牛客 - Greater and Greater

传送门：牛客 - Greater and Greater

思路分析

开两个$biset$，$ans$用来维护那一位可以作为子段的开头，tmp维护$A$中哪些可以对应当前$B_i$
由$tmp$可以推出目前哪些位置可以作为子段的开头（$tmp>>B_i->pos$）可以将数组从大到小排序，这样每次$tmp$就可以直接继承之前的了

AC代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 150010;

bitset<N>ans,tmp;
pair<int,int>a[N],b[N];

int cmp(pair<int,int>x,pair<int,int>y){
    return x.fi>y.fi;
}

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>a[i].fi;
        a[i].se=i;
    }
    for(int i=1;i<=m;i++){
        cin>>b[i].fi;
        b[i].se=i;
    }
    sort(a+1,a+1+n,cmp);
    sort(b+1,b+1+n,cmp);
    ans.set();
    tmp.reset();
    for(int i=1,j=1;i<=m;i++){
        while(j<=n && a[j].fi>=b[i].fi){
            tmp.set(a[j].se);
            j++;
        }
        ans &= tmp>>b[i].se;
    }
    cout<<ans.count()<<endl;




    return 0;
}