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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

const ll INF = 1ll<<60;

typedef pair<ll,ll> P;

int main() {
    ll n, x, d;
    cin >> n >> x >> d;
    
    if (d == 0) {
        if (x == 0) {
            cout << 1 << endl;
        } else {
            cout << n+1 << endl;
        }
        return 0;
    }
    
    map<int,vector<P>> mp;
    rep(i,n+1) {
        ll s = x*i;//kx
        ll l = (ll)i*(i-1)/2;//min
        ll r =  (ll)i*(n-i);//max
        s += l*d;
        //r -= l;//r'
        //負のmod対策
        mp[(s+(d<<30))%d].emplace_back(s,r);
    }
    
    ll ans = 0;
    for (auto p : mp) {
        auto a = p.second;
        int m = a.size();
        rep(i,m) a[i].first = (a[i].first-p.first)/d;
        vector<P> ev;
        rep(i,m) {
            ev.emplace_back(a[i].first, 1);
            ev.emplace_back(a[i].first+a[i].second+1, -1);
        }
        sort(ev.begin(), ev.end());
        ll cnt = 0, pre = -INF;
        for (auto e : ev) {
            ll len = e.first-pre;
            if (cnt) ans += len;
            cnt += e.second;
            pre = e.first;
        }
    }
    cout << ans << endl;
    return 0;
}