Adding Integers In A Vector - c++

Question

Ok, I have another challenge. I have a question that asks me to "Read a set of integers into a vector. Print the sum of each pair of adjacent elements . Change your program so that it prints out the sum of the first and last elements, followed by the sum of the second and second-to-last, and so on."
Currently I have managed to get it to add consecutive numbers, not exactly in pairs, but as close as I can get it.
The only operators I have learnt thus far for vectors, and what the question wants me to use are: v.empty(); v.size(); v.push_back(t); v[n]; v1 = v2; v1 = {a,b,c...}; v1 == v2; v1 != v2; <, <=, >, >=
Here is my code as it stands now:
#include <iostream>
#include <string>
#include <vector>
using std::string; using std::vector; using std::cout; using std::cin; using std::endl;
int main ()
{
vector<unsigned> numbers {5, 6, 7, 8, 9, 10, 11};
unsigned sum;
decltype(numbers.size()) count;
for (count = 0; count < numbers.size(); ++count){
sum = (numbers[count]++) + (numbers[count]);
cout << sum << " ";
}
}
So I can get it to add 5+6; 6+7; 7+8 etc... but as for each pair and all the rest, not 100% sure where to start. Any guidance will be greatly appreciated!

Answer 1

To print the sum of each pair of adjacent elements.
auto count = numbers.size();
for (auto i = 0; i < numbers.size()-1; i++)
{
std::cout << (numbers[i] + numbers[i+1]) << "\t";
}
prints out the sum of the first and last elements, followed by the sum of the second and second-to-last, and so on
for (auto i = 0; i < count/2; ++i)
{
std::cout << (numbers[i] + numbers[count - i -1]) << "\t";
}
Note: if vector has odd number, the middle one won't be calculated in second print
See sample output

Answer 2

Below two loops that print (1) sum of adjacent pairs and (2) the sum of opposite elements.
#include <cstddef>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int main ()
{
vector<unsigned> numbers {5, 6, 7, 8, 9, 10, 11};
auto N = numbers.size();
std::cout << "first part\n";
for (size_t i = 1; i < N; ++i) {
std::cout << numbers[i-1] + numbers[i] << "\n";
}
std::cout << "second part\n";
for (size_t i = 0; i < (N - 1 - i); ++i) {
std::cout << numbers[i] + numbers[N - 1 - i] << "\n";
}
// in case you have an odd number of elements
if (N % 2 == 1) std::cout << numbers[N / 2] << "\n";
}
Output on LiveWorkSpace

Answer 3

I think your current code won't work for different examples. You need to change numbers[count]++ to numbers[count+1] for it to work, and change the loop's end condition to size()-1.
As for your other question, you should be able to do it if you only loop through half the vector and add the numbers of v[i] + v[v.size() - i - 1]
Edit: Added -1 to v[v.size() - i - 1] as suggested by borisbn in the comments

Answer 4

Hi I tried this on VS and it worked
vector <int> num;
num.push_back(5);
num.push_back(6);
num.push_back(7);
num.push_back(8);
num.push_back(9);
num.push_back(10);
int count = 0;
for(;count < (num.size()-1);count++)
{
int sum =0;
sum = num[count] + num[count+1];
cout<<sum<<endl;
}
for(int c =0; c < (num.size()-1);c++)
{
int altrsum = num[c]+ num[(num.size()-1)-c];
cout<<altrsum<<endl;
}
Hope this helps.