[ABC265E] Warp

Problem Statement

Takahashi is at the origin of a two-dimensional plane.

Takahashi will repeat teleporting $N$ times. In each teleportation, he makes one of the following moves:

• Move from the current coordinates $(x,y)$ to $(x+A,y+B)$
• Move from the current coordinates $(x,y)$ to $(x+C,y+D)$
• Move from the current coordinates $(x,y)$ to $(x+E,y+F)$

There are obstacles on $M$ points $(X_1,Y_1),\ldots,(X_M,Y_M)$ on the plane; he cannot teleport to these coordinates.

How many paths are there resulting from the $N$ teleportations? Find the count modulo $998244353$.

Constraints

• $1 \leq N \leq 300$
• $0 \leq M \leq 10^5$
• $-10^9 \leq A,B,C,D,E,F \leq 10^9$
• $(A,B)$, $(C,D)$, and $(E,F)$ are distinct.
• $-10^9 \leq X_i,Y_i \leq 10^9$
• $(X_i,Y_i)\neq(0,0)$
• $(X_i,Y_i)$ are distinct.
• All values in input are integers.

---

Input

Input is given from Standard Input in the following format:

$N$ $M$
$A$ $B$ $C$ $D$ $E$ $F$
$X_1$ $Y_1$
$X_2$ $Y_2$
$\vdots$
$X_M$ $Y_M$

Output

Print the answer.

---

Sample Input 1

2 2
1 1 1 2 1 3
1 2
2 2

Sample Output 1

5

The following $5$ paths are possible:

• $(0,0)\to(1,1)\to(2,3)$
• $(0,0)\to(1,1)\to(2,4)$
• $(0,0)\to(1,3)\to(2,4)$
• $(0,0)\to(1,3)\to(2,5)$
• $(0,0)\to(1,3)\to(2,6)$

---

Sample Input 2

10 3
-1000000000 -1000000000 1000000000 1000000000 -1000000000 1000000000
-1000000000 -1000000000
1000000000 1000000000
-1000000000 1000000000

Sample Output 2

0

---

Sample Input 3

300 0
0 0 1 0 0 1

Sample Output 3

292172978

有三种步伐，发现我们在坐标系上很难 dp。因为只有三种移动方式，所以我们考虑 dp 每种操作了多少次。

为了 \(O(1)\) 判断，我们可以把所有不能走的坐标用哈希表存起来。

定义 \(dp_{i,j,k}\) 为第一中移动方式移动了 \(i\) 次，第二种移动方式移动了 \(j\) 次第三种移动方式移动了 \(k\) 次的情况下，有多少种方案。

首先在哈希表上查一下这样子到达的点是否可以走，如果可以走，\(dp_{i,j,k}=dp_{i-1,j,k}+dp_{i,j-1,k}+dp_{i,j,k-1}\)

#include<cstdio>
const int N=305,S=2e9+1,P=998244353,mod=1e7+3;
typedef long long LL;
int n,m,a,b,c,d,e,f,dp[N][N][N];
LL hs[mod],ans,x,y;
LL hsh(int x,int y)
{
	return 1LL*(x+S-1)*S+y+S;
}
void insert(LL a)
{
	for(int i=a%mod;;i++)
	{
		if(i==mod)
			i=0;
		if(!hs[i])
		{
			hs[i]=a;
			break;
		}
	}
}
int find(LL a)
{
	for(int i=a%mod;;i++)
	{
		if(i==mod)
			i=0;
		if(!hs[i])
			return 0;
		if(hs[i]==a)
			return 1;
	}
}
LL tx(int x,int y,int z)
{
	return 1LL*x*a+1LL*y*c+1LL*z*e;
}
LL ty(int x,int y,int z)
{
	return 1LL*x*b+1LL*y*d+1LL*z*f;
}
int main()
{
	scanf("%d%d%d%d%d%d%d%d",&n,&m,&a,&b,&c,&d,&e,&f);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		insert(hsh(x,y));
	}
	dp[0][0][0]=1;
	for(int i=0;i<=n;i++)
	{
		for(int j=0;j+i<=n;j++)
		{
			for(int k=0;k+j+i<=n;k++)
			{
				x=tx(i,j,k),y=ty(i,j,k);
				if(x<-1e9||x>1e9||y<-1e9||y>1e9||!find(hsh(x,y)))
				{
					if(i)
						dp[i][j][k]=dp[i-1][j][k];
					if(j)
						dp[i][j][k]=(dp[i][j][k]+dp[i][j-1][k])%P;
					if(k)
						dp[i][j][k]=(dp[i][j][k]+dp[i][j][k-1])%P;
				}
				if(i+j+k==n)
					ans=(ans+dp[i][j][k])%P;
//				printf("%lld %lld %d %d %d %d\n",x,y,i,j,k,dp[i][j][k]);
			}
		}
	}
	printf("%d",ans);
	return 0;
}