1018. Binary Prefix Divisible By 5

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

难度：easy
解题思路：本题解题思路比较容易想，这里只说测试点的问题，若直接计算加和，会发现num超过int表示范围，因为A.length上限是30000，所有计算的num最大值是2^30000-1.但是我们进一步分析，判断num是否被5整除，我们只用关心个位数能否整除5即可，因此计算的时候num只存放num的个位数就行了。
代码：

vector<bool> prefixesDivBy5(vector<int>& A) {
  int num=0; 
  int len=A.size();
  vector<bool> ans;
  for(int i=0;i<len;i++){
    num=(num*2+A[i])%10;
    ans.push_back(num);
  }
  return ans;
}