codeforces 671C Ultimate Weirdness of an Array 线段树+构造

题解上说的很清楚了，我照着写的，表示膜拜题解

然后时间复杂度我觉得应该是O(nlogn)，虽然常数略大，预处理和倒着扫，都是O(nlogn)

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 2e5+;
int mat[N],n,mx,tmp,lz[N<<],a[N<<];
LL c[N<<],sum,h[N];
vector<int>v[N];
void build(int rt,int l,int r){
    if(l==r){
        a[rt]=c[rt]=1ll*l;
        return;
    }
    int m=(l+r)>>;
    build(rt<<,l,m);
    build(rt<<|,m+,r);
    c[rt]=c[rt<<]+c[rt<<|];
    a[rt]=min(a[rt<<],a[rt<<|]);
}
void down(int rt,int l,int r){
   if(!lz[rt])return;
   int m=(l+r)>>;
   c[rt<<]=1ll*lz[rt]*(m-l+);
   c[rt<<|]=1ll*lz[rt]*(r-m);
   a[rt<<]=a[rt<<|]=a[rt];
   lz[rt<<]=lz[rt<<|]=lz[rt];
   lz[rt]=;
}
void modify(int rt,int l,int r,int x,int y){
   if(x<=l&&r<=y){
     c[rt]=1ll*tmp*(r-l+);
     a[rt]=lz[rt]=tmp;
     return;
   }
   down(rt,l,r);
   int m=(l+r)>>;
   if(x<=m)modify(rt<<,l,m,x,y);
   if(y>m)modify(rt<<|,m+,r,x,y);
   c[rt]=c[rt<<]+c[rt<<|];
   a[rt]=min(a[rt<<],a[rt<<|]);
}
int ask(int rt,int l,int r,int k){
    if(l==r)return l;
    int m=(l+r)>>;
    if(a[rt<<|]<k)ask(rt<<|,m+,r,k);
    else ask(rt<<,l,m,k);
}
int main(){
    scanf("%d",&n);
    for(int i=;i<=n;++i){
      int x;scanf("%d",&x);
      mat[x]=i;mx=max(mx,x);
    }
    for(int i=;i<=mx;++i){
      for(int j=i;j<=mx;j+=i)
        if(mat[j])
        v[i].push_back(mat[j]);
    }
    for(int i=;i<=mx;++i)
    if(v[i].size())sort(v[i].begin(),v[i].end());
    build(,,n);
    sum=1ll*n*(n+);
    for(int i=mx;i>=;--i){
      h[i]=sum-c[];
 
      int k=v[i].size();
      if(k<)continue;
      tmp=n+;
      if(v[i][]+<=n){
          modify(,,n,v[i][]+,n);
      }
      tmp=v[i][k-];
      if(a[]<tmp){
        int pos=ask(,,n,tmp);
        pos=min(pos,v[i][]);
        if(pos>v[i][])
        modify(,,n,v[i][]+,pos);
      }
      tmp=v[i][k-];
      if(a[]<tmp){
        int pos=ask(,,n,tmp);
        pos=min(pos,v[i][]);
        if(pos>)
        modify(,,n,,pos);
      }
    }
    LL ans=;
    for(int i=;i<=mx;++i)
        ans+=1ll*i*(h[i]-h[i-]);
    printf("%I64d\n",ans);
    return ;
}