Sum of an Array by Divide and Conquer Technique

#include<bits/stdc++.h>

#define mx 10001

#define s start

#define e end

using namespace std;

int a[mx], tree[mx*3];

void sum(int node, int s, int e)

{

    if(s == e)

    {

        tree[node] = a[s];

        return;

    }

    int left = node*2;

    int right = node*2 + 1;

    int mid = (s+e)/2;

    sum(left, s, mid);

    sum(right, mid+1, e);

    tree[node] = tree[left] + tree[right];

}

int main()

{

    int n;

    cout << "Enter Number of Array Element :: ";

    cin >> n;

    cout << "Enter Array Element :: ";

    for(int i=1;i<=n;i++)

        cin >> a[i];

    sum(1, 1, n);

    cout << "Summation is :: " << tree[1] << "\n";

    return 0;

}