Problem:
Score : 300 points

Problem Statement

You are given integers N and K. Find the number of triples (a,b,c) of positive integers not greater than N such that a+b,b+c and c+a are all multiples of K.
The order of a,b,c does matter, and some of them can be the same.

Constraints

1 \leq N,K \leq 2\times 10^5

N and K are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of triples (a,b,c) of positive integers not greater than N such that a+b,b+c and c+a are all multiples of K.

Sample Input 1

3 2

Sample Output 1

9

(1,1,1),(1,1,3),(1,3,1),(1,3,3),(2,2,2),(3,1,1),(3,1,3),(3,3,1) and (3,3,3) satisfy the condition.

Sample Input 2

5 3

Sample Output 2

1

Sample Input 3

31415 9265

Sample Output 3

27

Sample Input 4

35897 932

Sample Output 4

114191

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ABC
{
    class Program
    {
        static void Main(string[] args)
        {
            var split = Console.ReadLine().Split(' ');
            int N = int.Parse(split[0]);
            int K = int.Parse(split[1]);
            //準備完了
            long output = 0;
            for (int a = 1; a <= K; a++)
            {
                //何周する？3,2のときは1週+1  N/K周して、aがn%K以下だったらもう一周分あるとして足せばいい
                //bのパターンは何通り？ K-aが最初で、Nを超えないような整数は  (N-K+a)/K + 1　個
                //ほんでもし、2*aがKを超えないようならば、Cを選んで良くなるので、
                if (2 * a % K == 0)
                {
                    int additional_minusone = 0;
                    if (K == a)
                    {
                        additional_minusone = -1;
                    }
                    int additional_one = (N % K >= a) ? 1 : 0;
                    output += (additional_one + N / K) * (int)Math.Pow(((N - K + a) / K + 1+additional_minusone),2);
                }

            }

            /*
            for (int a_i = 1; a_i <= N; a_i++)
            {
                //bの初期値はつまりa_iと足してKの倍数になる最小の数
                for (int b_i = ((a_i / K) * (K) + K - a_i); b_i <= N; b_i += K)
                {
                    if (2 * a_i % K == 0)
                    {
                        output += (N - (K - (b_i % K))) / K + 1;
                       // Console.WriteLine(a_i + ","+ b_i + "が" + ((N - (K - (b_i % K))) / K + 1) + "個");
                    }



                } 
            }
            */
            Console.WriteLine(output);
        }

    }
}