CF 102697100 - Ohm Sweet Ohm

We have n positive resistors connected in parallel. If two resistors have resistances a and b, their equivalent resistance is [ R=frac{ab}{a+b}.

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CF 102697100 - Ohm Sweet Ohm

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Solution

Problem Understanding

We have n positive resistors connected in parallel. If two resistors have resistances a and b, their equivalent resistance is

[ R=\frac{ab}{a+b}. ]

For more than two resistors, the statement defines the result by repeatedly combining the equivalent resistance obtained so far with the next resistor. We need to print the final equivalent resistance as a decimal number, without manually rounding it. The official statement gives a one-second time limit and 256 MB memory limit.

The most useful way to reinterpret a parallel circuit is through conductance. The conductance of a resistor with resistance r is 1/r. Parallel conductances add, so for resistances r1, r2, ..., rn the final resistance is

[ R=\frac{1}{\frac1{r_1}+\frac1{r_2}+\cdots+\frac1{r_n}}. ]

The statement does not expose a numerical upper bound for n, so we should avoid relying on a particular maximum. A linear pass over all resistors is the appropriate target because the input itself contains n resistance values. Any approach that performs work proportional to every pair, every subset, or every ordering would grow much faster than the input size.

There are two edge cases that deserve attention. First, the smallest valid input has exactly two resistors. For example,

2
11 19

gives

6.966666666666667

A careless implementation that initializes the accumulated resistance to zero and immediately applies (a * b) / (a + b) would create an invalid intermediate value when the first resistor is processed. Starting from zero only makes sense if we use the reciprocal formulation.

Second, identical resistors are completely valid. For example,

3
10 10 10

has total conductance

[ \frac1{10}+\frac1{10}+\frac1{10}=\frac3{10}, ]

so the answer is

3.333333333333333

An implementation that assumes the resistance values must be distinct would reject a valid circuit. The input only says that the resistances are positive integers.

A third floating-point edge case is that the answer is generally not an integer and is not expected to be rounded to a fixed number of decimal places. For example, 11 and 19 produce a repeating decimal. Printing the ordinary Python floating-point representation is appropriate because the judge accepts a decimal approximation rather than requiring symbolic rational output.

Approaches

A direct simulation follows the recursive definition in the statement. Start with the first resistor and repeatedly combine the current equivalent resistance R with the next resistance x using

[ R\leftarrow\frac{Rx}{R+x}. ]

This is already a linear-time algorithm. After processing the first resistor, there are n-1 updates, and each update performs a constant number of arithmetic operations. Its total work is therefore O(n), so unlike many problems, the straightforward simulation is not actually too slow.

A truly brute-force approach could go further and try every ordering of the resistors, or every possible parenthesization of the pairwise combinations, before evaluating the result. There are n! possible orderings alone, and even with a fixed ordering there are C(n-1) possible binary parenthesizations, where C is a Catalan number. That produces vastly more work than the input size and is unnecessary because parallel resistance is mathematically independent of the order in which the resistors are combined.

The useful observation is stronger than merely saying that the operation is associative. Starting with

[ R=\frac{ab}{a+b}, ]

take its reciprocal:

[ \frac1R=\frac{a+b}{ab}=\frac1a+\frac1b. ]

This means that every parallel resistor contributes independently to the same sum. Instead of repeatedly constructing a new resistance and combining it with the next one, we can simply accumulate

[ S=\sum_{i=1}^{n}\frac1{r_i} ]

and return 1 / S.

The direct simulation and the reciprocal formulation are both O(n). The reciprocal formulation is preferable because it follows the physical structure of a parallel circuit directly, avoids repeated multiplication and division, and makes correctness especially easy to prove.

Approach Time Complexity Space Complexity Verdict
Brute Force over orderings/parenthesizations Superlinear, at least O(n!) O(n) Too slow
Direct sequential simulation O(n) O(1) Accepted
Reciprocal sum O(n) O(1) Accepted

Algorithm Walkthrough

  1. Read n and the n positive resistance values. The values themselves are all that is needed, so there is no need to construct a circuit graph or store any additional structure.
  2. Initialize a floating-point variable conductance to zero. This variable represents the total conductance accumulated from the resistors processed so far.
  3. For every resistance r, add 1.0 / r to conductance. A resistor with resistance r contributes exactly 1/r conductance, and conductances add for parallel components.
  4. After all resistors have been processed, compute 1.0 / conductance. The reciprocal of the total conductance is the equivalent resistance of the complete parallel circuit.
  5. Print the resulting floating-point value directly. The problem does not ask for a fixed number of digits or manual rounding, so Python's normal decimal representation is sufficient.

The invariant is that after processing the first k resistors, conductance equals

[ \sum_{i=1}^{k}\frac1{r_i}. ]

Initially this sum is zero, which is correct for no processed resistors. Adding 1/r when processing the next resistor extends the sum by exactly that resistor's conductance. After the final resistor, the accumulated value is the total parallel conductance, whose reciprocal is exactly the required equivalent resistance.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    resistances = list(map(int, input().split()))

    conductance = 0.0

    for r in resistances:
        conductance += 1.0 / r

    answer = 1.0 / conductance
    print(answer)

if __name__ == "__main__":
    solve()

The first two lines read the number of resistors and the resistance array. Since the input contains all resistances on the next line, a single split() is enough for the specified format.

conductance is initialized to 0.0, rather than 0, to make the intended floating-point arithmetic explicit. Each 1.0 / r is a floating-point value, so the accumulated sum retains the fractional conductance.

The final reciprocal is taken only after every resistor has contributed. Reversing the order and trying to repeatedly take reciprocals would change the mathematical expression, so the accumulation must happen in conductance space first.

Python integers do not overflow, and the resistance values are positive, so division by zero cannot occur. The problem has no multiple-test-case structure, so there is only one call to solve().

Worked Examples

Sample 1

The input is

9
37 23 56 85 41 99 33 22 11

The algorithm maintains the sum of reciprocal resistances.

Step Resistance r Added conductance 1/r Total conductance
1 37 0.0270270270... 0.0270270270...
2 23 0.0434782609... 0.0705052879...
3 56 0.0178571429... 0.0883624308...
4 85 0.0117647059... 0.1001271367...
5 41 0.0243902439... 0.1245173806...
6 99 0.0101010101... 0.1346183907...
7 33 0.0303030303... 0.1649214210...
8 22 0.0454545455... 0.2103759665...
9 11 0.0909090909... 0.3012850574...

The final answer is the reciprocal of the last column:

3.319115819885067

This trace demonstrates the central invariant. At every row, the accumulated value is exactly the conductance of the prefix of the resistor list.

Sample 2

The input is

2
11 19
Step Resistance r Added conductance 1/r Total conductance
1 11 0.0909090909... 0.0909090909...
2 19 0.0526315789... 0.1435406698...

Taking the reciprocal gives

6.966666666666667

This is also the smallest valid input size, so it confirms that the initialization handles the boundary correctly.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each resistor contributes exactly one reciprocal and one addition.
Space O(n) with the shown input storage, O(1) auxiliary The list stores the input resistances; the calculation itself uses only one accumulator.

The linear running time is optimal with respect to the input size because every resistance must be read. The calculation itself uses constant extra state. The official limits are one second and 256 MB, and this approach performs only one simple arithmetic update per resistor.

The list can also be avoided by processing the resistance values directly from the input tokens, but retaining the list makes the implementation straightforward and does not affect the asymptotic result.

Test Cases

The original statement provides two samples. Since it does not publish a numeric upper bound for n, the maximum-size test below uses 100000 resistors as a stress test rather than claiming that 100000 is the official maximum.

import sys
import io

def solve():
    input = sys.stdin.readline

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

    conductance = 0.0
    for r in resistances:
        conductance += 1.0 / r

    print(1.0 / conductance)

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 1
assert run(
    "9\n"
    "37 23 56 85 41 99 33 22 11\n"
) == "3.319115819885067\n"

# Provided sample 2
assert run(
    "2\n"
    "11 19\n"
) == "6.966666666666667\n"

# Minimum-size input
assert run(
    "2\n"
    "1 1\n"
) == "0.5\n"

# All equal values
assert run(
    "3\n"
    "10 10 10\n"
) == "3.333333333333333\n"

# Boundary-style case with very different resistance magnitudes
assert run(
    "4\n"
    "1 2 4 8\n"
) == "0.5333333333333333\n"

# Large stress case, 100000 equal resistors
large_input = "100000\n" + " ".join(["1"] * 100000) + "\n"
assert run(large_input) == "1e-05\n"
Test input Expected output What it validates
2 / 1 1 0.5 Minimum valid n and the basic parallel formula
3 / 10 10 10 3.333333333333333 Repeated equal values and floating-point output
4 / 1 2 4 8 0.5333333333333333 Several different reciprocal contributions
100000 resistors of value 1 1e-05 Linear-time behavior on a large input

The large test is particularly useful because the expected answer has a compact scientific-notation representation. Python prints 1e-05, which is still a valid decimal floating-point representation of the same value.

Edge Cases

For exactly two resistors,

2
11 19

the conductance accumulator starts at zero, receives 1/11, then receives 1/19, and finally becomes 1/(1/11 + 1/19). The result is 6.966666666666667. There is no special case for two resistors, which removes a common off-by-one risk.

For equal resistors,

3
10 10 10

the accumulator becomes 0.1, then 0.2, then 0.3. Its reciprocal is 3.333333333333333. Every resistor is processed independently, so duplicates require no special handling.

For very different values,

4
1 2 4 8

the conductance is

[ 1+\frac12+\frac14+\frac18=1.875, ]

so the equivalent resistance is

[ \frac1{1.875}=0.5333333333333333. ]

This catches implementations that accidentally add resistances instead of conductances.

Finally, consider a large collection of identical unit resistors:

100000
1 1 1 ... 1

with 100000 occurrences of 1. Every resistor contributes one unit of conductance, so the total is 100000 and the equivalent resistance is 0.00001. The algorithm performs exactly one update per resistor, so the running time grows linearly with the amount of input rather than with the number of resistor pairs.

The key idea behind every case is the same: for parallel components, resistance itself is not additive, but conductance is. Once the circuit is represented as a sum of reciprocals, the problem reduces to one linear scan.