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<(int)(n);i++)
#define rrep(i,n) for(int i=(int)(n-1);i>=0;i--)
#define all(x) (x).begin(),(x).end()
#define sz(x) int(x.size())
#define get_unique(x) x.erase(std::unique(all(x)), x.end());
typedef long long ll;
const int INF = 1e9;
const ll MOD = 1e9+7;
const ll LINF = 1e18;
const double PI=acos(-1);
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<class T>
vector<T> make_vec(size_t a){
    return vector<T>(a);
}
template<class T, class... Ts>
auto make_vec(size_t a, Ts... ts){
    return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}

int main() {
    ll n,x,d;
    cin>>n>>x>>d;

    if(d==0) {
        cout<<(x==0 ? 1 : n+1)<<endl;
        return 0;
    }

    map<ll, vector<pair<ll,ll>>> mp;
    for(ll k = 0; k <= n; k++) {
        ll l=k*(k-1)/2;
        ll r=k*(2*n-k-1)/2;
        ll c=k*x+d*l;

        mp[(c+d*INF)%d].emplace_back((c-(c+d*INF)%d)/d,1);
        mp[(c+d*INF)%d].emplace_back((c-(c+d*INF)%d)/d+r-l+1,-1);
    }

    ll ans=0;
    for(auto vv:mp) {
        auto ev=vv.second;
        sort(all(ev));
        ll cnt=0,pre=-LINF;
        for(auto p:ev) {
            if(cnt>0) ans+=p.first-pre;
            cnt+=p.second;
            pre=p.first;
        }
    }
    cout<<ans<<endl;
}