CF1567E. Non-Decreasing Dilemma(线段树)

// Problem: E. Non-Decreasing Dilemma
// Contest: Codeforces - Codeforces Round #742 (Div. 2)
// URL: https://codeforces.com/contest/1567/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
//#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
typedef pair<string,string>PSS;
#define I_int ll
inline ll read(){ll 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;}
 
inline void write(ll x){if (x < 0) x = ~x + 1, putchar('-');if (x > 9) write(x / 10);putchar(x % 10 + '0');}
 
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
#define x first
#define y second
ll ksm(ll a, ll b,ll mod){ll res = 1;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
 
const int maxn=2e5+100;
 
struct node{
    ll l,r;
    ll sum;
    ll cntl,cntr;
}tr[maxn*4];
ll n,a[maxn],m;

void pushup(int u){
    node l=tr[u<<1],r=tr[u<<1|1];
    tr[u].sum=l.sum+r.sum;
    if(a[l.r]<=a[r.l]) tr[u].sum+=l.cntr*r.cntl;
    
    if(l.cntl==l.r-l.l+1){
        tr[u].cntl=l.cntl;
        if(a[l.r]<=a[r.l]) tr[u].cntl+=r.cntl;
    }
    else tr[u].cntl=l.cntl;
    
    if(r.cntr==r.r-r.l+1){
        tr[u].cntr=r.cntr;
        if(a[l.r]<=a[r.l]) tr[u].cntr+=l.cntr;
    }
    else tr[u].cntr=r.cntr;
}

void build(int u,int l,int r){
    
    if(l==r){
        tr[u]={l,r,1,1,1};
        return ;
    }
    tr[u]={l,r,0,0,0};
    int mid=(l+r)/2;
    build(u<<1,l,mid);build(u<<1|1,mid+1,r);
    pushup(u);
}

void update(int u,int l,int r,int pos,int val){
    if(l==r&&l==pos){
        a[pos]=val;
        return ;
    }
    int mid=(l+r)/2;
    if(pos<=mid) update(u<<1,l,mid,pos,val);
    if(pos>mid) update(u<<1|1,mid+1,r,pos,val);
    pushup(u);
}

ll query(int u,int l,int r,int ql,int qr){
    if(l>=ql&&r<=qr) return tr[u].sum;
    int mid=(l+r)/2;
    ll ans=0;
    if(qr<=mid) return query(u<<1,l,mid,ql,qr);
    else if(ql>mid) return query(u<<1|1,mid+1,r,ql,qr);
    else{
        ans=query(u<<1,l,mid,ql,qr)+query(u<<1|1,mid+1,r,ql,qr);
        node ll=tr[u<<1],rr=tr[u<<1|1];
        if(a[ll.r]<=a[rr.l]) 
            ans+=min(ll.r-ql+1,ll.cntr)*min(qr-rr.l+1,rr.cntl);
        return ans;
    }
}

int main() {
    n=read;m=read;
    rep(i,1,n) a[i]=read;
    build(1,1,n);
    rep(i,1,m){
        int op=read,l=read,r=read;
        if(op==1){
            update(1,1,n,l,r);
        }
        else printf("%lld\n",query(1,1,n,l,r));
    }
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
    return 0;
}