Problème
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.
Contraintes
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.
Exemples
Entrée :
coins = [1, 2]
amount = 4
Sortie attendue :
expected = 3Entrée :
std::vector<long long> coins = {1, 2};
long long amount = 4;
Sortie attendue :
long long expected = 3;Entrée :
long[] coins = new long[] {1, 2};
long amount = 4;
Sortie attendue :
long expected = 3;Explication
The combinations are four 1s, two 1s with one 2, and two 2s.
Entrée :
coins = [2]
amount = 3
Sortie attendue :
expected = 0Entrée :
std::vector<long long> coins = {2};
long long amount = 3;
Sortie attendue :
long long expected = 0;Entrée :
long[] coins = new long[] {2};
long amount = 3;
Sortie attendue :
long expected = 0;Explication
Any sum of 2-valued coins is even, so 3 is impossible.
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
The order of coins does not matter, so combinations should be counted by denomination, not by permutation.
Indice 2
Build the number of ways for each amount incrementally, letting each coin contribute to all reachable totals.
Indice 3
Be careful with amount zero, which always has one valid way: choose no coins.
Complexité attendue
O(amount × number_of_coins) time, O(amount) extra space
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.