while(done.count(curr)){
        curr++;
    }

As you are searching in linear time

while(done.count(curr)){
    curr++;
}
.
.
.
if(curr-prev==a[i]){
    cout<<prev<<" ";
    done.insert(prev);
    prev=curr;
}

As you are updating curr to a smaller value and then recalculating curr in the next loop again, the time complexity in worst case can go to $$$O(n^2)$$$