CF 102697003 - Triangle Sum

CF 102697003 - Triangle Sum

Rating: -
Tags: -
Solve time: 48s
Verified: yes

Solution

Problem Understanding

The task is about a collection of equilateral triangles. Each triangle is described only by the length of one side. Since all three sides of an equilateral triangle are equal, the perimeter of a triangle with side length t is 3 * t. The goal is to add the perimeters of all given triangles and print the total value.

The input contains the number of triangles followed by their side lengths. The output is a single integer representing the sum of all triangle perimeters. The main observation is that the geometry disappears after using the equilateral property. The problem becomes a simple accumulation of values after multiplying each side length by three.

The constraints are small enough for a direct traversal of the input. Even if the number of triangles were very large, the only reasonable approach would still be linear because every side length contributes independently to the answer. Any approach that stores unnecessary information or compares triangles against each other would add work without using any property of the problem.

The main edge cases come from handling the arithmetic correctly. A single triangle must still be processed. For example, an input of:

1
5

has output:

15

A careless implementation might forget that the input value is a side length rather than a perimeter.

Another case is when all triangles have the same side length:

3
7
7
7

The output is:

63

The contribution of every triangle is independent, so every occurrence must be included. An implementation that only keeps the largest or smallest side length would fail here.

Large values are also a possible source of mistakes. The total perimeter can be much larger than an individual side length, so the implementation should rely on Python's integer arithmetic rather than manually limiting the result.

Approaches

The brute-force approach would be to treat each triangle as an object, calculate its three sides, compute the perimeter, and then add it to the answer. This is correct because every triangle contributes exactly its own perimeter and no triangle affects another one. However, creating unnecessary representations does not help because the only information needed from each triangle is its side length.

The useful observation is that an equilateral triangle always has perimeter equal to three times its side length. There is no need to reconstruct geometry or check any conditions. We can read each side length, multiply it by three, and add it immediately.

The brute-force idea works because the number of operations per triangle is constant, but it can be simplified further by removing all unnecessary steps. The structure of the problem lets us reduce the solution to a single pass over the input.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n)	O(n)	Accepted, but stores unnecessary data
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Read the number of triangles. Each following value represents one side length that contributes independently to the final answer.
2. For every side length t, add 3 * t to the running total. The factor of three comes directly from the perimeter formula of an equilateral triangle.
3. Print the accumulated sum after all triangles have been processed.

Why it works:

The invariant maintained during the scan is that after processing the first k triangles, the stored answer is exactly the sum of the perimeters of those k triangles. When the next side length is read, adding 3 * t adds precisely the missing contribution of that triangle. After the final triangle is processed, the invariant gives the required total perimeter.

Python Solution

import sys
input = sys.stdin.readline

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

    for _ in range(n):
        side = int(input())
        ans += 3 * side

    print(ans)

if __name__ == "__main__":
    solve()

The program keeps only the current answer because previous side lengths are no longer needed after their contribution has been added.

The multiplication is performed before adding to the answer so that each triangle contributes its full perimeter. Python integers automatically handle large sums, so there is no overflow concern.

The loop runs exactly once per triangle. There is no sorting, searching, or extra storage, which keeps the implementation aligned with the problem structure.

Worked Examples

For the input:

3
3
4
6

the execution is:

Triangle processed	Side length	Added value	Current answer
1	3	9	9
2	4	12	21
3	6	18	39

The final answer is 39, which matches the sum of the three individual perimeters.

For the input:

4
1
1
2
10

the execution is:

Triangle processed	Side length	Added value	Current answer
1	1	3	3
2	1	3	6
3	2	6	12
4	10	30	42

This example checks repeated values and confirms that every triangle is counted separately.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Every triangle side length is read and processed once
Space	O(1)	Only the running sum and current side length are stored

The solution performs a constant amount of work for each triangle, so it easily fits within the available limits.

Test Cases

import sys
import io

def solve_io(inp: str) -> str:
    old_stdin = sys.stdin
    sys.stdin = io.StringIO(inp)

    data = sys.stdin.read().split()

    sys.stdin = old_stdin

    if not data:
        return ""

    n = int(data[0])
    ans = 0

    for i in range(1, n + 1):
        ans += 3 * int(data[i])

    return str(ans)

# provided sample
assert solve_io("""3
3
4
6
""") == "39", "sample 1"

# single triangle
assert solve_io("""1
5
""") == "15", "single triangle"

# all equal values
assert solve_io("""3
7
7
7
""") == "63", "equal sides"

# large values
assert solve_io("""2
1000000000
1000000000
""") == "6000000000", "large arithmetic"

# mixed values
assert solve_io("""5
1
2
3
4
5
""") == "45", "mixed sides"

Test input	Expected output	What it validates
1 / 5	15	Minimum number of triangles
3 / 7 7 7	63	Repeated identical side lengths
2 / 1000000000 1000000000	6000000000	Large integer handling
5 / 1 2 3 4 5	45	Normal accumulation

Edge Cases

For one triangle with side length 5, the algorithm reads the value, calculates 3 * 5, and prints 15. It does not rely on having multiple triangles, so the smallest possible input is handled naturally.

For repeated equal triangles such as:

3
7
7
7

the algorithm performs the same calculation three times and accumulates 21 + 21 + 21, producing 63. This avoids mistakes caused by treating the input as a set of unique lengths.

For very large side lengths, such as:

2
1000000000
1000000000

the running sum becomes 6000000000. The calculation remains correct because Python integers expand automatically when the value exceeds normal machine integer sizes.