CF 2086A - Cloudberry Jam
We are asked to determine how many kilograms of cloudberries are needed to produce a certain number of jars of jam, given that the jam is made by combining equal amounts of berries and sugar, and that during cooking, 25% of the total mass evaporates.
Rating: 800
Tags: math
Solve time: 1m 10s
Verified: yes
Solution
Problem Understanding
We are asked to determine how many kilograms of cloudberries are needed to produce a certain number of jars of jam, given that the jam is made by combining equal amounts of berries and sugar, and that during cooking, 25% of the total mass evaporates. The input gives the number of jars for several test cases, and the output should be the number of kilograms of berries needed for each.
Restating in simpler terms, every jar of jam has a fixed final mass of 3 kilograms. To produce it, we mix an equal mass of berries and sugar, which together sum to more than 3 kg, because 25% evaporates. Algebraically, if we use x kg of berries and x kg of sugar, the final mass is 3x/2 after evaporation. We need this to equal 3 kg per jar. Solving 3x/2 = 3 gives x = 2 kg of berries per jar. Therefore, the problem reduces to multiplying the number of jars by 2 to get the required berries.
Constraints are mild: up to 10^4 test cases and up to 10^8 jars per test case. This immediately suggests that any solution must be O(1) per test case. Iterative or simulation-based approaches are unnecessary and would be far too slow. Edge cases involve the smallest number of jars (n = 1) and the largest possible number (n = 10^8), but the calculation remains simple multiplication, which will not overflow standard 64-bit integers.
A naive mistake would be to incorrectly account for the evaporation fraction or forget to multiply by the number of jars. For example, computing (3 * n) * 0.75 as a float and rounding could lead to off-by-one errors for large n. The correct approach is purely integer arithmetic: 2 * n.
Approaches
The brute-force approach would attempt to simulate making each jar by repeatedly mixing berries and sugar and applying the 25% evaporation. For n = 10^8 jars, this is completely impractical. Observing the problem mathematically reveals a direct formula: each jar requires exactly 2 kg of berries. Once this is identified, the optimal solution is simply to multiply the number of jars by 2.
The key insight is that the problem is linear: the evaporation factor is constant, and each jar is independent. There is no dependency between jars or diminishing returns. This reduces the solution from a potential iterative simulation to a single multiplication per test case. Floating-point operations are unnecessary; integer arithmetic suffices.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force Simulation | O(n) per test case | O(1) | Too slow for large n |
| Direct Multiplication | O(1) per test case | O(1) | Accepted |
Algorithm Walkthrough
- Read the number of test cases
t. - For each test case, read the number of jars
n. - Compute the number of berries required as
2 * n. - Print the result.
Why this works: each jar requires 2 kg of berries, derived from the formula final_mass = 3/4 * total_mass, solving total_mass / 2 * 2/3 = 1 for one jar yields 2 kg of berries. Since each jar is independent, the total is linear in n. Integer arithmetic ensures exact answers without rounding errors.
Python Solution
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
print(2 * n)
The solution uses fast I/O with sys.stdin.readline, although for this problem the standard input() would suffice. Each test case is handled independently, and multiplication by 2 is sufficient to account for the required berries per jar.
Worked Examples
| n | Computation | Result |
|---|---|---|
| 1 | 2 * 1 | 2 |
| 3 | 2 * 3 | 6 |
For n = 1, the single jar requires 2 kg of berries. For n = 3, three jars require 6 kg in total. The trace confirms that the formula 2 * n correctly computes the required mass in every case.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(t) | One multiplication per test case |
| Space | O(1) | No additional memory required per test case |
This is optimal given the input constraints, as we must read t test cases and output a result for each.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
output = io.StringIO()
sys.stdout = output
t = int(input())
for _ in range(t):
n = int(input())
print(2 * n)
sys.stdout = sys.__stdout__
return output.getvalue().strip()
# Provided samples
assert run("2\n1\n3\n") == "2\n6", "sample tests"
# Custom cases
assert run("1\n10\n") == "20", "ten jars"
assert run("1\n100000000\n") == "200000000", "max jars"
assert run("3\n1\n2\n5\n") == "2\n4\n10", "multiple small cases"
assert run("1\n0\n") == "0", "zero jars"
| Test input | Expected output | What it validates |
|---|---|---|
| 1 | 2 | minimum non-zero jars |
| 3 | 6 | small number of jars |
| 100000000 | 200000000 | maximum number of jars, checks large integer handling |
| 0 | 0 | edge case with zero jars |
| multiple small | 2, 4, 10 | multiple test cases |
Edge Cases
For n = 1, the calculation 2 * 1 = 2 correctly handles the smallest positive number of jars. For n = 10^8, 2 * 10^8 is within 64-bit integer limits, so no overflow occurs. The zero-jar case n = 0 produces 0, correctly reflecting that no berries are needed. These confirm the formula is robust across the problem domain.