【模板】带有直线限制的NE Latice Path计数

Pro：
从((0,0))出发走到((n,m))
给定(y=x+l)和(y=x+r)两条直线
要求不能穿过(可以接触)这两条直线
求方案数
(n,m,l,r<=1e6)
Sol：
参考了这篇blog
https://www.cnblogs.com/xzyxzy/p/9812585.html
注意预处理的范围要为原数据范围的10倍

#include<bits/stdc++.h>
#define N 11000000
#define db double
#define ll long long
#define ldb long double
#define ull unsigned long long
using namespace std;
const int h=3,ki=149,mo=1e9+7;
int mod(int x){return (x%mo+mo)%mo;}
int inc(int x,int k){x+=k;return x<mo?x:x-mo;}
int dec(int x,int k){x-=k;return x>=0?x:x+mo;}
int ksm(int x,int k)
{
	int ans=1;
	while(k){if(k&1)ans=1ll*ans*x%mo;k>>=1;x=1ll*x*x%mo;}
	return ans;
}
int inv(int x){return ksm(mod(x),mo-2);}
int read()
{
	char ch=0;int x=0,flag=1;
	while(!isdigit(ch)){ch=getchar();if(ch=='-')flag=-1;}
	while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0',ch=getchar();}
	return x*flag;
}
void write(int x)
{
	if(!x)return (void)putchar(48);
	if(x<0)putchar(45),x=-x;
	int len=0,p[20];
	while(x)p[++len]=x%10,x/=10;
	for(int i=len;i>=1;i--)putchar(p[i]+48);
}
const db eps=1e-7,inf=1e9+7,pi=acos(-1);
db Read(){db x;scanf("%lf",&x);return x;}
void Write(db x){printf("%lf",x);}
int fac[N],vac[N];
void prepare(int n)
{
	fac[0]=1;for(int i=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mo;
	vac[n]=inv(fac[n]);for(int i=n;i>=1;i--)vac[i-1]=1ll*vac[i]*i%mo; 
}
int C(int n,int m){return n>=m?1ll*fac[n]*vac[m]%mo*vac[n-m]%mo:0;}
void fold(int &x,int &y,int k){swap(x,y);x-=k;y+=k;}
int solve(int n,int m){return (n>=0&&m>=0)?C(n+m,n):0;}
int work(int n,int m,int l,int r)
{
	l--;r++;
	int x,y,ans=solve(n,m);
	
	x=n;y=m;
	while(true)
	{
		fold(x,y,l);if(x<0||y<0)break;ans=dec(ans,solve(x,y));
		fold(x,y,r);if(x<0||y<0)break;ans=inc(ans,solve(x,y));
	}
	
	x=n;y=m;
	while(true)
	{
		fold(x,y,r);if(x<0||y<0)break;ans=dec(ans,solve(x,y));
		fold(x,y,l);if(x<0||y<0)break;ans=inc(ans,solve(x,y));
	}
	return ans;
}
int main()
{
	prepare(1e7);
	int n=read(),m=read(),l=read(),r=read();
    write(work(n,m,-l,+r));
    return 0;
}