求助，后5个点全wa，long long也开了，还是wa

#include <bits/stdc++.h>
using namespace std;

#define ls(x) (x<<1)
#define rs(x) (x<<1)+1
#define ll long long
//--------快读快写 
inline ll read(){
    ll num=0,f=1LL;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1LL;c=getchar();}
    while(c>='0'&&c<='9'){num=num*10+(c-'0');c=getchar();}
    return num*f;
}

inline void write(ll x){
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
//--------快读快写 
const int N=1e6+50,sb=1e18;
ll n,q,a[N];

struct tree{
    ll l,r,val,lazy,lazy_,maxn;//lazy存增加的，lazy_存修改的 
}t[N<<2];

void pushup(ll x){
    t[x].val=t[ls(x)].val+t[rs(x)].val;
    t[x].maxn=max(t[ls(x)].maxn,t[rs(x)].maxn);//maxn设为两个子节点最大值的最大值 
    return;
}

void pushdown(ll x){
    if(t[x].lazy_!=sb){//如果是修改就直接跳进来，修改完就润 
        t[ls(x)].lazy_=t[x].lazy_; t[ls(x)].val=t[x].lazy_;
        t[rs(x)].lazy_=t[x].lazy_; t[rs(x)].val=t[x].lazy_;
        t[ls(x)].maxn=t[x].lazy_; t[rs(x)].maxn=t[x].lazy_;
        t[x].lazy_=sb; t[x].lazy=0;//重置lazy_和lazy，防止乱修改， 
        return;
    }
    t[ls(x)].val+=(t[ls(x)].r-t[ls(x)].l+1)*t[x].lazy;
    t[rs(x)].val+=(t[rs(x)].r-t[rs(x)].l+1)*t[x].lazy;
    t[ls(x)].lazy+=t[x].lazy;
    t[rs(x)].lazy+=t[x].lazy;
    t[ls(x)].maxn+=t[x].lazy;
    t[rs(x)].maxn+=t[x].lazy; 
    t[x].lazy=0;
    return;
    //正常的push_back,不同的是如果处于这个区间，直接给maxn加lazytag就行 
}

void build(ll l,ll r,ll x){
    t[x].l=l; t[x].r=r;
    t[x].lazy=0; t[x].lazy_=sb;//初始化两个lazytag 
    if(l==r){
        t[x].val=a[l]; t[x].maxn=a[l];
        //printf("%d %d %lld %lld\n",l,r,t[x].val,t[x].maxn);
        return;
    }
    ll mid=(r-l)/2+l;
    build(l,mid,ls(x)); build(mid+1,r,rs(x));
    pushup(x);
    return;
}

void alter(ll l,ll r,ll x,ll k){
    if(t[x].l>r || t[x].r<l) return;
    if(t[x].l>=l && t[x].r<=r){
        t[x].val=(t[x].r-t[x].l+1)*k;
        t[x].lazy_=k;
        t[x].lazy=0;
        t[x].maxn=k;
        return;
    }
    pushdown(x);
    alter(l,r,ls(x),k); alter(l,r,rs(x),k);
    pushup(x);
    return;
}

void add(ll l,ll r,ll x,ll k){
    if(t[x].l>r || t[x].r<l) return;
    if(t[x].l>=l && t[x].r<=r){
        t[x].val+=(t[x].r-t[x].l+1)*k;
        t[x].lazy+=k;
        t[x].maxn+=k; return;//同上push_down 
    }
    pushdown(x);
    add(l,r,ls(x),k); add(l,r,rs(x),k);
    pushup(x);
    return;
}

ll ask(ll l,ll r,ll x){
    if(t[x].l>r || t[x].r<l) return 0;
    if(t[x].l>=l && t[x].r<=r){
        return t[x].maxn;
    }
    pushdown(x);
    return max(ask(l,r,ls(x)),ask(l,r,rs(x)));//返回区间最大的maxn 
}

int main(){
    //freopen("1253_6.in","r",stdin);
    //freopen("1253_6.out","w",stdout);
    n=read(); q=read();

    for(ll i=1;i<=n;i++) a[i]=read();
    build(1,n,1);

    ll k,op,l,r;
    for(ll i=1;i<=q;i++){
        op=read(); l=read(); r=read();

        if(op==1){
            k=read();
            alter(l,r,1,k);
        }

        if(op==2){
            k=read();
            add(l,r,1,k);
        }

        if(op==3){
            write(ask(l,r,1));
            printf("\n");
        }
    }
    return 0;
}