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: Wrong Answer)

#include<bits/stdc++.h>
//ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//clock_t start=clock();clock_t end=clock();cout<<(double)(end-start)/CLOCKS_PER_SEC<<endl;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef pair<int,int> pii;
typedef pair<pii,int> ppii;
typedef pair<int,pii> pipi;
typedef pair<ll,ll> pll;
typedef pair<pll,ll> ppll;
typedef pair<ll,pll> plpl;
typedef tuple<ll,ll,ll> tl;
typedef pair<double,double> pdd;
ll mod=1000000007;
ll mod2=998244353;
ll mod3=1000003;
ll mod4=998244853;
ll mod5=1000000009;
ll inf=1LL<<60;
double pi=3.141592653589793238462643383279L;
double eps=1e-9;
#define rep(i,m,n) for(int i=m;i<n;i++)
#define rrep(i,n,m) for(int i=n;i>=m;i--)
#define srep(itr,st) for(auto itr=st.begin();itr!=st.end();itr++)
#define Max(a,b) a=max(a,b)
#define Min(a,b) a=min(a,b)
int dh[4]={1,-1,0,0};
int dw[4]={0,0,1,-1};
int ddh[8]={-1,-1,-1,0,0,1,1,1};
int ddw[8]={-1,0,1,-1,1,-1,0,1};
ll gcd(ll a,ll b){
    if(a<b)swap(a,b);
    if(b==0)return a;
    if(a%b==0)return b;
    return gcd(b,a%b);
}
ll Pow(ll n,ll k){
    ll ret=1;
    ll now=n;
    while(k>0){
        if(k&1)ret*=now;
        now*=now;
        k/=2;
    }
    return ret;
}
ll beki(ll n,ll k,ll md){
  ll ret=1;
  ll now=n;
  while(k>0){
    if(k%2==1){
      ret*=now;
      ret%=md;
    }
    now*=now;
    now%=md;
    k/=2;
  }
  return ret;
}
ll gyaku(ll n,ll md){
  return beki(n,md-2,md);
}
ll popcount(ll n){
    ll ret=0;
    ll u=n;
    while(u>0){
        ret+=u%2;
        u/=2;
        
    }
    return ret;
}
int main(){
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    
    ll n,x,d;cin>>n>>x>>d;
    if(d==0){
        cout<<n+1<<endl;
        return 0;
    }
    if(d<0){
        x=-x;
        d=-d;
    }
    map<ll,vector<pll>> mp;
    set<ll> st;
    rep(i,0,n+1){
        
        ll sl=i*(i-1)*d+2*i*x+1000000000000000,sr=i*(2*n-i-1)*d+2*i*x+1000000000000000;
        ll r=sl%(2*d);
        st.insert(r);
        sl/=2*d;sr/=2*d;
        if(sl>sr)swap(sl,sr);
        mp[r].push_back({sl,sr});
    }
    ll ans=0;
    /*for(auto e:mp[0]){
        cout<<e.first<<" "<<e.second<<endl;
    }*/
    srep(itr,st){
        ll now=*itr;
        sort(mp[now].begin(),mp[now].end());
        ll nr=mp[now][0].second,nl=mp[now][0].first;
        ans+=nr-nl+1;
        rep(i,1,mp[now].size()){
            //cout<<nr<<" "<<nl<<endl;
            //cout<<ans<<endl;
            ll nnr=mp[now][i].second,nnl=mp[now][i].first;
            //cout<<nnr<<" "<<nnl<<endl;
            if(nnl>nr){
                ans+=nnr-nnl+1;
                nl=nnl;
                nr=nnr;
            }
            else{
                if(nnr>nr){
                    ans+=nnr-nr;
                    nl=nnl;
                    nr=nnr;
                }
                else{
                    nl=nnl;
                }
            }
        }
    }
    cout<<ans<<endl;
}