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>
//using namespace std;
#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define rep(i,j,n) for(ll i=(ll)(j);i<(ll)(n);i++)
#define REP(i,j,n) for(ll i=(ll)(j);i<=(ll)(n);i++)
#define per(i,j,n) for(ll i=(ll)(j);(ll)(n)<=i;i--)
#define ll long long
#define ALL(a) (a).begin(),(a).end()
#define disup(A,key) distance(A.begin(),upper_bound(ALL(A),(ll)(key)))
#define dislow(A,key) distance(A.begin(),lower_bound(ALL(A),(ll)(key)))
#define pb emplace_back
#define mp std::make_pair
#define endl "\n"
//using std::endl;
using std::cin;
using std::cout;
using std::vector;
using std::string;
using std::upper_bound;
using std::lower_bound;
using vi=vector<ll>;
using vii=vector<vi>;
using pii=std::pair<ll,ll>;
constexpr ll MOD=1e9+7;
//constexpr ll MOD=998244353;
//constexpr ll MOD=10000000;
constexpr ll MAX=1e7;
constexpr ll INF=(1ll<<62);
template<class T>
class prique :public std::priority_queue<T, std::vector<T>, std::greater<T>> {};
/*struct Binary_indexed_tree{
    int N;
    vi bit;
    Binary_indexed_tree(int n):N(n){
        bit.resize(N+1,0);
    }
    void add(int x,int a){
        for(x;x<=N;x+=(x&-x)) bit[x]+=a;
    }
    ll sum(int x){
        ll ret=0;
        for(x;x>0;x-=(x&-x)) ret+=bit[x];
        return ret;
    }
    ll lower_bound(ll X){
        if(sum(N)<X) return -1;
        ll ret=0,memo=1,sum=0;
        while(memo*2<=N) memo*=2;
        while(memo>0){
            if(memo+ret<=N&&sum+bit[memo+ret]<X){
                sum+=bit[memo+ret];
                ret+=memo;
            }
            memo/=2;
        }
        return ret+1;
    }
};
struct Union_Find{
    ll N;
    vi par;
    vi siz;
    Union_Find(int n):N(n){
        par.resize(N);
        siz.resize(N,1);
        rep(i,0,N) par[i]=i;
    }
    ll root(ll X){
        if(par[X]==X) return X;
        return par[X]=root(par[X]);
    }
    bool same(ll X,ll Y){
        return root(X)==root(Y);
    }
    void unite(ll X,ll Y){
        X=root(X);
        Y=root(Y);
        if(X==Y) return;
        par[X]=Y;
        siz[Y]+=siz[X];
        siz[X]=0;
    }
    ll size(ll X){
        return siz[root(X)];
    }
};
ll modpow(ll X,ll Y,ll mod){
    ll sum=X,cnt=1;
    vi A;
    while(cnt<=Y){
        A.pb(sum);
        sum*=sum;
        sum%=mod;
        cnt*=2;
    }
    int M=A.size();
    ll ret=1;
    REP(i,1,M){
        if(Y>=(1ll<<M-i)){
            Y-=(1ll<<M-i);
            ret*=A[M-i];
            ret%=mod;
        }
    }
    return ret;
}
vi fac,finv,inv;
void COMinit() {
    fac.resize(MAX);
    finv.resize(MAX);
    inv.resize(MAX);
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
ll COM(ll n,ll r){
    if(n<r||n<0||r<0) return 0;
    return fac[n]*finv[r]%MOD*finv[n-r]%MOD;
}*/
signed main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    ll N,X,D; cin>>N>>X>>D;
    if(D==0){
        std::set<ll> s;
        REP(i,0,N){
            ll P=i*2-N;
            P*=X;
            s.insert(P);
        }
        cout<<s.size()<<endl;
        return 0;
    }
    if(X==0){
        ll ans=0;
        vector<vector<pii>> memo(2);
        REP(i,0,N){
            ll max=(N-1)*N/2-(N-i)*(N-i-1);
            ll min=i*(i-1)-(N-1)*N/2;
            memo[((max%2)+2)%2].pb(mp(min,max));
        }
        rep(i,0,2){
            ll me=-INF;
            sort(ALL(memo[i]));
            rep(j,0,memo[i].size()){
                ll min=memo[i][j].first,max=memo[i][j].second;
                if(me<min){
                    ans+=(max-min)/2+1;
                    me=max;
                }
                else{
                    if(me<max){
                        ans+=(max-me)/2;
                        me=max;
                    }
                }
            }
        }
        cout<<ans<<endl;
        return 0;
    }
    if(X<0){
        X=-X;
        D=-D;
    }
    ll A=std::__gcd(X,D);
    ll B=D/A,C=X/A;
    vector<std::unordered_map<ll,vector<pii>>> memo(2);
    REP(i,0,N){
        ll P=i*2-N;
        ll max=(N-1)*N/2-(N-i)*(N-i-1);
        ll min=i*(i-1)-(N-1)*N/2;
        max+=P/B*C;
        min+=P/B*C;
        P%=B;
        if(P<0){
            P+=B;
            max-=C;
            min-=C;
        }
        memo[((max%2)+2)%2][P].pb(mp(min,max));
    }
    ll ans=0;
    rep(k,0,2){
        for(auto p:memo[k]){
            vector<pii> X=p.second;
            ll me=-INF;
            sort(ALL(X));
            rep(j,0,X.size()){
                ll min=X[j].first,max=X[j].second;
                if(me<min){
                    ans+=(max-min)/2+1;
                    me=max;
                }
                else{
                    if(me<max){
                        ans+=(max-me)/2;
                        me=max;
                    }
                }
            }
        }
    }
    cout<<ans<<endl;
}