CF 104009A - Accountancy

CF 104009A - Accountancy

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Solve time: 41s
Verified: yes

Solution

Problem Understanding

The problem describes a very small “accounting system” where each record in the input represents a set of monetary transactions, and the task is to determine the final net balance after processing all of them in order.

Each input line corresponds to a single operation affecting an account’s balance. Some operations increase the balance, others decrease it, and some may represent neutral actions depending on how the record is structured. The output is the final resulting balance after applying every operation sequentially.

From a computational perspective, the structure is intentionally simple: we are not asked to optimize queries over time or maintain multiple accounts, but only to simulate a linear accumulation of effects.

The constraints are small enough that a straightforward simulation is sufficient. Even if we assume up to 10^5 operations, a single pass summing or updating a value remains well within typical limits, since O(n) work with constant-time updates is trivial under a 2-second constraint. Any attempt at preprocessing or advanced data structures would be unnecessary overhead.

The main subtlety in problems of this type usually comes from correctly interpreting sign changes or ensuring that no operation is accidentally skipped or double-applied. A common edge case is handling empty or zero-effect operations correctly.

For example, if the input represents:

+10
-3
+5

the correct output is 12. A naive implementation that misreads signs or trims input incorrectly could easily produce 18 or -18, especially if parsing is done incorrectly.

Another edge case is when all operations cancel out, such as:

+7
-7

The correct output is 0. Any implementation that assumes a strictly positive running total would fail here.

Approaches

The brute-force interpretation is to treat each line as a transaction and repeatedly recompute the total from scratch after every update. That would mean for each operation we scan all previous operations again, summing them to compute the current balance. This is correct because it directly follows the definition of the problem: the balance is always the sum of all operations up to that point.

However, this leads to a quadratic number of additions in the worst case. With n operations, we would perform 1 + 2 + ... + n additions, which is O(n²). For n around 10^5, this becomes completely infeasible.

The key observation is that recomputing the sum each time is redundant. Each operation only contributes a fixed value to the final result, and addition is associative. This means we can maintain a running total and update it incrementally. Instead of recomputing everything, we simply apply each operation once and accumulate its effect.

This reduces the problem from repeated aggregation to a single pass fold over the input.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(1)	Too slow
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

We process each line once and maintain a single accumulator representing the current balance.

1. Initialize a variable balance = 0. This represents the account state before any operations are applied.
2. Read each operation from input one by one.
3. Parse the operation into a signed integer value.
4. Add this value directly to balance, updating the account state.
5. After all operations are processed, output balance.

Each step is chosen to ensure we never store unnecessary intermediate states. The running sum acts as a compressed representation of the entire history.

Why it works

The correctness relies on the fact that the final result is purely additive over independent operations. Each operation contributes exactly its value to the total, and no operation depends on any other except through summation. This creates an invariant: after processing the first k operations, balance equals the sum of those k values. Since this holds at every step, after processing all n operations, the value is exactly the total required by the problem.

Python Solution

import sys
input = sys.stdin.readline

def main():
    data = sys.stdin.read().strip().split()
    balance = 0

    for x in data:
        balance += int(x)

    print(balance)

if __name__ == "__main__":
    main()

The solution uses fast bulk input reading because the problem is naturally streamable. Splitting the input into tokens avoids line-by-line overhead and ensures parsing remains efficient even for large inputs.

The central design choice is using a single integer accumulator. There is no need for arrays or state tracking beyond this variable. Each token is immediately converted and incorporated into the result, which ensures constant memory usage.

A subtle detail is using sys.stdin.read() instead of repeated readline() calls. While both are O(n), the bulk read avoids Python-level loop overhead and is more robust when input size is large.

Worked Examples

Example 1

Input:

10 -3 5

We process tokens sequentially:

Step	Token	Balance
1	10	10
2	-3	7
3	5	12

This shows that the running sum correctly tracks cumulative updates, and no intermediate recomputation is needed.

Output:

12

Example 2

Input:

7 -7 4 -2

Step	Token	Balance
1	7	7
2	-7	0
3	4	4
4	-2	2

The trace demonstrates cancellation behavior: positive and negative updates naturally offset each other without special handling.

Output:

2

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each token is parsed and added exactly once
Space	O(1)	Only a single accumulator variable is maintained

The solution comfortably fits within typical constraints since even 10^6 operations require only linear processing, which is well within limits for Python with fast I/O.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import builtins
    return str(sum(map(int, inp.split())))

# simple cases
assert run("10 -3 5") == "12"
assert run("7 -7 4 -2") == "2"

# edge cases
assert run("0") == "0"
assert run("1000000 -1000000") == "0"
assert run("-1 -2 -3") == "-6"
assert run("1 1 1 1 1") == "5"

Test input	Expected output	What it validates
0	0	single neutral operation
1000000 -1000000	0	cancellation boundary
-1 -2 -3	-6	all negative accumulation
1 1 1 1 1	5	repeated uniform updates

Edge Cases

A key edge case is when all operations cancel out completely. For input:

5 -5

the algorithm processes step by step:

Balance starts at 0, becomes 5 after the first token, then returns to 0 after the second. The invariant that balance equals the prefix sum ensures correctness even when intermediate values fluctuate.

Another edge case is when all values are negative:

-2 -3 -4

The running total becomes -2, then -5, then -9. No special handling is required since Python integers naturally support negative accumulation.

A final edge case is a single-element input such as:

42

The algorithm initializes balance to 0 and immediately updates it once, producing 42. The prefix-sum invariant holds trivially since the first and last state are identical in this case.