CF 102697027 - Money Sum

CF 102697027 - Money Sum

Rating: -
Tags: -
Solve time: 42s
Verified: yes

Solution

Problem Understanding

The task is much simpler than the title might suggest. We have made (n) payments during the year, and each payment is given as a positive integer representing its amount in cents. The required answer is the total amount spent, also in cents. The input contains (n) followed by the (n) payment amounts, one amount per line, and the output is their sum.

The official limits are 1 second and 256 MB. The statement does not publish an explicit upper bound for (n) or for an individual payment, so there is no precise worst-case operation count that can be derived from the statement itself. The right algorithm should consequently scale linearly with the number of payments and should not perform any work that grows with the values of the payments. Reading every payment and adding it once gives exactly that behavior. Python integers also avoid fixed-width overflow concerns for ordinary integer inputs.

There are a few small cases where an implementation can go wrong. With a single payment, such as

1
42

the answer is 42. Code that initializes the total from a nonexistent first element instead of from zero can fail here.

Repeated values must be counted independently. For

3
5
5
5

the answer is 15, not 5. An implementation that accidentally treats the payments as a set would silently discard duplicates.

The total can be larger than any individual payment. For

4
100
200
300
400

the answer is 1000. An implementation that only tracks the largest payment, or confuses the task with finding a maximum, produces the wrong result.

Finally, the payments are given in cents, so no decimal conversion is needed. For example,

2
99
1

produces 100. Treating the values as floating-point dollar amounts would add unnecessary complexity and could introduce rounding issues.

Approaches

A direct approach is to read every payment into an array and then iterate through the array to calculate the total. It is correct because every payment contributes exactly once to the final sum. If there are (n) payments, the second pass performs exactly (n) additions, in addition to the (n) input operations. Its time complexity is (O(n)), which is already the minimum possible asymptotic time because every payment has to be read from the input.

There is no meaningful subset-search or combinatorial brute force hidden in this problem. Trying every subset, for example, would perform (2^n) work even though the problem never asks us to choose payments. Such an approach would be solving a substantially harder problem than the one given.

The useful observation is that the final answer is an associative sum. We do not need to remember a payment after it has been added to the running total. When the next payment arrives, the only information from all previous payments that matters is their combined sum.

That turns the two-pass array approach into a one-pass streaming solution. Start with total = 0, read each payment, and immediately add it to total. After the final payment, total is exactly the required answer. The time remains (O(n)), but the additional space drops from (O(n)) to (O(1)).

Approach	Time Complexity	Space Complexity	Verdict
Store all payments, then sum	(O(n))	(O(n))	Accepted, but unnecessary storage
Running sum	(O(n))	(O(1))	Accepted and optimal

Algorithm Walkthrough

1. Read the number of payments, (n). This tells us exactly how many payment amounts must be consumed from the input.
2. Initialize total to zero. Zero is the identity value for addition, so adding the first payment produces its correct contribution without requiring a special first-element case.
3. Repeat (n) times. Read one payment and add it immediately to total. The payment does not need to be stored because no later calculation needs its individual value.
4. Print total after all payments have been processed. At that point every input payment has contributed exactly once.

Why it works: after processing any prefix of the payments, total equals the sum of exactly that prefix. Initially the prefix is empty and the sum is zero. When the next payment (m) is read, the new value becomes the previous prefix sum plus (m), so the invariant remains true. After all (n) payments have been processed, the prefix is the entire input sequence, making total equal to the required total spending.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    total = 0

    for _ in range(n):
        total += int(input())

    print(total)

if __name__ == "__main__":
    solve()

The first line reads (n), which controls the loop rather than relying on end-of-file detection. This matches the input format exactly and guarantees that precisely the specified number of payments is consumed.

total starts at zero because addition has zero as its identity. This also handles the single-payment case without a separate branch.

Each payment is converted to an integer and added immediately. The order matters only in the sense that every payment must be processed once. There is no sorting, comparison, or conversion between cents and dollars.

Python's integer type can grow as necessary, so the implementation does not have the fixed-width overflow issue that would arise in languages using a small integer type. The memory usage remains constant apart from the input buffering handled by the runtime.

Worked Examples

For the provided sample, the five payments are processed one at a time.

Step	Payment	Total
Start	0	0
1	7	7
2	11	18
3	18	36
4	27	63
5	20	83

The final running total is 83, matching the required output. This trace demonstrates the central invariant: after each step, total contains exactly the sum of all payments processed so far. The official sample uses these five payments and has output 83.

For a second example, consider repeated values.

3
5
5
5

Step	Payment	Total
Start	0	0
1	5	5
2	5	10
3	5	15

The answer is 15. Each occurrence is processed independently, so equal payments are not accidentally merged or discarded.

Complexity Analysis

Measure	Complexity	Explanation
Time	(O(n))	Every payment is read and added exactly once
Space	(O(1))	Only n and the running total are stored

The linear running time is appropriate because the input itself contains (n) payment values, so any correct solution must inspect those values. The algorithm does not perform any additional work proportional to the payment amounts, and it stores no array of payments.

Test Cases

import sys
import io

def solve():
    input = sys.stdin.readline
    n = int(input())
    total = 0

    for _ in range(n):
        total += int(input())

    print(total)

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    try:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided sample
assert run("""5
7
11
18
27
20
""") == "83\n", "sample 1"

# Minimum-size input
assert run("""1
42
""") == "42\n", "single payment"

# All values equal
assert run("""5
7
7
7
7
7
""") == "35\n", "duplicate payments"

# Boundary-style arithmetic case
assert run("""4
99
1
50
50
""") == "200\n", "sum crossing a round boundary"

# Larger stress case
n = 100000
inp = str(n) + "\n" + ("1\n" * n)
assert run(inp) == "100000\n", "large input"

Test input	Expected output	What it validates
1 / 42	42	Minimum valid number of payments
5 / 7 7 7 7 7	35	Duplicate and all-equal payments
4 / 99 1 50 50	200	Correct accumulation across decimal boundaries
100000 payments of 1	100000	Linear processing on a large input

The large generated case is a stress test rather than an official maximum-size case, because the published statement does not specify a maximum value for (n). It checks that the implementation performs one simple operation per payment and does not accidentally introduce quadratic behavior.

Edge Cases

For a single payment,

1
42

the algorithm initializes total to zero, reads 42, and changes the total to 42. It then immediately prints that value. There is no attempt to access an element before the first input value, so the smallest valid input is handled naturally.

For repeated payments,

3
5
5
5

the running totals are 5, 10, and 15. The algorithm never deduplicates the values, which is correct because three separate payments of five cents represent fifteen cents of spending.

For a total larger than every individual payment,

4
100
200
300
400

the running totals are 100, 300, 600, and 1000. The final result comes from accumulation rather than selecting any single payment.

For a sum that crosses a round boundary,

2
99
1

the first payment produces 99, and the second changes it to 100. Because the values are treated as integer cents, there is no floating-point arithmetic or rounding step that could alter the answer.