the solution of CountNonDivisible by Codility

question:https://codility.com/programmers/lessons/9

To solve this question , I get each element's divsors which appearing in input Array A using Sieve of Eratosthenes method. Time complexity is O(nlogn);

Then  we iterate array A to get the ith non-divsors by A.size() - count(element) for element in divsor[A[i]] in divsors.  Time complexity is O(n*?

); ?

represent the average of divsors

this method unsatisfy the time requirement , for two test case get TIMEOUT error.  NEED IMPROVE IT LATER.

code:

#include <algorithm>
#include <map>
//this method not fast enough
vector<int> solution(vector<int> &A) {
    // write your code in C++11
    map<int,int> dic;
    map<int,vector<int> > divsors;
    int size = A.size();
    int max = *max_element(A.begin(),A.end());
    for(int i=0; i<size; i++){
        dic[A[i]]++;
        if(divsors.count(A[i])==0){
            vector<int> vec(1,1);
            divsors.insert(make_pair(A[i],vec));
        }
    }

    for(int i=2; i<= max; i++){
        int element = i;
        while(element <=max){
            if(divsors.count(element)!=0 &&  find(divsors[element].begin(),divsors[element].end(),i)==divsors[element].end()){
                divsors[element].push_back(i);
            }
            element+=i;
        }
    }
    vector<int > res;
    for(int i=0; i<size; i++){
        vector<int> t = divsors[A[i]];
        int cnt = size;
        for(int j=0; j<t.size(); j++){
            cnt -= dic[t[j]];
        }
        res.push_back(cnt);
    }
    return res;
}