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;}