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;
}
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