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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

///////////////////////////////////////////////////////////////////////////////

#include <bits/stdc++.h>
#include <unistd.h>
#include <time.h>
#include <sys/time.h>

using namespace std;

///////////////////////////////////////////////////////////////////////////////

#define DEBUG 0

#define pb push_back
#define V vector
#define M unordered_map
#define S static
#define PI  (3.1415926535)

#define rep(i,n) for(ll i=0LL; i<n; ++i)
#define srep(i,s,n) for(ll i=s; i<n; ++i)
#define rrep(i,n) for(ll i=n-1LL; i>=0LL; --i)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define CIN(x) do { \
        assert(!cin.eof()); \
        cin >> x; \
        assert(!cin.fail()); \
} while(0);
#if DEBUG
#define debug_print(...) _debug_print(__VA_ARGS__)
#define debug_printf(...) printf(__VA_ARGS__)
#define debug_print_time _debug_print_time
#else // DEBUG
#define debug_print(...)
#define debug_printf(...)
#define debug_print_time
#endif // DEBUG

typedef long long ll;
typedef unsigned long long ull;
typedef tuple<ll, ll> t2;
typedef tuple<ll, ll, ll> t3;
typedef tuple<ll, ll, ll, ll> t4;
typedef tuple<ll, ll, ll, ll, ll> t5;
template <typename T>
using priority_queue_incr = priority_queue<T, V<T>, greater<T> >;

///////////////////////////////////////////////////////////////////////////////

template <typename TT, typename T>
T get_m(M<TT, T> &m, TT k, T default_value)
{
    if (m.find(k) == m.end()) return m[k] = default_value;
    return m[k];
}

template <typename TT, typename T>
void incr_m(M<TT, T> &m, TT k)
{
    if (m.find(k) == m.end()) m[k] = 0;
    m[k]++;
}

void llin(ll &a)
{
    CIN(a);
}

void llinl1(V<ll> &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a;
        CIN(a);
        v.push_back(a);
    }
}

void llinl2(V<t2> &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a, b;
        CIN(a >> b);
        v.push_back(t2(a, b));
    }
}

void llinl3(V<t3> &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a, b, c;
        CIN(a >> b >> c);
        v.push_back(t3(a, b, c));
    }
}

void llinl4(V<t4> &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a, b, c, d;
        CIN(a >> b >> c >> d);
        v.push_back(t4(a, b, c, d));
    }
}

void llina(V<ll> &v, ll count)
{
    llinl1(v, count);
}

template <typename T>
void sort(V<T> &v)
{
    sort(v.begin(), v.end());
}

template <typename T>
void sort_reverse(V<T> &v)
{
    sort(v.begin(), v.end(), greater<T>());
}

void _debug_print(auto x)
{
    cout << x << endl;
}

void _debug_print(const t2 &x)
{
    ll x1 = get<0>(x);
    ll x2 = get<1>(x);
    cout << "-- " << x1 << " -- " << x2 << endl;
}

void _debug_print(const t3 &x)
{
    ll x1 = get<0>(x);
    ll x2 = get<1>(x);
    ll x3 = get<2>(x);
    cout << "-- " << x1 << " -- " << x2 << " -- " << x3 << endl;
}

void _debug_print(const t4 &x)
{
    ll x1 = get<0>(x);
    ll x2 = get<1>(x);
    ll x3 = get<2>(x);
    ll x4 = get<3>(x);
    cout << "-- " << x1 << " -- " << x2
         << " -- " << x3 << " -- " << x4 << endl;
}

template <typename T>
void _debug_print(T xarray[], ll n)
{
    rep (i, n) _debug_print(xarray[i]);
}

template <typename T>
void _debug_print(const V<T> &xlist)
{
    for (auto x : xlist) {
        cout << "-- ";
        _debug_print(x);
    }
}

template <typename T>
void _debug_print(const set<T> &xset)
{
    for (auto x : xset) {
        cout << "-- ";
        _debug_print(x);
    }
}

template <typename TT, typename T>
void _debug_print(const M<TT, T> &xlist)
{
    for (auto x : xlist) {
        TT k = x.first;
        T v = x.second;
        cout << "====" << endl;
        cout << "K=";
        _debug_print(k);
        cout << "V=";
        _debug_print(v);
    }
}

void _debug_print_time(const char *prefix)
{
    struct timeval tv;
    gettimeofday(&tv, NULL);
    struct tm *tm = localtime(&tv.tv_sec);
    printf("-- %s %02d:%02d:%02d.%06ld\n",
           prefix, tm->tm_hour, tm->tm_min, tm->tm_sec, tv.tv_usec);
}






ll gcm(ll x, ll y)
{
    while (y != 0) {
        ll z = x % y;
        x = y;
        y = z;
    }
    return x;
}

void get_divisors(V<ll> &retlist, ll x)
{
    for (ll i = 1LL; i < sqrt(x) + 3LL; ++i) {
        if (x % i == 0LL) {
            retlist.push_back(i);
            retlist.push_back(x / i);
        }
    }
}

// returns factors and 1
void get_factors(V<ll> &retlist, ll x)
{
    retlist.pb(1LL);
    for (ll i = 2LL; i < (ll)(sqrt(x)) + 3LL; ++i) {
        while (x % i == 0LL) {
            retlist.pb(i);
            x /= i;
        }
    }
    retlist.pb(x);
}

bool is_prime(ll x)
{
    V<ll> factors, factors2;

    get_factors(factors, x);

    for (auto factor : factors) {
        if (factor > 1) factors2.pb(factor);
    }

    return factors2.size() == 1 && x == factors2[0];
}

ull combination(ll x, ll y)
{
    if (y > x / 2LL) y = x - y;

    ull ret = 1LL;
    for (ll i = 0LL; i < y; ++i) {
        ret *= x--;
        ret /= (i + 1LL);
    }

    return ret;
}










///////////////////////////////////////////////////////////////////////////////

void _main();
int main()
{
    cout << setprecision(12);
#if !DEBUG
    ios::sync_with_stdio(false);
    cin.tie(0);
#endif

    _main();
    return 0;
}

void _main()
{
    ll n; llin(n);
    ll x; llin(x);
    ll d; llin(d);

    if (!d) {
        if (x) cout << n+1 << endl;
        else cout << 1 << endl;
        return;
    }
    if (!x) {
        cout << n * (n-1LL) / 2LL + 1LL << endl;
        return;
    }

    ll div = gcm(abs(x), abs(d));
    x /= div;
    d /= div;

    if (d < 0) {
        x = -x;
        d = -d;
    }

    debug_printf("x=%lld d=%lld\n", x, d);

    V<ll> alist;
    ll val = x;
    rep (i, n) {
        alist.pb(val);
        val += d;
    }

    ll dd = abs(d);
    S priority_queue_incr<t2> ranges[200005];
    ll minval = 0;
    ll maxval = 0;

    rep (cnt, n+1) {
        ll idx = cnt % dd;

        ranges[idx].push(t2(minval / dd, 1));
        ranges[idx].push(t2(maxval / dd + 1LL, -1));

        if (cnt < n) minval += alist[cnt];
        if (cnt < n) maxval += alist[n-cnt-1];
    }

    ll ans = 0;

    rep (idx, 200005) {
        if (ranges[idx].size()) debug_printf("-------- idx=%lld\n", idx);
        ll valid = 0;
        ll prev = -1e18;
        while (ranges[idx].size()) {
            auto item = ranges[idx].top(); ranges[idx].pop();
            ll val = get<0>(item);
            ll diff = get<1>(item);
            ll prev_valid = valid;
            valid += diff;
            if (!prev_valid && valid) {
                debug_printf("xx1 %lld\n", val);
                prev = val;
            }
            if (prev_valid && !valid) {
                debug_printf("xx2 %lld\n", val);
                ans += val - prev;
                prev = -1e18;
            }
        }
    }

    cout << ans << endl;
}

///////////////////////////////////////////////////////////////////////////////