G - RMQ and RAQ (cugb2023-s)

Write a program which manipulates a sequence AA = {a_0, a_1, ..., a_{n-1}a0,a1,...,an−1} with the following operations:

add(s, t, x)add(s,t,x) : add xx to a_s, a_{s+1}, ..., a_tas,as+1,...,at.
find(s, t)find(s,t) : report the minimum value in a_s, a_{s+1}, ..., a_tas,as+1,...,at.

Note that the initial values of a_i ( i = 0, 1, ..., n-1 )ai(i=0,1,...,n−1) are 0.

Input

nn qq
query_1query1
query_2query2
:
query_qqueryq

In the first line, nn (the number of elements in AA) and qq (the number of queries) are given. Then, iith query query_iqueryi is given in the following format:

0 ss tt xx

1 ss tt

The first digit represents the type of the query. '0' denotes add(s, t, x)add(s,t,x) and '1' denotes find(s, t)find(s,t).

Output

For each findfind query, print the minimum value.

Constraints

1 ≤ n ≤ 1000001≤n≤100000
1 ≤ q ≤ 1000001≤q≤100000
0 ≤ s ≤ t < n0≤s≤t<n
-1000 ≤ x ≤ 1000−1000≤x≤1000

Sample Input 1

Sample Output 1

-2
0
1
-1

#include <iostream>
#include <cmath>
using namespace std;
#define ll long long
#define inf 1e31+5
const int N=1e5+5;
ll a[N];
struct node{int l,r;ll lazy;ll minn;} tr[N<<2];//维护
void pushup(int u)
{tr[u].minn=min(tr[u<<1].minn,tr[u<<1|1].minn);} //建树void build(int u,int l,int r){tr[u].l=l;tr[u].r=r;tr[u].lazy=0;if(l==r){tr[u].minn=a[l];return ;}int mid=l+r>>1;build(u<<1,l,mid);build(u<<1|1,mid+1,r);pushup(u);}
void pushdown(int u)
{if(tr[u].lazy){tr[u<<1].lazy+=tr[u].lazy;tr[u<<1].minn=tr[u<<1].minn+tr[u].lazy;tr[u<<1|1].lazy+=tr[u].lazy;tr[u<<1|1].minn=tr[u].lazy+tr[u<<1|1].minn;tr[u].lazy=0;}
}
//区间修改
void modify(int u,int l,int r,ll k)
{if(l<=tr[u].l&&r>=tr[u].r){tr[u].lazy+=k;tr[u].minn=k+tr[u].minn;return;}pushdown(u);int mid=tr[u].l+tr[u].r>>1;if(l<=mid)modify(u<<1,l,r,k);if(r>mid)modify(u<<1|1,l,r,k);pushup(u);} //区间查询ll query(int u,int l,int r){if(l<=tr[u].l&&r>=tr[u].r){return tr[u].minn;}pushdown(u);int mid=tr[u].l+tr[u].r>>1;ll res=inf;if(l<=mid){res=min(res,query(u<<1,l,r));}if(r>mid){res=min(res,query(u<<1|1,l,r));}return res;} int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);int n,m;cin>>n>>m;for(int i=1;i<=n+5;i++){a[i]=0;}build(1,1,n);for(int i=1;i<=m;i++){int t;cin>>t;if(t==0){int l,r;ll k;cin>>l>>r>>k;modify(1,l+1,r+1,k);//注意下标从1开始}else{int l,r;cin>>l>>r;cout<<query(1,l+1,r+1)<<endl;}}}