Problem:
Score : 600 points

Problem Statement

We have an integer sequence A of length N, where A_1 = X, A_{i+1} = A_i + D (1 \leq i < N ) holds.

Takahashi will take some (possibly all or none) of the elements in this sequence, and Aoki will take all of the others.

Let S and T be the sum of the numbers taken by Takahashi and Aoki, respectively. How many possible values of S - T are there?

Constraints

-10^8 \leq X, D \leq 10^8

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

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N X D

Output

Print the number of possible values of S - T.

Sample Input 1

3 4 2

Sample Output 1

8

A is (4, 6, 8).

There are eight ways for (Takahashi, Aoki) to take the elements: ((), (4, 6, 8)), ((4), (6, 8)), ((6), (4, 8)), ((8), (4, 6))), ((4, 6), (8))), ((4, 8), (6))), ((6, 8), (4))), and ((4, 6, 8), ()).

The values of S - T in these ways are -18, -10, -6, -2, 2, 6, 10, and 18, respectively, so there are eight possible values of S - T.

Sample Input 2

2 3 -3

Sample Output 2

2

A is (3, 0). There are two possible values of S - T: -3 and 3.

Sample Input 3

100 14 20

Sample Output 3

49805

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

using System;
using System.Collections.Generic;
using System.Linq;

class F
{
	struct LR
	{
		public long L, R;
	}

	static void Main()
	{
		var h = Console.ReadLine().Split().Select(long.Parse).ToArray();
		long n = h[0], x = h[1], d = h[2];

		if (d == 0) { Console.WriteLine(x == 0 ? 1 : n + 1); return; }
		if (d < 0) { d = -d; x = -x; }

		var map = new Dictionary<long, List<LR>>();
		for (long i = 0; i <= n; i++)
		{
			var l = i * (i - 1) / 2;
			var r = n * (n - 1) / 2 - (n - i) * (n - i - 1) / 2;

			var a = i * x;
			var m = a % d;
			if (m < 0) m += d;
			a = (a - m) / d;

			if (!map.ContainsKey(m)) map[m] = new List<LR>();
			map[m].Add(new LR { L = a + l, R = a + r });
		}

		var c = 0L;
		foreach (var lrs in map.Values)
		{
			var M = long.MinValue;
			foreach (var lr in lrs.OrderBy(s => s.L))
			{
				c += Math.Max(0, lr.R - Math.Max(M, lr.L - 1));
				M = Math.Max(M, lr.R);
			}
		}
		Console.WriteLine(c);
	}
}