CF 104619J - Java Warriors

Rating: -
Tags: -
Solve time: 41s
Verified: yes

Solution

Problem Understanding

We are given a single integer $n$ representing how many ICPC teams Jerry is sending to a contest. Each team requires a fixed registration fee of 4000 dollars. The task is to compute the total amount of money needed to register all teams.

The input does not involve any structure beyond this single count, and the output is just the total cost after scaling by the per-team fee.

The constraint $n \le 20$ means the input size is extremely small. This immediately rules out any concern about efficiency, overflow in typical integer ranges is also not a problem in Python, and even in fixed-width integer languages the maximum value $20 \cdot 4000 = 80000$ is trivial.

The main edge case to be aware of is the smallest possible input. Since $n$ is a positive integer, the smallest case is $n = 1$, which should produce 4000. Any off-by-one thinking such as treating $n$ as zero-based or mistakenly adding an extra constant would immediately show up on such a case.

Approaches

The brute-force interpretation would literally simulate the registration process team by team, adding 4000 for each team. This would look like initializing a total to zero and looping $n$ times, incrementing by 4000 each time. This is correct because each team contributes an independent fixed cost, so accumulation matches multiplication.

This approach runs in $O(n)$, which is already negligible for $n \le 20$. However, the structure of the problem makes the repeated addition unnecessary. Since every term in the sum is identical, the entire computation collapses into a direct multiplication $4000 \cdot n$. The observation is that we are summing a constant sequence, not processing varying inputs, so we can replace iteration with a single arithmetic operation.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (loop accumulation)	$O(n)$	$O(1)$	Accepted
Optimal (direct multiplication)	$O(1)$	$O(1)$	Accepted

Algorithm Walkthrough

We compute the total cost in the most direct way possible.

Read the integer $n$ from input. This represents the number of teams being registered.
Compute the total cost as $n \times 4000$. This step directly applies the definition of cost per team without simulation.
Output the computed value as the final answer.

Why it works

Each team contributes exactly 4000 independently of all others. The total cost is therefore the sum of $n$ identical terms, which is mathematically equivalent to multiplication. Since there are no conditional rules, discounts, or interactions between teams, no additional logic is required beyond scaling the count.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input().strip())
    print(n * 4000)

if __name__ == "__main__":
    solve()

The solution reads a single integer and immediately multiplies it by 4000. The use of strip() ensures that any trailing newline does not interfere with integer parsing. There are no loops, no conditionals, and no risk of boundary issues because the computation is a single arithmetic expression.

Worked Examples

Since the statement provides no concrete numeric samples, we construct representative cases.

Example 1

Input:

Step	n	Computation	Total
Read input	1	-	0
Multiply	1	1 × 4000	4000

Output:

This confirms that a single team produces exactly one fee unit.

Example 2

Input:

Step	n	Computation	Total
Read input	5	-	0
Multiply	5	5 × 4000	20000

Output:

This demonstrates linear scaling: each additional team adds another 4000 to the total.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(1)$	Only one multiplication and one input read are performed
Space	$O(1)$	No additional data structures are used

The constraints allow far more complex solutions than necessary, but the constant-time computation is the most direct fit and easily satisfies any time limit.

Test Cases

import sys, io

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

    def solve():
        n = int(sys.stdin.readline().strip())
        print(n * 4000)

    solve()
    return out.getvalue().strip()

# basic cases
assert run("1\n") == "4000"
assert run("2\n") == "8000"

# minimum boundary
assert run("1\n") == "4000"

# maximum constraint
assert run("20\n") == "80000"

# linear scaling check
assert run("10\n") == "40000"

Test input	Expected output	What it validates
1	4000	minimum valid case
20	80000	upper bound scaling
10	40000	correctness of linear multiplication
2	8000	basic arithmetic consistency

Edge Cases

The only meaningful edge case is the smallest input, since there are no structural variations or special rules.

For input:

The algorithm reads $n = 1$, computes $1 \times 4000 = 4000$, and outputs 4000. There is no iteration, so there is no risk of off-by-one loop errors or initialization mistakes. The correctness follows directly from the single multiplication step, which already matches the definition of the cost model.