力扣 LeetCode 1018. 可被 5 整除的二进制前缀

令sum为N_i%5的的余数，那么N_i+1的sum为N_i的sum*2+A[i+1]，然后利用该式子即可求出答案

class Solution {
     
public:
    vector<bool> prefixesDivBy5(vector<int>& A) {
     
        vector<bool>res;
        int index=0;
        int sum=0;
        while(index<A.size()){
     
            sum=sum*2%10;
            sum+=A[index];
            if(sum%5==0){
     
                res.push_back(true);
            }
            else{
     
                res.push_back(false);
            }
            index++;
        }
        return res;
    }
};