[leetcode] 1018. Binary Prefix Divisible By 5

Description

Given an array A of 0s and 1s, consider N_i: the i-th subarray from A[0] to A[i] interpreted as a binary number (from most-significant-bit to least-significant-bit.)

Return a list of booleans answer, where answer[i] is true if and only if N_i is divisible by 5.

Example 1:

Input: [0,1,1]
Output: [true,false,false]
Explanation: 
The input numbers in binary are 0, 01, 011; which are 0, 1, and 3 in base-10.  Only the first number is divisible by 5, so answer[0] is true.

Example 2:

Input: [1,1,1]
Output: [false,false,false]

Example 3:

Input: [0,1,1,1,1,1]
Output: [true,false,false,false,true,false]

Example 4:

Input: [1,1,1,0,1]
Output: [false,false,false,false,false]

Note:

1. 1 <= A.length <= 30000.
2. A[i] is 0 or 1.

分析

题目的意思是：给定一个数组，判断一第i个数为结尾的二进制数能够被5整除，返回一个bool数组。这道题我想着暴力破解一下，发现A的长度能够达到30000，按照直觉这个地方会存在溢出，结果却没有溢出，python太厉害了啊，膜拜一下，如果不溢出那就能够暴力破解了哈。这道题出得有点失败啊，
更好的解法因该是：
https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/800137/C%2B%2B-simple-Transition-graph-99-fast

有机会学学这个能够被5整除的状态转移图

代码

class Solution:
    def prefixesDivBy5(self, A: List[int]) -> List[bool]:
        n=len(A)
        prev=0
        res=[]
        for i in range(n):
            prev=prev<<1
            cur=prev+A[i]
            if(cur%5==0):
                res.append(True)
            else:
                res.append(False)
            prev=cur
        return res

参考文献

[LeetCode] python easy to understand and 100% memory, runtime beats 70.53 %