CF 102697013 - Resistors in Parallel
The problem asks for the effective resistance of two resistors connected in parallel. The two input values are the resistance values of the individual components, measured in ohms.
CF 102697013 - Resistors in Parallel
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Solve time: 1m
Verified: yes
Solution
Problem Understanding
The problem asks for the effective resistance of two resistors connected in parallel. The two input values are the resistance values of the individual components, measured in ohms. The output is the resulting equivalent resistance, which is calculated using the parallel resistor formula:
$$R = \frac{a \times b}{a + b}$$
The task is a direct application of this formula from circuit theory. The Codeforces archive version of this problem provides two positive integer resistances as input and asks for the resulting decimal value.
The constraints are small enough that no algorithmic optimization is required. Since only two numbers are processed, the amount of work is constant regardless of the input size. Even extremely large resistance values only affect the size of arithmetic operations, not the number of operations performed. A simulation, search, or iterative method would add unnecessary complexity.
The main implementation risks come from handling the arithmetic expression correctly. Integer division is a common mistake because the answer is usually not an integer. For example, with input:
80
70
the correct output is approximately:
37.333333333333336
A careless integer-only calculation would truncate the fractional part and produce an incorrect answer.
Another edge case is when both resistors have the same resistance. For example:
10
10
The correct output is:
5.0
The result is half of either resistor, and code that assumes the answer must be different from the input values could fail.
A final case is when one resistance is much larger than the other:
1
1000
The correct output is:
0.999000999000999
The parallel resistance is slightly smaller than the smaller resistor. An approach that incorrectly treats parallel resistances like series resistances would output 1001, which is the opposite behavior.
Approaches
The brute-force approach is to look for some kind of circuit simulation or manually combine resistors step by step. Such an approach is unnecessary here because the circuit contains exactly two resistors and the mathematical relationship is already given. Any simulation would still need to perform the same multiplication and division, while adding extra work.
The key observation is that the entire problem is reduced to evaluating a single formula. Parallel resistors combine by multiplying the two resistances and dividing by their sum. Once this relationship is recognized, the problem becomes a constant-time arithmetic calculation.
The important implementation detail is preserving the fractional result. In Python, normal division using / already produces a floating-point number, so the formula can be written directly.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(1) | O(1) | Unnecessary |
| Optimal | O(1) | O(1) | Accepted |
Algorithm Walkthrough
- Read the two resistance values from the input. They represent the individual resistors that will be combined.
- Multiply the two resistances together. This gives the numerator of the parallel resistance formula.
- Add the two resistances together. This gives the denominator of the formula.
- Divide the multiplication result by the sum and print the resulting decimal value. Floating-point division is needed because the equivalent resistance is not always an integer.
Why it works:
The formula for two parallel resistors is derived from the fact that the reciprocal of the total resistance equals the sum of the reciprocals of the individual resistances. Simplifying that equation gives:
$$R = \frac{a \times b}{a+b}$$
The algorithm performs exactly this calculation, so the produced value is the physical equivalent resistance of the two components.
Python Solution
import sys
input = sys.stdin.readline
def solve():
a = int(input())
b = int(input())
ans = (a * b) / (a + b)
print(ans)
if __name__ == "__main__":
solve()
The program first converts both input lines into integers because the resistances are whole numbers. The multiplication is performed before the division so that the numerator matches the mathematical formula exactly.
Python's / operator is used instead of //. The floor division operator would discard the fractional part and fail on cases where the equivalent resistance is not an integer.
Python integers do not overflow, so multiplying the two resistance values does not require any special handling. The final division produces the required decimal representation.
Worked Examples
For the first example:
60
40
the calculation proceeds as follows.
| Step | a | b | Formula Result |
|---|---|---|---|
| Initial values | 60 | 40 | |
| Multiply resistances | 60 | 40 | 2400 |
| Add resistances | 60 | 40 | 100 |
| Divide | 60 | 40 | 24.0 |
The result is 24.0, showing that the algorithm directly evaluates the parallel resistance equation.
For the second example:
80
70
the calculation is:
| Step | a | b | Formula Result |
|---|---|---|---|
| Initial values | 80 | 70 | |
| Multiply resistances | 80 | 70 | 5600 |
| Add resistances | 80 | 70 | 150 |
| Divide | 80 | 70 | 37.333333333333336 |
This example demonstrates why floating-point division is required. The answer cannot be represented as an integer.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(1) | Only one multiplication, one addition, and one division are performed. |
| Space | O(1) | The solution stores only the two input values and the answer. |
The solution fits easily within the given limits because it does not depend on the size of the resistance values in terms of algorithmic operations.
Test Cases
import sys
import io
def run(inp: str) -> str:
old_stdin = sys.stdin
old_stdout = sys.stdout
sys.stdin = io.StringIO(inp)
sys.stdout = io.StringIO()
a = int(sys.stdin.readline())
b = int(sys.stdin.readline())
print((a * b) / (a + b))
result = sys.stdout.getvalue()
sys.stdin = old_stdin
sys.stdout = old_stdout
return result
assert run("60\n40\n") == "24.0\n", "sample 1"
assert run("80\n70\n") == "37.333333333333336\n", "sample 2"
assert run("1\n1\n") == "0.5\n", "equal minimum values"
assert run("10\n10\n") == "5.0\n", "equal resistors"
assert run("1\n1000\n") == "0.999000999000999\n", "large imbalance"
| Test input | Expected output | What it validates |
|---|---|---|
60\n40\n |
24.0 |
Provided sample calculation |
80\n70\n |
37.333333333333336 |
Fractional output handling |
1\n1\n |
0.5 |
Smallest values and floating-point division |
10\n10\n |
5.0 |
Equal resistance behavior |
1\n1000\n |
0.999000999000999 |
Large difference between resistors |
Edge Cases
For equal resistances:
10
10
the algorithm computes:
$$\frac{10 \times 10}{10+10}=\frac{100}{20}=5$$
The multiplication and division steps work without any special case handling.
For a non-integer result:
80
70
the numerator is 5600 and the denominator is 150, giving approximately 37.333333333333336. Because the code uses floating-point division, it keeps the decimal part instead of truncating it.
For a highly unbalanced pair:
1
1000
the formula gives:
$$\frac{1 \times 1000}{1+1000}=\frac{1000}{1001}$$
which is slightly less than 1. The algorithm handles this naturally because it does not rely on assumptions about the relative sizes of the resistors.