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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "debug.h"
#else
#define dump(...)
#endif
#define endl '\n'
#define int long long
#define FOR(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define REP(i, x) for (int i = 0; i < (int)(x); i++)
#define REPS(i, x) for (int i = 1; i <= (int)(x); i++)
#define RREP(i, x) for (int i = ((int)(x)-1); i >= 0; i--)
#define RREPS(i, x) for (int i = ((int)(x)); i > 0; i--)
#define INF 2147483647
#define LLINF 9223372036854775807LL
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define BIT(n) (1LL << (n))
#define TOUPPER(s) transform(s.begin(), s.end(), s.begin(), ::toupper)
#define TOLOWER(s) transform(s.begin(), s.end(), s.begin(), ::tolower)
#define FILLZERO(s, n)                      \
    do {                                    \
        ostringstream os;                   \
        os << setw(n) << setfill('0') << s; \
        s = os.str();                       \
    } while (0);
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end()) //sortしてからつかうこと
#define INIT                               \
    std::cin.tie(nullptr);                 \
    std::ios_base::sync_with_stdio(false); \
    cout << std::fixed << std::setprecision(10);

using ll = long long;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using pi = pair<int, int>;
template <typename T>
using Pq = priority_queue<T>;
template <typename T>
using pQ = priority_queue<T, vector<T>, greater<T>>;

template <typename T>
vector<T> make_vec(size_t a, T n) { return vector<T>(a, n); }
template <typename T, typename... Ts>
auto make_vec(size_t a, Ts... ts) { return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...)); }
template <class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
    for (T &x : vec)
        is >> x;
    return is;
}

constexpr ll MOD = 1e9 + 7;
int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
ll A, B, C, D, E, F, G, H, N, M, L, K, P, Q, R, W, X, Y, Z;
string S, T;
ll ans = 0;

signed main() {
    INIT;
    cin >> N >> X >> D;
    if (D == 0) {
        cout << N + 1 << endl;
        return 0;
    } else if (D < 0) {
        X += (N - 1) * D;
        D *= -1;
    }
    map<int, vector<pi>> mp;
    REP(i, N + 1) {
        int m = (i * X) % D;
        int base = (i * X - m) / D;
        int l = i * (1LL + i) / 2 - i + base;
        int r = i * (2LL * N - i + 1) / 2 - i + base;
        mp[m].push_back({l, r + 1});
    }
    for (auto &[_, v] : mp) {
        v.push_back({LLINF, LLINF});
        v.push_back({-LLINF, -LLINF});
        sort(ALL(v));
        dump(_, v);
        int l = 0, r = 0;
        REP(i, v.size()) {
            if (r <= v[i].first) {
                ans += r - l;
                l = v[i].first;
                r = v[i].second;
            } else {
                chmax(r, v[i].second);
            }
        }
        // vi mem(1e5);
        // REP(i, v.size()) {
        //     mem[v[i].first]++;
        //     mem[v[i].second]--;
        // }
        // REP(i, mem.size() - 1) {
        //     mem[i + 1] += mem[i];
        // }
        // REP(i, mem.size()) {
        //     if (mem[i] > 0)
        //         ans++;
        // }
    }
    cout << ans << endl;
}