Topic links: atcoder
Consider all the holes axes will be divided into multiple sections, each robot is no matter how he can not move out of this range, it is up to each robot may only fall into two holes
For the leftmost and rightmost no holes robots, apparently out of their hole program unique, so we do not consider it for the rest of the robots, we use a tuple \ ((l_i, r_i) \ ) indicates that it to its distance from the nearest left / right of the hole, it is clear that a robot which fall into the hole and the location of the sequence of operations to reach only the most left / right.
All tuples on a flat surface, uses them to some point in the operation leftmost / rightmost position history of the sequence can be used to achieve only an up and right fold lines. FIG. At this point polyline will fall below the hole on the right, at the point above the fold line will fall into the hole on the left.
We noticed only when the broken line turning down a place in the corner of the marked point will cause a change in the final program. So we can write \ (f_i \) represents the last point of the polyline after \ (i \) , and the number of programs finally broken line to go to the right, as a turning point only in \ (i \) lower left, so difficult to obtain transfer equation
\ [f_i = 1 + \ sum_ {x_j <x_i, y_j <y_i} f_j \]
The more classic forms, using the Fenwick tree maintenance \ (f \) can pay attention to the second dimension should avoid sorted in descending order with the same \ (x \) and smaller \ (y \) location will present generating value transfer
#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
const int N=10000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define fir first
#define sec second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define maxd 1000000007
#define eps 1e-8
int n,m,tot=0,val[100100],rob[100100],hol[100100],tot1=0;
ll f[100100];
pii p[100100];
struct fenwick_tree{
int a[100100];
void modify(int x,int p)
{
for (int i=p;i<=tot;i+=lowbit(i)) a[i]=(a[i]+x)%maxd;
}
int query(int p)
{
int ans=0;
for (int i=p;i;i-=lowbit(i)) ans=(ans+a[i])%maxd;
return ans;
}
}tr;
int read()
{
int x=0,f=1;char ch=getchar();
while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
return x*f;
}
bool cmp(pii p,pii q) {return ((p.fir<q.fir) || ((p.fir==q.fir) && (p.sec>q.sec)));}
int main()
{
n=read();m=read();
rep(i,1,n) rob[i]=read();
rep(i,1,m) hol[i]=read();
rep(i,1,n)
{
if ((rob[i]<=hol[1]) || (rob[i]>=hol[m])) continue;
int pos=lower_bound(hol+1,hol+1+m,rob[i])-hol;
if (hol[pos]==rob[i]) continue;
p[++tot1]=mp(rob[i]-hol[pos-1],hol[pos]-rob[i]);
val[++tot]=hol[pos]-rob[i];
}
//rep(i,1,n) cout << p[i].fir << " " << p[i].sec << endl;
sort(val+1,val+1+tot);
tot=unique(val+1,val+1+tot)-val-1;
rep(i,1,tot1) p[i].sec=lower_bound(val+1,val+1+tot,p[i].sec)-val;
sort(p+1,p+1+tot1,cmp);
tot1=unique(p+1,p+1+tot1)-p-1;
//cout << endl;
//rep(i,1,tot1) cout << p[i].fir << " " << p[i].sec << endl;
rep(i,1,tot1)
{
f[i]=tr.query(p[i].sec-1)+1;
tr.modify(f[i],p[i].sec);
}
//rep(i,1,tot1) cout << f[i] << " ";cout << endl;
ll ans=1;
rep(i,1,tot1) ans=(ans+f[i])%maxd;
printf("%lld",ans);
return 0;
}