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;

// A, B, S
// B = S - A
// X, X+D, X+2D, X+3D, ... X+(N-1)D
// X+iD = nX+sD ( s = \sigma ^ n)
// (m-n)X + (s-t)D = 0


int main(void) {
    long n, x, d;
    cin >> n >> x >> d;
    if(d==0) {cout << n << endl; return 0;}
    if(d < 0) x = -x, d = -d;

    map<long, vector<pair<long,long>>> v;

    for(long c=0; c<=n; c++) {
        long a = c*(c-1)/2;
        long b = c*(n-1 + n-c)/2;

        v[(1000*d+(c*x)%d)%d].push_back({c*x/d + a, c*x/d + b});
    }

    long r = 0;
    for(auto &w: v) {
        auto &l = w.second;
        sort(l.begin(), l.end());
        long b = -1e18;
        long a = b+1;
        for(auto &x: l) {
            if(x.first > b) {
                r += b-a+1;
                a = x.first;
                b = x.second;
            } else {
                b = x.second;
            }
        }
        r += b-a+1;
    }
    cout << r << endl;
}