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 Rust (1.42.0) (judge verdict: Accepted)

fn main() {
    let (n, x, d) = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        let mut iter = buf.split_whitespace();
        (
            iter.next().unwrap().parse::<usize>().unwrap(),
            iter.next().unwrap().parse::<i64>().unwrap(),
            iter.next().unwrap().parse::<i64>().unwrap(),
        )
    };

    if d == 0 {
        if x == 0 {
            println!("1");
        } else {
            println!("{}", n + 1);
        }
        return;
    }

    let x = if d < 0 { -x } else { x };
    let d = if d < 0 { -d } else { d };

    let mut map = std::collections::HashMap::new();
    for k in 0..=n {
        let n = n as i64;
        let k = k as i64;
        let key = k * x % d;
        let range = {
            [
                k * x + (k - 1) * k / 2 * d,
                k * x + ((n - k) + (n - 1)) * k / 2 * d,
            ]
        };
        (*map.entry(key).or_insert(Vec::new())).push(range);
    }

    let mut ans = 0;
    for value in map.values_mut() {
        value.sort();
        ans -= 1;
        let mut left = std::i64::MIN;
        let mut right = std::i64::MIN;
        for range in value {
            if range[0] <= right {
                right = std::cmp::max(right, range[1]);
            } else {
                ans += (right - left) / d + 1;
                left = range[0];
                right = range[1];
            }
        }
        ans += (right - left) / d + 1;
    }
    println!("{}", ans);
}