蒟蒻线段树20pts求调

#include<bits/stdc++.h>
#define int long long
#define INT_MIN LLONG_MIN
using namespace std;
int n,m,tree[5000005],a[1000005],op,a1,a2,a3,lazytag1[5000005],lazytag2[2000005]; 
void pushup(int k)
{
    tree[k] = max(tree[k * 2],tree[k * 2 + 1]);
    lazytag2[k] = INT_MIN;
}
void add1(int k,int v)
{
    lazytag1[k] += v;
    tree[k] += lazytag1[k];
}
void add2(int k,int v)
{
    lazytag2[k] = v;
    tree[k] = lazytag2[k];
}
void pushdown2(int k)
{
    if(lazytag2[k] == INT_MIN)return;
    lazytag1[k] = 0;
    add2(k * 2,lazytag2[k]);
    add2(k * 2 + 1,lazytag2[k]);
    lazytag2[k] = INT_MIN;
}
void pushdown1(int k)
{
    pushdown2(k);
    if(lazytag1[k] == 0)return;
    add1(k * 2,lazytag1[k]);
    add1(k * 2 + 1,lazytag1[k]);
    lazytag1[k] = 0;
}
void build(int k,int l,int r)
{
    if(l == r)
    {
        tree[k] = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(k * 2,l,mid);
    build(k * 2 + 1,mid + 1,r);
    pushup(k);
}
void update1(int k,int l,int r,int x,int y,int v)
{
    if(l > y || r < x)return;
    if(l >= x && r <= y)
    {
        //lazytag2[k] = 1e9 + 1;
        pushdown2(k);
        lazytag1[k] += v;
        tree[k] += v;
        return;
    }
    int  mid = (l + r) >> 1;
    pushdown1(k);
    update1(k * 2,l,mid,x,y,v);
    update1(k * 2 + 1,mid + 1,r,x,y,v);
    pushup(k);
}
void update2(int k,int l,int r,int x,int y,int v)
{
    if(l > y || r < x)return;
    if(l >= x && r <= y)
    {
        lazytag1[k] = 0;
        lazytag2[k] = v;
        tree[k] = v;
        return;
    }
    int  mid = (l + r) >> 1;
    pushdown2(k);
    update2(k * 2,l,mid,x,y,v);
    update2(k * 2 + 1,mid + 1,r,x,y,v);
    pushup(k);
}
int ask(int k,int l,int r,int x,int y)
{
    if(l > y || r < x)return INT_MIN;
    if(l >= x && r <= y)return tree[k];
    int mid = (l + r) >> 1;
    pushdown1(k);
    //pushdown2(k);
    return max(ask(k * 2,l,mid,x,y),ask(k * 2 + 1,mid + 1,r,x,y));
}
signed main()
{
    //freopen(".in","r",stdin);
    //freopen(".out","w",stdout);
    cin >> n >> m;
    for(int i = 1;i <= n;i ++)cin >> a[i];
    build(1,1,n);
    for(int i = 1;i <= m;i ++)
    {
        cin >> op;
        if(op == 1)
        {
            cin >> a1 >> a2 >> a3;
            update2(1,1,n,a1,a2,a3);
        }
        if(op == 2)
        {
            cin >> a1 >> a2 >> a3;
            update1(1,1,n,a1,a2,a3);
        }
        if(op == 3)
        {
            cin >> a1 >> a2;
            cout << ask(1,1,n,a1,a2) << endl;
        }
    }
    return 0;
}