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4A. Watermelon – CodeForces Solution

Solution in__

Python:


w = int(input())  # Reading the input
if w % 2 == 0 and w > 2:
    print("YES")
else:
    print("NO")
Python

C#:


using System;

class Program
{
    static void Main()
    {
        int w = int.Parse(Console.ReadLine());
        if (w % 2 == 0 && w > 2)
        {
            Console.WriteLine("YES");
        }
        else
        {
            Console.WriteLine("NO");
        }
    }
}
C#

Go:


package main

import (
    "fmt"
)

func main() {
    var w int
    fmt.Scan(&w)
    if w%2 == 0 && w > 2 {
        fmt.Println("YES")
    } else {
        fmt.Println("NO")
    }
}
Go

Ruby:


w = gets.to_i
if w.even? && w > 2
  puts "YES"
else
  puts "NO"
end
Ruby

Rust:


use std::io;

fn main() {
    let mut input = String::new();
    io::stdin().read_line(&mut input).unwrap();
    let w: i32 = input.trim().parse().unwrap();
    
    if w % 2 == 0 && w > 2 {
        println!("YES");
    } else {
        println!("NO");
    }
}
Rust

Kotlin:


fun main() {
    val w = readLine()!!.toInt()
    if (w % 2 == 0 && w > 2) {
        println("YES")
    } else {
        println("NO")
    }
}
Kotlin

These solutions are straightforward, where we check if the weight w is an even number and greater than 2 to ensure the watermelon can be split into two even parts. If true, the output is “YES”; otherwise, it’s “NO”.

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CodeForces

April 7, 2024