Hi everyone 👋🏻

Here is solution of #Day2 of #30daysofcode by Newton School & 30 Days Of Code 🚀

Problem name: Edward and Maths Competition

Link: https://my.newtonschool.co/playground/code/0vb6uf9tra6m

✅Approach: Solved using simple maths concept. In series 1,2,...n, we've n/2 even numbers and (n/2)+1 odd numbers. So we'll multiply these to get the number of possible pairs.

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How to Submit Code: youtu.be/lhtgBI9wMCY

  #include <bits/stdc++.h>
  using namespace std;
  int main() {
      int n;
      cin >> n;
      int even = n/2;
      int odd = n-even;
      cout << even*odd;
      return 0;
  }