优化这段代码 n, k = map(int, input().split()) a = list(map(int, input().split())) dp = [[[0]* 7 for _ in range(k + 1)] for _ in range(n + 1)] for i in range(n + 1) : dp[i][0][0] = 1 for i in range(1, n + 1) : for j in range(1, k + 1) : for r in range(7) : dp[i][j][r] = dp[i - 1][j][r] if r >= a[i - 1] % 7 : dp[i][j][r] += dp[i - 1][j - 1][(r - a[i - 1] % 7) % 7] ans = sum(dp[n][k][0::7]) print(ans)
时间: 2023-10-17 14:05:14 浏览: 65
可以将代码进行一些优化,避免重复计算。具体而言,可以将三重循环中的r从0到6分别计算dp[i][j][r]的值,然后再利用dp[i][j][r]计算dp[i][j][(r+a[i-1])%7]的值。这样可以避免重复计算,提高代码的效率。
另外,可以使用一维数组代替二维数组,减少空间复杂度。
修改后的代码如下:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
dp = [0] * 7
dp[0] = 1
for i in range(1, n+1):
for j in range(k, 0, -1):
for r in range(7):
dp[(r+a[i-1])%7] += dp[r]
dp[a[i-1]%7] += 1
print(dp[0] - 1)
```
代码中,dp[r]表示长度为j-1的子序列中和模7的余数为r的方案数。每次更新dp数组时,先更新长度为j的子序列中和模7的余数为a[i-1]%7的方案数,然后再更新长度为j-1的子序列中的方案数,最后计算长度为k且和为7的倍数的子序列个数即可。
需要注意的是,最后计算答案时,要将长度为0的子序列排除,即dp[0]-1为所求。
相关问题
list1=map(int(input().split()))
list1=map(int,input().split())的含义是将用户输入的一行数字字符串按空格分割成多个数字,并将这些数字转化为整型,然后将它们存储在一个迭代器对象中。这个迭代器对象可以通过list()函数转化为列表,或者直接在循环中使用。下面是一个示例代码:
```python
list1 = list(map(int, input().split()))
print(list1)
```
用户输入:1 2 3 4
输出:[1, 2, 3, 4]
n, k, x = map(int, input().split())
Sorry, as an AI language model, I cannot take input from the user. However, I can explain to you what the code does.
The code uses the input function to take input from the user. The input function takes a string as input and returns the string. The split function is used to split the string into a list of strings, which are then converted to integers using the map function.
The variables n, k, and x are assigned the first three elements of the list. The variables represent the number of elements in the sequence, the number of elements to be deleted, and the value of x respectively.