문제
The decorator pattern adds behaviour by wrapping an object in layers. Each layer here changes the price of the layer inside it.
Start from base_price and apply the decorators in order:
discount p: reduce the price by p percent.tax p: increase the price by p percent.shipping a: add a fixed amount a.coupon a: subtract a, but never go below 0.
Implement price_decorators(base_price, decorators).
Input
base_price: The starting price.
decorators: The layers, in the order they wrap the price. Amounts may have decimals.
Output
Return the final price. Answers within 1e-6 are accepted.
제약 조건
0 ≤ base_price ≤ 1,000,000
0 ≤ number of decorators ≤ 100; 0 ≤ p ≤ 100, 0 ≤ a ≤ 100,000
예시
입력:
base_price = 100
decorators = ["discount 10", "tax 8", "shipping 5"]
예상 출력:
expected = 102.2입력:
long long base_price = 100;
std::vector<std::string> decorators = {"discount 10", "tax 8", "shipping 5"};
예상 출력:
double expected = 102.2;입력:
long base_price = 100;
string[] decorators = new string[] {"discount 10", "tax 8", "shipping 5"};
예상 출력:
double expected = 102.2;설명
100 − 10% = 90, plus 8% tax = 97.2, plus 5 shipping = 102.2.
입력:
base_price = 20
decorators = ["coupon 25", "shipping 4"]
예상 출력:
expected = 4.0입력:
long long base_price = 20;
std::vector<std::string> decorators = {"coupon 25", "shipping 4"};
예상 출력:
double expected = 4.0;입력:
long base_price = 20;
string[] decorators = new string[] {"coupon 25", "shipping 4"};
예상 출력:
double expected = 4.0;설명
The coupon would take the price below zero, so it stops at 0; shipping then makes it 4.
접근 방법
먼저 스스로 풀어 본 뒤 힌트를 하나씩 열어 보세요.
힌트 1
Apply the layers strictly in the given order.
힌트 2
Percentages are taken from the current price, not the base price.
힌트 3
A coupon can never push the price below 0.
예상 복잡도
O(number of decorators) time, O(1) extra space
참고 자료
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