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

解题思路

本题很简单，主要考虑一下溢出情况。

(n * 2 + m) % 5 = ((n % 5) * 2 + m) % 5

实现代码

// Runtime: 3 ms, faster than 97.49% of Java online submissions for Binary Prefix Divisible By 5.
// Memory Usage: 40.2 MB, less than 100.00% of Java online submissions for Binary Prefix Divisible By 5.
class Solution {
    public List<Boolean> prefixesDivBy5(int[] A) {
        List<Boolean> result = new ArrayList<>(A.length);
        int num = 0;
        for (int a : A) {
            num = (num << 1 | a) % 5;
            if (num == 0) {
                result.add(true);
            } else {
                result.add(false);
            }
        }
        return result;
    }
}