BZOJ 3745

题解：

分治好题

首先暴力显然rmq可以做到n^2

比较容易想到是以最值分治，这样在数据随机复杂度是nlogn，不随机还是n^2的

以最值分治只有做多与较小区间复杂度相同才是nlogn的

而这题里我们直接分治

想清楚再搞个暴力对拍还是比较好写的

代码：

#include <bits/stdc++.h>
using namespace std;
#define rint register int
#define IL inline
#define rep(i,h,t) for (rint i=h;i<=t;i++)
#define dep(i,t,h) for (rint i=t;i>=h;i--) 
#define ll long long
#define mid ((h+t)>>1)
const int N=6e5;
const int mo=1e9;
int a[N],sum[N],sum2[N],sum3[N],sum4[N],sum5[N],sum6[N];
ll ans=0;
void fz(int h,int t)
{
  if (h==t)
  {
    ans=(ans+1ll*a[h]*a[h])%mo;
    return;
  }
  int mina=a[mid],maxa=a[mid];
  sum[mid+1]=sum2[mid+1]=sum3[mid+1]=sum4[mid+1]=0;
  dep(i,mid,h)
  {
    mina=min(a[i],mina);
    maxa=max(a[i],maxa);
    sum[i]=(sum[i+1]+(1ll*mina*maxa)%mo*(mid-i+1))%mo;
    sum2[i]=(sum2[i+1]+1ll*mina*maxa)%mo;
    sum3[i]=(sum3[i+1]+1ll*(mid-i+1)*mina)%mo;
    sum4[i]=(sum4[i+1]+mina)%mo;
    sum5[i]=(sum5[i+1]+1ll*(mid-i+1)*maxa)%mo;
    sum6[i]=(sum6[i+1]+maxa)%mo;
  }
  mina=a[mid],maxa=a[mid];
  int pos1=mid,pos2=mid;
  rep(i,mid,t)
  {
    mina=min(a[i],mina);
    maxa=max(a[i],maxa);
    while (a[pos1]>=mina&&pos1>=h) pos1--; pos1++;
    while (a[pos2]<=maxa&&pos2>=h) pos2--; pos2++;
    if (pos2<pos1)
    {
       ans+=(1ll*(2*i-mid-pos1+2)*(mid-pos1+1)/2)%mo*mina%mo*maxa%mo;
       ans+=(sum3[pos2]-sum3[pos1]+1ll*(i-mid)*(sum4[pos2]-sum4[pos1]))%mo*maxa%mo;
       ans+=(sum[h]-sum[pos2]+1ll*(i-mid)*(sum2[h]-sum2[pos2]))%mo;
       ans%=mo;
    } else
    {
      ans+=(1ll*(2*i-mid-pos2+2)*(mid-pos2+1)/2)%mo*mina%mo*maxa%mo;
      ans+=(sum5[pos1]-sum5[pos2]+1ll*(i-mid)*(sum6[pos1]-sum6[pos2]))%mo*mina%mo;
      ans+=(sum[h]-sum[pos1]+1ll*(i-mid)*(sum2[h]-sum2[pos1]))%mo;
      ans%=mo;
    }
  }
  if (h<=mid-1) fz(h,mid-1);
  if (mid+1<=t) fz(mid+1,t);
}
int main()
{
  freopen("1.in","r",stdin);
  freopen("1.out","w",stdout);
  ios::sync_with_stdio(false);
  int n;
  cin>>n;
  rep(i,1,n) cin>>a[i];
  fz(1,n);
  cout<<ans<<endl; 
  return 0;
}