英文:
Anyway to make this code goes faster in python?
问题
以下是您提供的代码的翻译部分：
from sys import setrecursionlimit
setrecursionlimit(10 ** 6)
MOD = 1000000007
dp = [0] * 1000001
def DP(dp, left):
if(dp[left] != 0):
return dp[left]
for i in range(1, 7):
if(left - i >= 0):
dp[left] += DP(dp, left - i)
dp[left] %= MOD
return dp[left]
n = int(input())
dp[0] = 1
print(DP(dp, n))
请注意，这段代码的翻译并没有进行性能优化。如果您希望提高代码的运行速度，可以考虑使用迭代而不是递归来实现动态规划，这通常会更快。此外，您还可以尝试使用其他算法来解决问题，以提高性能。
英文:
from sys import setrecursionlimit
setrecursionlimit(10 ** 6)
MOD = 1000000007
dp = [0] * 1000001
def DP(dp, left):
if(dp[left] != 0):
return dp[left]
for i in range(1, 7):
if(left - i >= 0):
dp[left] += DP(dp, left - i)
dp[left] %= MOD
return dp[left]
n = int(input())
dp[0] = 1
print(DP(dp, n))
Is there anyway to make this code goes faster? It can run when n is around 100000 but it was slowdown when n=654321. The task is to count the way to combining 1, 2, 3, 4, 5, 6 (a dice's faces) to make n. Example: when n=3, all the way to create n is 4: 1+1+1; 2+1; 1+2; 3. I was trying to use Dynamic Programming skill to solve but timeout. Thanks for your help!
答案1
得分: 0
让我们尝试使用自底向上的表格法，我们迭代地首先计算较小问题的值，然后使用这些值来计算该值。
def DP(n):
MOD = 1000000007
dp = [0] * (n + 1)
dp[0] = 1
for i in range(1, n+1):
for j in range(1, 7):
if i >= j:
dp[i] += dp[i - j]
dp[i] %= MOD
return dp[n]
n = int(input())
print(DP(n))
英文:
Lets try to use a bottom-up, tabulation method, we iteratively compute the values for smaller problems first and then use these values to compute the value.
def DP(n):
MOD = 1000000007
dp = [0] * (n + 1)
dp[0] = 1
for i in range(1, n+1):
for j in range(1, 7):
if i >= j:
dp[i] += dp[i - j]
dp[i] %= MOD
return dp[n]
n = int(input())
print(DP(n))
答案2
得分: 0
内部循环可以通过使用前缀和进行优化。使用长度为 n+1 的数字列表来存储前缀和。
def DP(n):
MOD = 1000000007
dp = [0] * (n + 1)
dp[0] = 1
for i in range(1, n+1):
for j in range(1, min(i, 6) + 1):
dp[i] = (dp[i] + dp[i-j]) % MOD
return dp[n]
n = int(input())
print(DP(n))
英文:
The inner loop can be optimized by using a prefix sums. A list of numbers with list length of n+1 to store the prefix sums.
def DP(n):
MOD = 1000000007
dp = [0] * (n + 1)
dp[0] = 1
for i in range(1, n+1):
for j in range(1, min(i, 6) + 1):
dp[i] = (dp[i] + dp[i-j]) % MOD
return dp[n]
int(input())
print(DP(n))
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