# 1018. Binary Prefix Divisible By 5 可被 5 整除的二进制前缀

@TOC

## # 题目描述

Given an array `A` of `0`s and `1`s, 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`

## # 解题方法

``````((a +b)mod p × c) mod p = ((a × c) mod p + (b × c) mod p) mod p
(a×b) mod c=((a mod c) * (b mod c)) mod c
(a+b) mod c=((a mod c)+ (b mod c)) mod c
(a-b) mod c=((a mod c)- (b mod c)) mod c
``````

Python代码如下：

``````class Solution(object):
def prefixesDivBy5(self, A):
"""
:type A: List[int]
:rtype: List[bool]
"""
res = []
prefix = 0
for a in A:
prefix = (prefix * 2 + a) % 5
res.append(prefix == 0)
return res
``````

## # 日期

2019 年 4 月 5 日 —— 清明节休息一下～