Approach

# 陣列向左移就是*=2
Appending a digit to the right of a binary number is equivalent to multiplying the current value by 2.

# 小心溢位
Be careful with overflow.

Time Complexity

# n = length of nums
O(n)

Space Complexity

# n = length of nums
O(n)

Code

class Solution {
public:
    vector<bool> prefixesDivBy5(vector<int>& nums) {
        int l = nums.size(), sum = 0;
        vector<bool> ans;

        for (int i = 0 ; i < l ; i++) {
            sum %= 5;
            sum *= 2;
            sum += nums[i];
            if (sum % 5 == 0) {
                ans.push_back(true);
            }
            else {
                ans.push_back(false);
            }
        }

        return ans;
    }
};