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>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> P;
typedef pair<int,int> Pi;
typedef pair<double,double> Pd;
#define rep(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define fi first
#define se second
#define PI (acos(-1))

template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}
template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}
template<typename T> ostream& operator<<(ostream& s,const complex<T>& d) {return s<<"("<<d.real()<<", "<<d.imag()<< ")";}
template<typename T1, typename T2> ostream& operator<<(ostream& s,const pair<T1,T2>& d) {return s<<"("<<d.first<<", "<<d.second<<")";}
template<typename T> ostream& operator<<(ostream& s, const vector<T>& d){
	int len=d.size();
	rep(i,len){
		s<<d[i];if(i<len-1) s<<" ";
	}
	return s;
}
template<typename T> ostream& operator<<(ostream& s,const vector<vector<T>>& d){
	int len=d.size();
	rep(i,len){
		s<<d[i]<<endl;
	}
	return s;
}
template<typename T> ostream& operator<<(ostream& s,const set<T>& v){
	s<<"{ ";
	for(auto itr=v.begin();itr!=v.end();++itr) {
		if (itr!=v.begin()) {s<< ", ";}
		s<<(*itr);
	}
	s<<" }";
	return s;
}
template<typename T1, typename T2> ostream& operator<<(ostream& s,const map<T1,T2>& m){
	s<<"{"<<endl;
	for(auto itr=m.begin();itr!=m.end();++itr){
		s<<" "<<(*itr).first<<" : "<<(*itr).second<<endl;
	}
	s<<"}"<<endl;
	return s;
}

const ll mod=1e9+7;
const ll inf=1e18;

struct edge{
	ll to,id;
};

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	ll x,d;
	cin>>n>>x>>d;
	if(d==0){
		if(x==0){
			cout<<1<<endl;
		}else{
			cout<<n+1<<endl;
		}
		return 0;
	}
	if(d<0){
		d*=-1;
		x*=-1;
	}
	map<int,vector<P>> mp;
	rep(k,n+1){//高橋君が取る数の個数を決め打ち
		ll l=k*(k-1)/2;
		ll r=k*(2*n-k-1)/2;
		ll c=k*x+l*d;
		r-=l;
		mp[(c%d+d)%d].emplace_back(c,r);
	}
	ll ans=0;
	for(auto itr:mp){
		vector<P> now=itr.se;
		int m=now.size();
		rep(i,m){
			now[i].fi=(now[i].fi-itr.fi)/d;
		}
		vector<P> ev;
		rep(i,m){
			ev.emplace_back(now[i].fi,1);
			ev.emplace_back(now[i].fi+now[i].se+1,-1);
		}
		sort(begin(ev),end(ev));
		ll cnt=0,pre=-inf;
		rep(i,2*m){
			ll len=ev[i].fi-pre;
			if(cnt) ans+=len;
			cnt+=ev[i].se;
			pre=ev[i].fi;
		}
	}
	cout<<ans<<endl;
}