CF 102697066 - Neural Network Weight Count

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Solve time: 48s
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Solution

Problem Understanding

A neural network is represented only by its layers. If two consecutive layers contain a and b nodes, every node in the first layer connects to every node in the second layer. Each such connection is one weight.

The input gives the number of layers N, followed by the number of nodes in each layer. The task is to count every connection between consecutive layers and print the total. The statement specifies a 1 second time limit and 256 MB of memory, but does not provide explicit numeric bounds for N or the layer sizes. Since we only need to inspect each layer once, the natural linear solution is comfortably within the limit for any practical input size.

For a pair of consecutive layers with a and b nodes, there are exactly a×b connections. For example, with layers containing 5,10,5 nodes, the first pair contributes 5×10=50 weights and the second contributes 10×5=50, giving 100.

There are two small cases that are easy to mishandle. If there is only one layer, there are no connections at all, so an input such as

has output

A careless implementation that multiplies the number of nodes by itself would incorrectly produce 49. The network only connects different consecutive layers.

Another common mistake is to forget the final pair. For

32 3 4

the correct output is

because the first pair contributes 2×3=6 and the second contributes 3×4=12. Summing only the first pair would produce 6, while the correct total is 18.

Approaches

The direct brute-force interpretation would be to explicitly consider every node in a layer and every node in the following layer, adding one to the answer for each pair. This is correct because every pair of nodes across consecutive layers represents exactly one weight. For layers containing a and b nodes, that process performs ab operations.

The problem is that we do not need to distinguish individual connections. Every node in the first layer connects to every node in the second, so the number of connections is known immediately as the product ab. The observation that turns the simulation into a counting problem lets us process each adjacent pair with one multiplication instead of iterating through all of its connections.

With N layers, there are exactly N−1 adjacent layer pairs. We simply add

L 0 L 1 +L 1 L 2 +⋯+L N−2 L N−1 .

The brute-force method can require a large number of operations if two consecutive layers are large, while the optimal method always performs only N−1 multiplications and additions.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(∑L i L i+1 )	O(1)	Too slow for large layers
Optimal	O(N)	O(N)	Accepted

The optimal implementation can also be made O(1) extra space by processing the layer sizes as a stream, but storing the input list is simple and already well within the stated memory limit.

Algorithm Walkthrough

Read the number of layers N and the N node counts. We need the counts of consecutive layers because only adjacent layers have connections.
Initialize the answer to zero. No connections have been counted yet.
For every pair of consecutive layer sizes L i and L i+1 , add L i L i+1 to the answer. This product counts every connection between those two layers exactly once.
Print the accumulated answer. All connections in the network occur between one of these N−1 consecutive pairs, so nothing else needs to be counted.

Why it works: after processing the first k adjacent pairs, the answer equals the exact number of weights whose two endpoints lie in those pairs. Each pair is disjoint from the others as a set of connections because a weight belongs to exactly one pair of consecutive layers. After the final pair is processed, every possible weight has been counted exactly once, so the resulting sum is the required total.

Python Solution

Pythonimport sysinput = sys.stdin.readline

def solve():    n = int(input())    layers = list(map(int, input().split()))
    answer = 0
    for i in range(n - 1):        answer += layers[i] * layers[i + 1]
    print(answer)

if __name__ == "__main__":    solve()

The first line reads the number of layers. The second line contains their node counts, so the list layers gives direct access to every adjacent pair.

The loop stops at n - 2 as its final index because layers[i + 1] must still exist. Equivalently, there are exactly n - 1 adjacent pairs. When n is one, the loop executes zero times and the answer remains zero, which correctly represents a network with no connections.

Python integers can grow beyond the size of fixed-width 32-bit integers, so the multiplication and accumulated sum do not require any special overflow handling in Python.

Worked Examples

For the provided sample,

35 10 5

the execution is:

i	Current layer	Next layer	Added weights	Answer
0	5	10	50	50
1	10	5	50	100

The first transition contains 5×10=50 connections. The second contains 10×5=50. Their sum is 100, matching the sample output.

A second example with only two layers is:

24 7

i	Current layer	Next layer	Added weights	Answer
0	4	7	28	28

There is only one transition, so every possible connection is between these two layers. The output is 28. This demonstrates that the loop correctly handles the smallest network that can actually contain weights.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(N)	Each of the N−1 adjacent layer pairs is processed once.
Space	O(N)	The node counts are stored in a list.

The algorithm performs one constant-time multiplication and addition for each adjacent pair of layers. Even if the number of layers is large, the work grows linearly rather than with the number of individual neural-network connections. The memory usage is also small because only the layer sizes are stored.

Test Cases

Python# helper: run solution on input string, return output stringimport sysimport io

def solve():    n = int(input())    layers = list(map(int, input().split()))
    answer = 0    for i in range(n - 1):        answer += layers[i] * layers[i + 1]
    print(answer)

def run(inp: str) -> str:    global input    old_stdin = sys.stdin    old_input = input
    sys.stdin = io.StringIO(inp)    input = sys.stdin.readline
    out = io.StringIO()    old_stdout = sys.stdout    sys.stdout = out
    try:        solve()        return out.getvalue()    finally:        sys.stdin = old_stdin        sys.stdout = old_stdout        input = old_input

# Provided sampleassert run("3\n5 10 5\n") == "100\n", "sample 1"
# Minimum-size network: one layer has no connections.assert run("1\n7\n") == "0\n", "single layer"
# Two layers: exactly one complete bipartite set of connections.assert run("2\n4 7\n") == "28\n", "two layers"
# All layer sizes equal.assert run("5\n3 3 3 3 3\n") == "36\n", "all equal"
# Boundary between every adjacent pair is counted.assert run("4\n1 2 1 2\n") == "8\n", "alternating sizes"
# Large values, checking that multiplication is handled correctly.assert run("3\n100000 100000 100000\n") == "20000000000\n", "large values"

Test input	Expected output	What it validates
`1 / 7`	`0`	A single layer has no adjacent pair.
`2 / 4 7`	`28`	Exactly one layer transition is counted.
`5 / 3 3 3 3 3`	`36`	Every adjacent pair has the same contribution.
`4 / 1 2 1 2`	`8`	Both boundaries and the final pair are included.
`3 / 100000 100000 100000`	`20000000000`	Large products and accumulated values.

Edge Cases

For a single layer, the input

produces 0. The algorithm initializes answer to zero and the loop has range range(0), so no multiplication is performed. This matches the graph structure because there is no second layer to connect to.

For the missing-final-pair mistake, consider

32 3 4

The algorithm first adds 2×3=6, giving an intermediate answer of 6. It then processes the final pair and adds 3×4=12, producing 18. Every transition is processed exactly once, including the boundary between the last two layers.

For two layers,

24 7

there is only one iteration. The product 4×7=28 directly counts all possible connections, so the algorithm does not accidentally expect at least two transitions.

For equal layer sizes,

53 3 3 3 3

each of the four transitions contributes 3×3=9. The final result is 9+9+9+9=36. The algorithm treats every transition independently, so repeated layer sizes require no special handling.

For large layer sizes,

3100000 100000 100000

each adjacent pair contributes 10 10, and the final answer is 20000000000. The calculation is still constant work per pair, and Python's integer representation safely handles the resulting value.