Luogu p1816 loyalty solution

Luogu P1816 loyal Answer key

Topic link ：P1816 loyal

The question ： Section min.

BIT The writing method is just for fun （ Actually not recommended ）

There are some differences in inquiry ,

For example, if x-lowbit(x) Less than $l$ , Then you cannot directly x-=lowbit(x)

Because this will put $[l,r]$ Get something else in , So this part deals directly with violence

Unclear complexity , It should still be $O(\log n)$

Code ：

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iomanip>
#include <random>
using namespace std;
#define int long long
#define INF 0x3f3f3f3f3f3f3f3f
#define N (int)(1e6+15)

int n,m,a[N],tree[N];
int lowbit(int x){
    return x&(-x);}
void update(int x,int v)
{
    
    for(; x<=n; x+=lowbit(x))
        tree[x]=min(tree[x],v);
}
int qMin(int l,int r)
{
    
    int x=r,mn=INF;
    for(; x>=l;)
    {
    
        if(x-lowbit(x)>l)
        {
    
            mn=min(mn,tree[x]);
            x-=lowbit(x);
        }
        else mn=min(mn,a[x]),--x;
    }
    return mn;
}
signed main()
{
    
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    // freopen("check.in","r",stdin);
    // freopen("check.out","w",stdout);
    memset(tree,0x3f,sizeof(tree));
    cin >> n >> m;
    for(int i=1; i<=n; i++)
    {
    
        cin >> a[i];
        update(i,a[i]);
    }
    for(int i=1,l,r; i<=m; i++)
    {
    
        cin >> l >> r;
        cout << qMin(l,r) << ' ';
    }
    return 0;
}

Reprint please explain the source