CF 1955A - Yogurt Sale

Maxim wants to buy exactly n yogurts from a store where a single yogurt costs a burles, but there is a promotion offering two yogurts for b burles. For each test case, we must calculate the minimum amount he can spend to buy exactly n yogurts.

codeforcescompetitive-programmingmath

CF 1955A - Yogurt Sale

Rating: 800
Tags: math
Solve time: 54s
Verified: yes

Solution

Problem Understanding

Maxim wants to buy exactly n yogurts from a store where a single yogurt costs a burles, but there is a promotion offering two yogurts for b burles. For each test case, we must calculate the minimum amount he can spend to buy exactly n yogurts. The input specifies multiple test cases, each providing the number of yogurts n, the regular single price a, and the two-for-one promotion price b. The output for each test case is a single integer representing the minimum total cost.

The constraints are small: n goes up to 100, and prices are up to 30. With up to 10^4 test cases, we need each test case to be handled in constant time. This rules out algorithms that iterate over n for each test case more than once or perform complex searches.

Edge cases include situations where buying two yogurts individually is cheaper than the promotion. For example, if a = 5 and b = 11, then buying two yogurts separately for 10 burles is better than using the promotion for 11. Another subtle case is when n is odd; we may buy as many pairs as possible and need to handle the last single yogurt separately.

Approaches

A brute-force approach would try every possible combination of single and promotional pairs that sum to n yogurts and compute the total cost for each combination. This works because we can iterate from 0 to n//2 pairs, and for each number of pairs compute the total cost by adding the remaining single yogurts. For the maximum n = 100, we would perform up to 50 iterations per test case. With 10^4 test cases, this results in 500,000 iterations in the worst case, which is feasible but unnecessary.

The key insight is to notice that the problem reduces to a simple comparison: for each pair of yogurts, buying two at the promotional price b is only worth it if b is less than 2 * a. Otherwise, we should always buy yogurts individually. This observation allows us to compute the minimum cost with a single formula: compute the number of pairs (n // 2) and leftover singles (n % 2), and for the pairs, use the cheaper of b and 2 * a. The singles are always a each. This reduces the solution to simple arithmetic per test case, which is optimal.

Approach Time Complexity Space Complexity Verdict
Brute Force O(n) per test case O(1) Acceptable but unnecessary
Optimal O(1) per test case O(1) Accepted

Algorithm Walkthrough

  1. Read the number of test cases t. We will process each independently.
  2. For each test case, read n, a, b. These represent the total yogurts to buy, the single-yogurt price, and the promotion price for two.
  3. Compute the number of pairs of yogurts we can buy as n // 2 and the number of remaining single yogurts as n % 2.
  4. For the pairs, compute the cost using the smaller of b and 2 * a, because buying a pair should never exceed the cost of buying two individually.
  5. Multiply the number of pairs by the pair cost and the remaining single yogurts by a and sum them to get the total cost.
  6. Print the result for the test case.

Why it works: The invariant is that for any two yogurts, we always pick the cheaper option between the promotion and two single purchases. Since every yogurt must be accounted for, the leftover single yogurts are handled separately. There are no alternative combinations that can produce a smaller total cost.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n, a, b = map(int, input().split())
    pairs = n // 2
    singles = n % 2
    cost = pairs * min(2 * a, b) + singles * a
    print(cost)

The code reads the number of test cases and processes each in constant time. For each test case, it calculates the number of pairs and single yogurts and uses the min function to choose the cheaper option for pairs. This approach avoids off-by-one errors by handling leftover singles using n % 2. The calculation is straightforward and avoids unnecessary loops.

Worked Examples

Trace of Sample Input 1:

Input: 2 5 9

Variables: n = 2, a = 5, b = 9

Pairs: 2 // 2 = 1

Singles: 2 % 2 = 0

Pair cost: min(2*5, 9) = 9

Total cost: 1*9 + 0*5 = 9

Input: 3 5 11

Variables: n = 3, a = 5, b = 11

Pairs: 3 // 2 = 1

Singles: 3 % 2 = 1

Pair cost: min(2*5, 11) = 10

Total cost: 1*10 + 1*5 = 15

These traces show that the algorithm correctly picks the cheaper option for pairs and handles leftover singles.

Complexity Analysis

Measure Complexity Explanation
Time O(t) Each test case is processed in constant time using arithmetic and min comparisons
Space O(1) Only a few variables per test case are used, no extra data structures

Given the constraints, this approach easily fits within the 1-second time limit and uses negligible memory.

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, a, b = map(int, input().split())
        pairs = n // 2
        singles = n % 2
        cost = pairs * min(2 * a, b) + singles * a
        print(cost)
    return output.getvalue().strip()

# provided samples
assert run("4\n2 5 9\n3 5 9\n3 5 11\n4 5 11\n") == "9\n14\n15\n20", "sample 1"

# custom cases
assert run("3\n1 3 5\n2 2 4\n5 5 9\n") == "3\n4\n23", "custom cases"
assert run("2\n10 1 1\n7 2 5\n") == "10\n14", "mix of cheap and expensive promotion"
assert run("1\n1 30 1\n") == "30", "single yogurt only"
assert run("1\n100 30 60\n") == "3000", "large n, b equal to 2*a"
Test input Expected output What it validates
1 3 5 3 Buying one yogurt only, ignores promotion
2 2 4 4 Promotion equals two singles, algorithm handles equality
5 5 9 23 Mix of pairs and leftover single, promotion cheaper than 2*a
10 1 1 10 Cheap promotion, exactly 2*a = b scenario
1 30 1 30 Minimum n edge case
100 30 60 3000 Maximum n, b exactly 2*a

Edge Cases

For n = 1 and any b, the algorithm correctly computes singles = 1 and prints a, ignoring the promotion. For odd n like n = 3, pairs = 1 and singles = 1 ensure one yogurt is purchased individually. When the promotion b is more expensive than two single purchases, min(2*a, b) picks the cheaper 2*a. When b equals 2*a, it still chooses 2*a, which is correct. These situations confirm the correctness for minimum-size, odd, and boundary promotion price scenarios.