传送门

题意简述:你要从(0,0)(0,0)(0,0)走到(ex,ey)(ex,ey)(ex,ey),每次可以从(x,y)(x,y)(x,y)走到(x+ax,y+ay)(x+ax,y+ay)(x+ax,y+ay)或者(x+bx,y+by)(x+bx,y+by)(x+bx,y+by),其中有nnn个障碍点问方案数,所有出现的值的绝对值≤500\le500≤500


思路:从(0,0)(0,0)(0,0)到每个障碍点需要用几次(x+ax,y+ay)(x+ax,y+ay)(x+ax,y+ay)和几次(x+bx,y+by)(x+bx,y+by)(x+bx,y+by)是可以直接算出来的,我们把这两个算出来的参数看成新的横纵坐标,然后在上面做常规的棋盘问题容斥dpdpdp即可。

代码:

#include<bits/stdc++.h>
#define ri register int
#define fi first
#define se second
using namespace std;
inline int read(){
	int ans=0,w=1;
	char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')w=-1;ch=getchar();}
	while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
	return ans*w;
}
typedef pair<int,int> pii;
typedef long long ll;
const int mod=1e9+7,N=505;
inline int add(const int&a,const int&b){return a+b>=mod?a+b-mod:a+b;}
inline int dec(const int&a,const int&b){return a>=b?a-b:a-b+mod;}
inline int mul(const int&a,const int&b){return (ll)a*b%mod;}
int ex,ey,n,ax,ay,bx,by,tot=0,f[N],lim;
vector<int>fac,ifac;
pii p[N],up;
inline pii calc(int x,int y){
	int t1=ax*by-bx*ay,t2=x*by-y*bx,A=-1,B=-1;
	if(t2%t1==0)A=t2/t1;
	t1=bx*ay-ax*by,t2=x*ay-y*ax;
	if(t2%t1==0)B=t2/t1;
	return pii(A,B);
}
inline void init(){
	fac.resize(lim+5),ifac.resize(lim+5);
	fac[0]=fac[1]=ifac[0]=ifac[1]=1;
	for(ri i=2;i<=lim;++i)fac[i]=mul(fac[i-1],i),ifac[i]=mul(ifac[mod%i],mod-mod/i);
	for(ri i=2;i<=lim;++i)ifac[i]=mul(ifac[i-1],ifac[i]);
}
inline int C(const int&a,const int&b){return a>=b?mul(mul(fac[a],ifac[b]),ifac[a-b]):0;}
int main(){
	ex=read(),ey=read(),n=read(),ax=read(),ay=read(),bx=read(),by=read();
	p[tot=1]=up=calc(ex,ey);
	if(up.fi<0||up.se<0)return puts("0"),0;
	for(ri i=1,x,y;i<=n;++i){
		x=read(),y=read();
		pii tmp=calc(x,y);
		if(tmp.fi>=0&&tmp.se>=0&&tmp.fi<=up.fi&&tmp.se<=up.se)p[++tot]=tmp;
	}
	sort(p+1,p+tot+1),n=unique(p+1,p+tot+1)-p-1;
	for(ri i=1;i<=n;++i)lim=max(lim,p[i].fi+p[i].se);
	init();
	for(ri i=1;i<=n;++i){
		f[i]=C(p[i].fi+p[i].se,p[i].fi);
		for(ri j=1;j<i;++j)f[i]=dec(f[i],mul(f[j],C(p[i].fi+p[i].se-p[j].fi-p[j].se,p[i].fi-p[j].fi)));
	}
	cout<<f[n];
	return 0;
}
05-10 23:41
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