CF 105940D - ASZoo Animals

CF 105940D - ASZoo Animals

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Solve time: 41s
Verified: yes

Solution

Problem Understanding

The task describes a zoo where animals are grouped into several kinds, and each kind has a known number of individual animals. The input gives a list where each value corresponds to how many animals belong to one specific kind. The goal is to determine the total number of animals across the entire zoo.

Interpreted more concretely, each element of the array is already a count. There is no hidden structure, no interaction between kinds, and no constraints that depend on ordering or relationships. The only operation required is aggregating all these counts into a single final value.

The constraints are small, with at most 100 kinds and each kind contributing at most 100 animals. This immediately places the computation in a regime where even the most straightforward linear scan is more than sufficient. Any algorithm that is linear in the number of kinds is optimal in practice, and even repeated passes over the data would not pose any performance issue.

Edge cases are minimal but still worth making explicit.

If there is only one kind, for example input 1 followed by 7, the answer is simply 7. Any implementation that mistakenly initializes an accumulator incorrectly or skips single-element cases would fail here.

If all values are identical, such as 3 1 1 1, the correct output is 3. A common mistake in summation problems is accidentally multiplying or misinterpreting the structure as something more complex, but here it remains a pure aggregation.

A corner case that sometimes reveals bugs is when all values are at their maximum, such as 100 repeated 100 times. The correct output is 10000, and this tests whether the implementation correctly accumulates without overflow issues in languages with fixed-width integers. In Python this is not a concern, but it is still a conceptual boundary worth keeping in mind.

Approaches

A brute-force interpretation would still be to iterate over every value and accumulate a running sum. There is no meaningful alternative formulation because the problem does not introduce constraints like selection, transformation, or dependency between elements. Even if one attempted to simulate “counting animals per family”, that simulation collapses immediately into addition since each family contributes independently.

The brute-force method therefore is already optimal. The only “inefficiency” one could imagine is repeatedly summing the array from scratch for each query, but since there are no queries, even that concern does not arise. The entire computation is a single pass over the input.

The key observation is that the problem is fundamentally a reduction of a list of independent quantities into a single scalar. Once this is recognized, the solution reduces to computing a prefix sum over the entire array or simply maintaining a running total while reading input.

Approach	Time Complexity	Space Complexity	Verdict
Direct summation	O(n)	O(1)	Accepted
Any alternative formulation	O(n)	O(1)	Accepted but unnecessary

Algorithm Walkthrough

The optimal algorithm is essentially a streaming accumulation over the input values.

1. Read the integer n, which represents how many kinds of animals exist. This determines how many numbers will follow.
2. Initialize an accumulator variable total to zero. This variable represents the cumulative number of animals seen so far.
3. Iterate over the n values in the input. For each value a_i, add it directly to total. This step works because each a_i represents an independent contribution with no overlap or interaction.
4. After processing all values, output total. At this point, every family has been accounted for exactly once, so the accumulator holds the final answer.

Why it works

Each input value is a disjoint count of animals belonging to a specific category. Since no animal belongs to more than one category and no transformation occurs between categories, the total population is exactly the sum of all category sizes. The algorithm maintains an invariant after processing the i-th value: total equals the sum of the first i family sizes. When the loop finishes, this invariant extends to all n values, guaranteeing correctness.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
arr = list(map(int, input().split()))

total = 0
for x in arr:
    total += x

print(total)

The solution reads the list in one pass and maintains a running sum. There is no need for extra data structures because no intermediate queries or transformations are required.

A subtle implementation detail is ensuring that input parsing correctly reads all n values even if they are split across lines. Using split() on the second line is sufficient under standard input formatting for this problem.

Worked Examples

Example 1

Input:

4
3 1 2 5

Step	Current value	Running total
1	3	3
2	1	4
3	2	6
4	5	11

The accumulation shows how each family contributes independently, and the final result is 11.

Example 2

Input:

3
1 1 1

Step	Current value	Running total
1	1	1
2	1	2
3	1	3

This case highlights that identical contributions do not change the logic; repeated uniform values still accumulate linearly.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each of the n values is read once and added to a running total
Space	O(1)	Only a single accumulator variable is used

Given the constraint that n is at most 100, this solution runs in negligible time. Even in much larger constraints, a single linear scan remains optimal because every input value must be read at least once.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = int(input())
    arr = list(map(int, input().split()))
    print(sum(arr))
    return str(sum(arr))

# provided samples
assert run("4\n3 1 2 5\n") == "11", "sample 1"
assert run("1\n6\n") == "6", "sample 2"
assert run("3\n1 1 1\n") == "3", "sample 3"

# custom cases
assert run("2\n100 100\n") == "200", "small max values"
assert run("1\n42\n") == "42", "single element"
assert run("5\n0 0 0 0 0\n") == "0", "all zero-like case (conceptual boundary)"
assert run("4\n1 2 3 4\n") == "10", "increasing sequence"

Test input	Expected output	What it validates
1\n42\n	42	Single-element edge case
5\n0 0 0 0 0\n	0	Neutral accumulation behavior
4\n1 2 3 4\n	10	General correctness on typical input
2\n100 100\n	200	Maximum-value accumulation stability

Edge Cases

A single family case such as 1 followed by 6 exercises the base case where the loop runs exactly once. The accumulator starts at zero, receives one update, and immediately becomes the final answer, confirming that no special-case logic is required for small inputs.

A uniform distribution such as 1 1 1 1 1 ensures that repeated identical updates behave consistently. The running sum increases monotonically by one at each step, and the final value matches the number of elements, confirming that no overwriting or resetting occurs during iteration.

A maximum-density case like 100 repeated 100 times checks whether accumulation handles larger intermediate sums correctly. The running total grows to 10000 without any need for modular arithmetic or overflow handling in Python, reinforcing that the solution relies purely on straightforward arithmetic aggregation.