CF 104599A - Tall Bots Short Bots

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Solve time: 57s
Verified: yes

Solution

Problem Understanding

We are given a collection of robot friends, each described only by a single integer value representing its height. Bob wants to build a single “mega robot” by stacking all of them vertically. The task is simply to determine the total height of this stack, which is the sum of all individual robot heights.

The input consists of an integer N followed by N integers. Each integer contributes directly to the final result, and nothing else affects the structure of the problem. There is no ordering constraint, no interaction between values, and no transformation applied to individual heights. The output is a single integer equal to the total accumulated height.

From the constraints, N is at most 1000 and each height is at most 1,000,000. This means the sum can reach up to 10^9, which is well within the range of standard 32-bit signed integers, though using a 64-bit type is still a safe habit in general-purpose solutions.

The problem is intentionally minimal, so the main risks are not algorithmic but implementation-related. A few subtle edge cases still exist.

One edge case is when there is only one robot. The input would look like:

1
42

The correct output is 42. A buggy pattern here would be code that assumes at least two values and initializes the sum incorrectly from an empty or partially filled structure.

Another edge case is when all robots have the maximum height:

3
1000000 1000000 1000000

The correct output is 3000000. A careless implementation using a fixed-width 32-bit integer in a language with strict overflow behavior could fail here, though Python naturally avoids this issue.

A third situation is when all values are zero, which is not explicitly in constraints but is logically allowed if interpreted loosely. The sum should still correctly be zero. This can expose logic that mistakenly initializes the accumulator with a nonzero default or skips values under a condition like if h_i.

Approaches

The structure of the problem suggests no dependencies between elements, so any correct solution must combine all values into a single aggregate result. The most direct approach is to iterate through the list and accumulate a running sum.

A brute-force mindset might imagine recomputing partial sums in different ways, such as repeatedly summing prefixes or recomputing totals from scratch for validation. For example, one could compute the sum for each prefix and repeatedly combine results, or even simulate stacking by repeatedly merging partial stacks. This still ultimately reduces to summing all elements, but with unnecessary repeated work. In the worst case, recomputing sums from scratch for each element leads to O(N^2) operations, which is still tiny for N ≤ 1000, but it is structurally wasteful and not the intended direction.

The key observation is that stacking does not introduce any interaction effects. The height of the final structure is invariant under order and grouping, meaning the problem is purely additive. Once this is recognized, the entire task collapses into a single pass accumulation.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Repeated Summation	O(N^2)	O(1)	Acceptable but unnecessary
Single Pass Accumulation	O(N)	O(1)	Accepted

Algorithm Walkthrough

Read the integer N, which determines how many heights will follow. This defines how many values we must aggregate.
Initialize a variable total to zero. This variable represents the running sum of all robot heights seen so far.
Iterate through the N input values one by one. For each value h, add it to total. This step directly reflects the stacking operation described in the problem: each robot contributes its full height to the final structure.
After processing all values, output total. At this point, every robot has been included exactly once, so the accumulated value is the final height of the mega robot.

Why it works

The algorithm relies on the fact that the final structure is formed by stacking all robots without modification or overlap. Each robot contributes independently to the total height, and no interaction changes its value. Because addition is associative and commutative, the order of accumulation does not matter, and every valid stacking corresponds to the same sum. The running total maintains an invariant: after processing k elements, total equals the sum of the first k heights. When k reaches N, the invariant guarantees that total equals the sum of all heights.

Python Solution

import sys
input = sys.stdin.readline

def main():
    n = int(input().strip())
    total = 0

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

    print(total)

if __name__ == "__main__":
    main()

The solution reads input using fast I/O to avoid overhead even though constraints are small. The accumulator total is updated in a single pass, which corresponds exactly to the conceptual stacking process. Each line is parsed and immediately incorporated, ensuring constant memory usage.

A subtle implementation detail is using strip() when reading integers from input lines. While not strictly necessary in most cases, it prevents issues with trailing whitespace. Another detail is keeping the accumulator separate from input parsing, which avoids accidental reuse of loop variables or partial overwrites.

Worked Examples

Example 1

Input:

5
7 2 12 4 8

We process each value sequentially.

Step	Current Value	Total
1	7	7
2	2	9
3	12	21
4	4	25
5	8	33

After processing all values, the final total is 33.

This trace confirms that the algorithm correctly accumulates contributions incrementally without needing to revisit earlier values.

Example 2

Input:

4
1 1 1 1

Step	Current Value	Total
1	1	1
2	1	2
3	1	3
4	1	4

The final result is 4, which matches the expectation for uniform inputs and confirms that repeated identical contributions are handled correctly.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(N)	Each height is read once and added once to the accumulator
Space	O(1)	Only a single running total is maintained regardless of input size

Given N ≤ 1000, the solution runs in negligible time, far below any practical limit. Memory usage remains constant since no list storage is required.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    input = sys.stdin.readline

    n = int(sys.stdin.readline().strip())
    total = 0
    for _ in range(n):
        total += int(sys.stdin.readline().strip())
    return str(total)

# provided sample
assert run("5\n7 2 12 4 8\n") == "33"

# single element
assert run("1\n42\n") == "42"

# all equal
assert run("4\n5 5 5 5\n") == "20"

# maximum values
assert run("3\n1000000 1000000 1000000\n") == "3000000"

# includes zeroes
assert run("5\n0 0 0 0 0\n") == "0"

Test input	Expected output	What it validates
single element	42	minimal input handling
all equal	20	repeated accumulation correctness
max values	3000000	large sum safety
all zeros	0	neutral element handling

Edge Cases

For a single robot input like 1 followed by 42, the loop runs exactly once. The accumulator starts at zero, becomes 42 after the first iteration, and is printed directly. This confirms that initialization does not depend on having multiple elements.

For maximum values such as 1000000 1000000 1000000, each iteration adds a large number, but since Python integers are unbounded, no overflow occurs. The running total evolves cleanly from 0 to 3,000,000, confirming that repeated addition scales safely within constraints.

For all-zero inputs, each iteration leaves the accumulator unchanged. The invariant that total equals the sum of processed elements holds trivially at every step, and the final output remains zero without any special casing.