Problem:
Score : 100 points

Problem Statement

Find the number of ways to choose a pair of an even number and an odd number from the positive integers between 1 and K (inclusive). The order does not matter.

Constraints

2\leq K\leq 100

K is an integer.

Input

Input is given from Standard Input in the following format:

K

Output

Print the number of ways to choose a pair of an even number and an odd number from the positive integers between 1 and K (inclusive).

Sample Input 1

3

Sample Output 1

2

Two pairs can be chosen: (2,1) and (2,3).

Sample Input 2

6

Sample Output 2

9

Sample Input 3

11

Sample Output 3

30

Sample Input 4

50

Sample Output 4

625

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using System.Numerics;
using static System.Console;

class Program {
    static Scanner sc = new Scanner();

    internal static void Main(string[] args) {
        var N = sc.nextInt();
        var div = N / 2;
        var hoge = Math.Floor((double)div);
        WriteLine((N - hoge) * hoge);
    }
}



class Scanner {
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner() {
        s = new string[0];
        i = 0;
    }

    public string next() {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt() {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0) {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

    public long nextLong() {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0) {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

    public double nextDouble() {
        return double.Parse(next());
    }

    public double[] ArrayDouble(int N, double add = 0) {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}