state to the final state.

We start at 1 part and end at $n$ parts, so each operation increases the count by 1. The task is to determine the number of operations required to reach the final count.

The input can be interpreted as the final number of parts $n$, and the output is the number of key presses needed to reach it.

Since each operation contributes exactly one increment, the relationship between initial and final state is linear and deterministic.

There are no hidden dependencies or branching states. The process is strictly monotonic, so the only possible ambiguity would come from incorrect handling of boundary cases such as $n = 1$.

A naive interpretation mistake would be to output $n$ instead of $n - 1$, treating the final state as if it includes an extra implicit operation. For example, if $n = 1$, no presses are made, so the correct answer must be 0. Any solution that forgets the initial state would incorrectly output 1.

Another possible mistake is assuming that pressing Enter produces the initial part as well as increments it, which would incorrectly shift all results by one.

Approaches

The brute-force interpretation would simulate the process directly. We start with a counter at 1 and repeatedly increment it until it reaches $n$, counting how many increments we performed. This is correct because it exactly follows the described process.

However, this simulation performs one step per increment, so in the worst case it executes $n - 1$ operations. If $n$ is large, for example $10^9$, this becomes infeasible due to linear runtime.

The key observation is that the process does not branch or depend on intermediate structure. Every operation has identical effect: increase by one. This means the final state is determined purely by the difference between start and end.

So instead of simulating transitions, we directly compute how many increments are needed to go from 1 to $n$, which is simply $n - 1$.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Simulation	$O(n)$	$O(1)$	Too slow
Direct Formula	$O(1)$	$O(1)$	Accepted

Algorithm Walkthrough

1. Read the final number of parts $n$

We first take the input value, which represents the final state of the system after all operations have been applied.

2. Recognize the initial state

The process always starts from exactly one part. This starting point is fixed and does not depend on input.

3. Compute the difference

Since each operation increases the count by exactly one, the number of operations required is the difference between the final and initial values, which is $n - 1$.

4. Output the result

We print the computed difference as the answer.

Why it works

The system evolves in a strictly linear way: each operation increases the state by exactly one, and no operation ever decreases or modifies the increment size. This creates a one-to-one mapping between operations and increments in the part count. Since the initial state is fixed at 1, reaching $n$ requires exactly $n - 1$ increments. No alternative sequence of operations exists, so this count is both necessary and sufficient.

Python Solution

import sys
input = sys.stdin.readline

def main():
    n = int(input().strip())
    print(n - 1)

if __name__ == "__main__":
    main()

The solution reads a single integer and outputs the difference from the fixed starting value. The subtraction directly encodes the number of increments needed to transform 1 into $n$.

The only subtle point is ensuring correct handling of $n = 1$. In that case, the result correctly becomes 0, matching the fact that no operations are needed.

Worked Examples

Example 1

Input:

We track the computation:

Step	n	Computation	Output
1	5	5 - 1	4

This demonstrates a standard case where multiple increments are required.

The output 4 corresponds to the exact number of transitions needed to go from 1 to 5.

Example 2

Input:

Step	n	Computation	Output
1	1	1 - 1	0

This confirms the boundary case where the system already starts at the final state.

No operations are required, and the formula naturally handles this without special casing.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(1)$	Only one subtraction and one output operation are performed
Space	$O(1)$	No additional data structures are used

The solution easily fits within any reasonable constraints since it performs constant-time arithmetic regardless of input size.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import builtins
    input_backup = builtins.input
    builtins.input = lambda: sys.stdin.readline()
    from contextlib import redirect_stdout
    out = io.StringIO()
    import sys as _sys

    def solve():
        n = int(_sys.stdin.readline().strip())
        print(n - 1)

    with redirect_stdout(out):
        solve()

    builtins.input = input_backup
    return out.getvalue().strip()

# provided samples
assert run("5\n") == "4"
assert run("1\n") == "0"

# custom cases
assert run("2\n") == "1", "minimum increment case"
assert run("10\n") == "9", "small linear growth"
assert run("1000000\n") == "999999", "large input stress case"
assert run("3\n") == "2", "simple mid case"

Test input	Expected output	What it validates
1	0	minimal boundary case
2	1	smallest non-trivial increment
10	9	general correctness
1000000	999999	large value handling

Edge Cases

Case: $n = 1$

Input:

The algorithm computes $1 - 1 = 0$. This reflects that the system starts already at the required state. No increments occur, so the output is correct.

Case: smallest increment

Input:

The computation is $2 - 1 = 1$. Starting from 1 part, exactly one operation is needed to reach 2 parts. The subtraction directly matches the single transition.

Case: large input

Input:

1000000000

The algorithm performs a single arithmetic operation, yielding $999999999$. Since no iteration is involved, performance remains constant regardless of magnitude, and correctness follows from the same invariant that each operation corresponds to one unit increase.