문제
Use any number of each supplied denomination to make amount exactly. Count combinations, not orderings: two arrangements using the same counts of each coin are the same way. Denominations are distinct and positive.
Implement coin_sums(amount, coins).
Input
amount: The total amount to make.
coins: The available distinct denominations.
Output
Return the number of unordered combinations.
제약 조건
0 ≤ amount ≤ 200. There are 0 through 10 distinct denominations, each from 1 through 200. All counts fit signed 64-bit. Amount 0 has one way: choose no coins.
예시
입력:
coins = [1, 2]
amount = 4
예상 출력:
expected = 3입력:
std::vector<long long> coins = {1, 2};
long long amount = 4;
예상 출력:
long long expected = 3;입력:
long[] coins = new long[] {1, 2};
long amount = 4;
예상 출력:
long expected = 3;설명
The combinations are four 1s, two 1s with one 2, and two 2s.
입력:
coins = [2]
amount = 3
예상 출력:
expected = 0입력:
std::vector<long long> coins = {2};
long long amount = 3;
예상 출력:
long long expected = 0;입력:
long[] coins = new long[] {2};
long amount = 3;
예상 출력:
long expected = 0;설명
Any sum of 2-valued coins is even, so 3 is impossible.
접근 방법
먼저 스스로 풀어 본 뒤 힌트를 하나씩 열어 보세요.
힌트 1
The order of coins does not matter, so combinations should be counted by denomination, not by permutation.
힌트 2
Build the number of ways for each amount incrementally, letting each coin contribute to all reachable totals.
힌트 3
Be careful with amount zero, which always has one valid way: choose no coins.
예상 복잡도
O(amount × number_of_coins) time, O(amount) extra space
참고 자료
이 문제는 CodeUtility에 맞게 수정되었습니다. 위의 설명과 제한 조건을 따르세요.