The simple problem of leetcode: dismantling bombs

subject

You have a bomb to dismantle , Pressed for time ！ Your agent will give you a length of n Of loop Array code And a key k .
To get the right password , You need to replace every number . All the numbers will meanwhile Be replaced .
If k > 0 , Will be the first i For numbers Next k The sum of the numbers replaces .
If k < 0 , Will be the first i For numbers Before k The sum of the numbers replaces .
If k == 0 , Will be the first i For numbers 0 Replace .
because code It's cyclical , code[n-1] The next element is code[0] , And code[0] The first element is code[n-1] .
Here you are. loop Array code And integer keys k , Please return the decrypted results to dismantle the bomb ！
Example 1：
Input ：code = [5,7,1,4], k = 3
Output ：[12,10,16,13]
explain ： Every number is followed by 3 The sum of the numbers replaces . The decrypted password is [7+1+4, 1+4+5, 4+5+7, 5+7+1]. Notice that arrays are concatenated circularly .
Example 2：
Input ：code = [1,2,3,4], k = 0
Output ：[0,0,0,0]
explain ： When k by 0 when , All the numbers are 0 Replace .
Example 3：
Input ：code = [2,4,9,3], k = -2
Output ：[12,5,6,13]
explain ： The decrypted password is [3+9, 2+3, 4+2, 9+4] . Notice that arrays are concatenated circularly . If k It's a negative number , So, and for Before The number of .
Tips ：
n == code.length
1 <= n <= 100
1 <= code[i] <= 100
-(n - 1) <= k <= n - 1
source ： Power button （LeetCode）

Their thinking

   This question mainly focuses on the access of circular arrays , The calculation of cyclic numbers is a relatively basic content in number theory , This kind of circular access often occurs in the circular queue . In this question, we only need to answer code Each element in can be modified one by one . notes ： The modified element does not participate in the new operation , So we need to open up a new space to store the answers .

class Solution:
    def decrypt(self, code: List[int], k: int) -> List[int]:
        n=len(code)
        CODE=[0]*n
        if k==0:
            return CODE
        elif k>0:
            for i in range(n):
                s=0
                for j in range(k):
                    s+=code[(i+j+1)%n]
                CODE[i]=s
        else:
            for i in range(n):
                s=0
                for j in range(-k):
                    s+=code[(i+n-j-1)%n]
                CODE[i]=s
        return CODE