迭代器算法

Question

I'm working right now with iterators arithmetic operations and stack on small problem .
I need to make a Sum of first and last element of vector<int> followed by second and last element of vector<int> , third and last element of vector<int>
Example:
Input numbers by user
1 2 3 4 5 6 7 8 9
Output should be
10 11 12 13 14 15 16 17
In general the code should do addition like that
1+9 2+9 3+9 4+9 5+9 6+9 7+9 ......

So basically i need the actual code for this arithmetic operation using iterator with member functions *.begin() , *.end() only ! I've try many ways but nothing coming in my head how to do this operation only with .begin() and .end() . I found other member functions but this functions is explained in STD library, not in basic knowledge level. So i need help to make code with only begin() and end() member functions if possible.

Code i got so far

int main()
{  

vector<int> numset;
int num_input;
auto beg=numset.begin(), end=numset.end();
while (cin>>num_input)
{
    numset.push_back(num_input);
}
for (auto it = numset.begin()+1; it !=numset.end(); ++it)
{
    // *it=*it+1+nuset.end(); -- Wrong  X
            // *it+=(end-beg)/2;      -- Totally wrong(and totally stupid) X
            // *it + numset.back()   -- can't use other member functions X 
    //////// I've stack here dont know what code need //////

              cout<<*it<<endl;    
}

Thank you for your time.

Answer 1

The operation you perform is *it+*(it-1) . (It might help to add more parentheses and spaces in your code.) That adds two adjacent elements from the sequence.

The last element in the sequence is numset.back() . So try *it + numset.back() instead. And there's no need to start with the second element, since you do want to print the sum of the first and last elements. If you don't want to print the sum of the last element with itself, you should stop at end() - 1 , though.