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 (1.15.1) (judge verdict: Wrong Answer)

// https://atcoder.jp/contests/abc147/tasks/abc147_f
//
#![allow(unused_imports)]
use std::io::*;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };

    ($iter:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, [ next / $t:tt ]) => {
        {
            let len = read_value!($iter, usize);
            (0..len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
        }
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_macros)]
macro_rules! dvec {
    ($t:expr ; $len:expr) => {
        vec![$t; $len]
    };

    ($t:expr ; $len:expr, $($rest:expr),*) => {
        vec![dvec!($t; $($rest),*); $len]
    };
}

#[allow(unused_macros)]
macro_rules! ifv {
    ($t:expr, $a:expr, $b: expr) => {
        if $t { $a } else { $b }
    }
}

#[allow(unused_macros)]
macro_rules! fill {
    ($t:expr, $v:expr) => {
        for i in 0..$t.len() {
            $t[i] = $v;
        }
    };
}

#[allow(unused_macros)]
macro_rules! join {
    ($t:expr, $glue:expr) => {
        $t.into_iter().map(|w| w.to_string()).collect::<Vec<_>>().join($glue)
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        println!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

type Line = (i64, i64); // first, num

fn nth(x: i64, d: i64, n: i64) -> i64 {
    x + d * (n-1)
}

fn solve(n: i64, x: i64, d: i64) -> i64 {
    if d == 0 {
        if x == 0 {
            return 1i64;
        }
        return n+1;
    } else if d < 0 {
        return solve(n, -x, -d);
    }

    let mut rec = vec![];
    for k in 1..n+1 {
        let mut from = (nth(x, d, 1) + nth(x, d, k)) * k / 2;
        let mut to = (nth(x, d, n-k+1) + nth(x, d, n)) * k / 2;
        rec.push((from, to));
    }
    rec.sort_by_key(|&(f, t)| (mo(f, d), f));
    // debug!(rec);

    let mut ans = 0;
    let mut fr = 0;
    while (fr as i64) < n {
        let mut to = fr+1;
        let mut lastf = rec[fr].0;
        let mut lastt = rec[fr].1;
        while (to as i64) < n && mo(rec[fr].0, d) == mo(rec[to].0, d) {
            if lastf <= rec[to].0 && rec[to].0 <= lastt {
                lastt = max(lastt, rec[to].1);
            } else {
                ans += (lastt - lastf) / d + 1;
                lastf = rec[to].0;
                lastt = rec[to].1;
            }
            to += 1;
        }
        ans += (lastt - lastf) / d + 1;
        fr = to;
    }
    return ans+1;
}

fn mo(a: i64, d: i64) -> i64 {
    if a < 0 {
        (d-mo(-a, d)) % d
    } else {
        a % d
    }
}

fn main() {
    input! {
        n: i64, x: i64, d: i64
    };

    println!("{}", solve(n, x, d));
}