[20200729NOIP提高组模拟T2]学数数——坎坷

题目大意：

​ 古今序列者，长皆(n)矣.今有一序列，其长亦若此，名之曰(a).世人皆知其连续子序列之数为(frac{n(n+1)}{2})矣.现有一士，欲取之最大值于各连续子序列也.今用此(frac{n(n+1)}{2})值，构建新序列.序列,操作之本源也.于是生操作几许.予君一数(k)，试问新序列中大于/小于/等于k之数有几何？

solution：

​ 调了好久好久好久的bug(泪).此题不难想到线段树维护.首先将序列离散，然后开两棵线段树维护每个位置(x)左右小于等于(a[x])的连续序列长(count[x][0],count[x][1]),为防止重复,左闭右开即可.不难发现,该节点对新序列的贡献为((count[x][0]+1) cdot (count[x][1]+1)),然后对于查询操作,再开一棵线段树维新序列即可.

code:

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<map>
#define R register
#define next kdjadskfj
#define debug puts("mlg")
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
inline ll read();
inline void write(ll x);
inline void writeln(ll x);
inline void writesp(ll x);
ll n,Q;
ll a[610000],b[620000];
ll q;
ll dat[6100000];
ll type[610000],que[610000];
ll Dat[6100000],used[620000];
ll Count[620000][2];
inline void Update(ll p,ll l,ll r,ll k){
	if(l==r){Dat[p]=used[k];return;}
	ll mid=l+r>>1;
	if(k<=mid) Update(p<<1,l,mid,k);
	else Update(p<<1|1,mid+1,r,k);
	Dat[p]=max(Dat[p<<1],Dat[p<<1|1]);
}

inline ll Query(ll p,ll l,ll r,ll u,ll v){
//	if(u>v) return 0;
	if(!Dat[p]||(u<=l&&r<=v)) return Dat[p];
	ll mid=l+r>>1,Ans=0;
	if(u<=mid) Ans=max(Ans,Query(p<<1,l,mid,u,v));
	if(v>mid) Ans=max(Ans,Query(p<<1|1,mid+1,r,u,v));
	return Ans;
}

inline void update(ll p,ll l,ll r,ll k,ll val){
	if(l==r){dat[p]+=val;return;}
	ll mid=l+r>>1;
	if(k<=mid) update(p<<1,l,mid,k,val);
	else update(p<<1|1,mid+1,r,k,val);
	dat[p]=dat[p<<1]+dat[p<<1|1];
}

inline ll query(ll p,ll l,ll r,ll u,ll v){
	if(!dat[p]||(u<=l&&r<=v)) return dat[p];
	ll mid=l+r>>1,Ans=0;
	if(u<=mid) Ans+=query(p<<1,l,mid,u,v);
	if(v>mid) Ans+=query(p<<1|1,mid+1,r,u,v);
	return Ans;
}

inline void solve2(){
	for(R ll i=1;i<=n;i++){
//		update(1,1,q,a[i],i-Query(1,1,q,a[i]+1,q)-1);
		Count[i][0]=i-Query(1,1,q,a[i],q)-1;
		used[a[i]]=i;
		Update(1,1,q,a[i]);
	}
		memset(used,0,sizeof used);
		memset(Dat,0,sizeof Dat);
	for(R ll i=n;i>=1;i--){
//		update(1,1,q,a[i],(n-i+1)-Query(1,1,q,a[i]+1,q)-1);
		Count[i][1]=(n-i+1)-Query(1,1,q,a[i]+1,q)-1;
		used[a[i]]=n-i+1;
		Update(1,1,q,a[i]);
	}
	for(R ll i=1;i<=n;i++){
		update(1,1,q,a[i],(Count[i][0]+1)*(Count[i][1]+1));
	}
}

inline void solve1(){
	for(R ll i=1;i<=n;i++){
		ll maxn=0;
		for(R ll j=i;j<=n;j++){
			maxn=max(maxn,a[j]);
			update(1,1,q,maxn,1);
		}
	}
}

inline void work(){
	for(R ll i=1;i<=Q;i++){
		if(type[i]==1){
			if(que[i]==q) writeln(0);
			else writeln(query(1,1,q,que[i]+1,q));
			continue;	
		}
		if(type[i]==2){
			writeln(query(1,1,q,que[i],que[i]));
			continue;
		}
		if(type[i]==3){
			if(que[i]==1) writeln(0);
			else writeln(query(1,1,q,1,que[i]-1));
			continue;
		}
	}
}

int main(){
	freopen("jxthree.in","r",stdin);
	freopen("jxthree.out","w",stdout);
	n=read();Q=read();
	for(R ll i=1;i<=n;i++) a[i]=b[i]=read();
	for(R ll i=1;i<=Q;i++){
		char wn=getchar();
		while(wn!='>'&&wn!='<'&&wn!='=') wn=getchar();
		type[i]=((wn=='>')?1:((wn=='=')?2:3));
		b[n+i]=que[i]=read();
	}
	sort(b+1,b+n+Q+1);
	q=unique(b+1,b+Q+n+1)-b-1;
	for(R ll i=1;i<=n;i++){
		a[i]=lower_bound(b+1,b+q+1,a[i])-b;
	}
	for(R ll i=1;i<=Q;i++){
		que[i]=lower_bound(b+1,b+q+1,que[i])-b;
	}
//	if(n<=5000){
//		solve1();
//		work();
//		return 0;
//	}
	solve2();
	work();
}
inline ll read(){ll x=0,t=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') t=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*t;}
inline void write(ll x){if(x<0){putchar('-');x=-x;}if(x<=9){putchar(x+'0');return;}write(x/10);putchar(x%10+'0');}
inline void writesp(ll x){write(x);putchar(' ');}
inline void writeln(ll x){write(x);putchar('
');}