See section divisible operations, intuition is not except too many times to become full \ (1 \) .
But now there plus operation.
I do not know why, when a node \ (\ lfloor \ frac {mx } {d} \ rfloor = \ lfloor \ frac {mn} {d} \ rfloor \) time interval becomes an assignment, or continue recursive complex of wrong, but \ (\ lfloor \ frac {mx } {d} \ rfloor-mx = \ lfloor \ frac {mn} {d} \ rfloor-mn \) into sections on the right added complexity? ? ?
The following two were using.
(Seeking to prove complexity ......)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=400040;
#define ls o<<1
#define rs o<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define ROF(i,a,b) for(int i=(a);i>=(b);i--)
#define MEM(x,v) memset(x,v,sizeof(x))
inline int read(){
int x=0,f=0;char ch=getchar();
while(ch<'0' || ch>'9') f|=ch=='-',ch=getchar();
while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
return f?-x:x;
}
int n,q,a[maxn];
ll sum[maxn],add[maxn],cov[maxn],mx[maxn],mn[maxn];
inline void pushup(int o){
sum[o]=sum[ls]+sum[rs];
mx[o]=max(mx[ls],mx[rs]);
mn[o]=min(mn[ls],mn[rs]);
}
inline void setadd(int o,int l,int r,ll v){
sum[o]+=(r-l+1)*v;
mx[o]+=v;
mn[o]+=v;
add[o]+=v;
}
inline void setcov(int o,int l,int r,ll v){
sum[o]=(r-l+1)*v;
mx[o]=mn[o]=cov[o]=v;
add[o]=0;
}
inline void pushdown(int o,int l,int r){
int mid=(l+r)>>1;
if(cov[o]!=1e18){
setcov(lson,cov[o]);
setcov(rson,cov[o]);
cov[o]=1e18;
}
if(add[o]){
setadd(lson,add[o]);
setadd(rson,add[o]);
add[o]=0;
}
}
void build(int o,int l,int r){
cov[o]=1e18;
if(l==r) return void(mx[o]=mn[o]=sum[o]=a[l]);
int mid=(l+r)>>1;
build(lson);build(rson);
pushup(o);
}
void update_add(int o,int l,int r,int ql,int qr,int v){
if(l>=ql && r<=qr) return setadd(o,l,r,v);
int mid=(l+r)>>1;
pushdown(o,l,r);
if(mid>=ql) update_add(lson,ql,qr,v);
if(mid<qr) update_add(rson,ql,qr,v);
pushup(o);
}
inline ll ddiv(ll x,ll y){
if(x>=0) return x/y;
else return -((-x+y-1)/y);
}
void update_div(int o,int l,int r,int ql,int qr,int v){
if(l>=ql && r<=qr){
if(ddiv(mx[o],v)==ddiv(mn[o],v)) return setcov(o,l,r,ddiv(mx[o],v));
if(ddiv(mx[o],v)-mx[o]==ddiv(mn[o],v)-mn[o]) return setadd(o,l,r,ddiv(mx[o],v)-mx[o]);
}
int mid=(l+r)>>1;
pushdown(o,l,r);
if(mid>=ql) update_div(lson,ql,qr,v);
if(mid<qr) update_div(rson,ql,qr,v);
pushup(o);
}
ll query_sum(int o,int l,int r,int ql,int qr){
if(l>=ql && r<=qr) return sum[o];
int mid=(l+r)>>1;
pushdown(o,l,r);
if(mid<ql) return query_sum(rson,ql,qr);
if(mid>=qr) return query_sum(lson,ql,qr);
return query_sum(lson,ql,qr)+query_sum(rson,ql,qr);
}
ll query_min(int o,int l,int r,int ql,int qr){
if(l>=ql && r<=qr) return mn[o];
int mid=(l+r)>>1;
pushdown(o,l,r);
if(mid<ql) return query_min(rson,ql,qr);
if(mid>=qr) return query_min(lson,ql,qr);
return min(query_min(lson,ql,qr),query_min(rson,ql,qr));
}
int main(){
n=read();q=read();
FOR(i,1,n) a[i]=read();
build(1,1,n);
while(q--){
int op=read(),l=read()+1,r=read()+1;
if(op==1) update_add(1,1,n,l,r,read());
if(op==2) update_div(1,1,n,l,r,read());
if(op==3) printf("%lld\n",query_min(1,1,n,l,r));
if(op==4) printf("%lld\n",query_sum(1,1,n,l,r));
}
}