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)

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#include <bits/stdc++.h>
using namespace std;
#define fi			first
#define se			second
#define fz(i,a,b)	for(int i=a;i<=b;i++)
#define fd(i,a,b)	for(int i=a;i>=b;i--)
#define foreach(it,v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define put(x)		putchar(x)
#define eoln        put('\n')
#define space		put(' ')
#define y1			y_chenxiaoyan_1
#define y0			y_chenxiaoyan_0
#define int long long
inline int read(){
	int x=0,neg=1;char c=getchar();
	while(!isdigit(c)){
		if(c=='-')	neg=-1;
		c=getchar();
	}
	while(isdigit(c))	x=x*10+c-'0',c=getchar();
	return x*neg;
}
inline void print(int x){
	if(x<0){
		putchar('-');
		print(abs(x));
		return;
	}
	if(x<=9)	putchar(x+'0');
	else{
		print(x/10);
		putchar(x%10+'0');
	}
}
inline int qpow(int x,int e,int _MOD){
	int ans=1;
	while(e){
		if(e&1)	ans=ans*x%_MOD;
		x=x*x%_MOD;
		e>>=1;
	}
	return ans;
}
int n=read(),x=read(),d=read(),a[200005],pre[200005],suf[200005];
map<int,set<pair<int,int> > > segs;
signed main(){
	a[1]=x;
	fz(i,2,n)	a[i]=a[i-1]+d;
	if(d==0){
		if(x==0)	puts("1");
		else		print(n);
		return 0;
	}
	fz(i,1,n)	pre[i]=pre[i-1]+a[i];
	fd(i,n,1)	suf[i]=suf[i+1]+a[i];
	fz(i,0,n){//x*i+d*y
		int mod=(x*i+abs(d)*(1ll<<30))%abs(d);
		int l,r;
		if(d>0)
			l=pre[i],r=suf[n-i+1];
		else
			l=suf[n-i+1],r=pre[i];
		int _l=(l-mod)/d,_r=(r-mod)/d;
		segs[mod].insert({_l,_r});
	}
	vector<int> used;
	foreach(it,segs)	used.push_back(it->first);
	int ans=0;
	foreach(it,used){
		int val=*it,r=-1ll<<45;
		foreach(seg,segs[val]){
			int L=seg->first,R=seg->second;
			if(r<L)			ans+=(R-L+1),r=R;
			else if(r<=R)	ans+=R-r,r=R;
		}
	}
	cout<<ans<<endl;
	return 0;
}
/*1
3 4 2
*/
/*2
2 3 -3
*/
/*3
100 14 20
*/