Problem A. Binomial Coefficient
Let's use the Memoization principle here. Let dp[i][j] be a Binomial Coefficient C(i,j). First, dp initialized with None.
Given dp[n][k]:
- if it is None, calculate it as
c(n-1, k-1) + c(n-1, k) - if it is a base case:
k==0 or k==n, thendp[n][k] = 1 - else return
dp[n][k]
Note that structure dp depends on n, and you must use it for defined n.
123456789101112131415n = 200 dp = [[None for _ in range(n+1)] for _ in range(n+1)] def c(n, k): if k==0 or k==n: dp[n][k] = 1 if dp[n][k] == None: dp[n][k] = c(n-1, k)+c(n-1, k-1) return dp[n][k] print(c(3, 2)) print(c(10, 4)) print(c(11, 5)) print(c(144, 7))
Was alles duidelijk?
Bedankt voor je feedback!
Sectie 3. Hoofdstuk 1
single
Vraag AI
Vraag AI
Vraag wat u wilt of probeer een van de voorgestelde vragen om onze chat te starten.
Suggested prompts:
Vat dit hoofdstuk samen
Explain code
Explain why doesn't solve task
Awesome!
Completion rate improved to 8.33Problem A. Binomial Coefficient
Veeg om het menu te tonen
Let's use the Memoization principle here. Let dp[i][j] be a Binomial Coefficient C(i,j). First, dp initialized with None.
Given dp[n][k]:
- if it is None, calculate it as
c(n-1, k-1) + c(n-1, k) - if it is a base case:
k==0 or k==n, thendp[n][k] = 1 - else return
dp[n][k]
Note that structure dp depends on n, and you must use it for defined n.
123456789101112131415n = 200 dp = [[None for _ in range(n+1)] for _ in range(n+1)] def c(n, k): if k==0 or k==n: dp[n][k] = 1 if dp[n][k] == None: dp[n][k] = c(n-1, k)+c(n-1, k-1) return dp[n][k] print(c(3, 2)) print(c(10, 4)) print(c(11, 5)) print(c(144, 7))
Schakel over naar desktop voor praktijkervaringGa verder vanaf waar je bent met een van de onderstaande optiesWas alles duidelijk?
Bedankt voor je feedback!
Awesome!
Completion rate improved to 8.33Sectie 3. Hoofdstuk 1
single
