📘 Day 7 – Teacher's Day Challenge | Watermelon Division Logic

🗓️ Date: July 31, 2025
🔢 Problems Solved Today: 1
🎯 Focus Topic: Conditional Logic, Input/Output, Edge Cases

💻 Platform: Codeforces

---

🔗 Table of Contents

• Problem 1 – Watermelon

• Daily Summary

---

🧩 Problem 1 – Watermelon

🔗 Codeforces Problem – Watermelon #4A
📚 Difficulty: Easy
🧠 Concepts: Even/Odd Numbers, Conditionals, Edge Case Handling

---

📄 Problem Statement (Summary):
Given a watermelon of weight w (1 ≤ w ≤ 100), determine if it can be split into two parts, both with even positive weights (not necessarily equal). Print "YES" or "NO".

---

💡 Approach:

• Input the weight w.

• Check two conditions:

  • w should be greater than 2 (to avoid cases like 1 or 2 which can’t be split into two even parts).

  • w should be even (since even + even = even).

• If both conditions are true, print "YES"; else, print "NO".

---

🧪 Code (C++):

#include<iostream>
using namespace std;
int main(){
    int w;
    cin >> w;
    if(w > 2 && w % 2 == 0)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
    return 0;
}

---

📸 Submission Screenshot:

---

✅ Key Takeaways:

• Always handle edge cases like smallest numbers (1 and 2).

• Use % operator for evenness check.

• w > 2 is critical since 2 is even but not splittable into two positive even parts.

• Simple if-else logic is enough for many Codeforces problems.

---

📈 Daily Summary

---

🏷️ Tags:
#codeforces #conditionals #cpp #evennumbers #logicbuilding #edgecases #43DaysChallenge