The Child and Sequence (modulo segment tree)

The meaning of problems: segment tree modulo operation, to maintain two arrays, a maximum current maintenance interval, maintaining a current interval and, if the maximum value is less than MOD, there is no longer a modulo down; single-point update interval to find normal operation ~ ;;;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
const int maxn=1e5+10;
const int inf=0x3f3f3f3f;
typedef long long ll;
ll a[maxn],tree[maxn*4],lazy[maxn*4];
ll maxtree[maxn*4];
void build(int l,int r,int now){
    if(l==r){
        tree[now]=maxtree[now]=a[l];
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,now<<1);
    build(mid+1,r,now<<1|1);
    tree[now]=tree[now<<1]+tree[now<<1|1];
    maxtree[now]=max(maxtree[now<<1],maxtree[now<<1|1]);
}
void o3(int l,int r,int now,int k,ll s){
    if(l==r){
        tree[now]=s;
        maxtree[now]=s;
        return;
    }
    int mid=(l+r)>>1;
    if(mid>=k)o3(l,mid,now<<1,k,s);
    else o3(mid+1,r,now<<1|1,k,s);
    tree[now]=tree[now<<1]+tree[now<<1|1];
    maxtree[now]=max(maxtree[now<<1],maxtree[now<<1|1]);
}
void o2(int l,int r,int now,int L,int R,int s){
    if(s>maxtree[now])return;
    if(l==r){
        tree[now]%=s;
        maxtree[now]%=s;
        return;
    }
    int mid=(l+r)>>1;
    if(mid>=L)o2(l,mid,now<<1,L,R,s);
    if(mid<R)o2(mid+1,r,now<<1|1,L,R,s);
    tree[now]=tree[now<<1]+tree[now<<1|1];
    maxtree[now]=max(maxtree[now<<1],maxtree[now<<1|1]);
}
ll o1(int l,int r,int now,int L,int R){
    if(l>=L&&r<=R){
        return tree[now];
    }
    int mid=(l+r)>>1;
    ll sum=0;
    if(L<=mid)sum+=o1(l,mid,now<<1,L,R);
    if(R>mid)sum+=o1(mid+1,r,now<<1|1,L,R);
    return sum;
}
int main(){
    ios::sync_with_stdio(false);
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++)cin>>a[i];
    build(1,n,1);
    while(m--){
        int op,x,y;ll c;
        cin>>op;
        if(op==1){
            cin>>x>>y;
            cout<<o1(1,n,1,x,y)<<endl;
        }
        else if(op==2){
            cin>>x>>y>>c;
            o2(1,n,1,x,y,c);
        }
        else {
            cin>>x>>c;
            o3(1,n,1,x,c);
        }
    }



    return 0;
}

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