传送门

题意：

思路：

考虑将贡献分开来算，先计算最大值，再算个最小值，之后答案就是$((max+min)/2)/(len\ast (len+1)/2)$。
这是一个原题，直接封装两个结构体跑两次答案即可。
在线做法且复杂度$O(nlogn)$，吊打标程 。
当然还有线段树 + 单调栈的写法，目前没看懂，看懂再补。

// Problem: P3246 [HNOI2016]序列
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3246
// Memory Limit: 500 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=400010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n,m;
int a[N];
int stk[N],top;
LL suf[N],pre[N];
LL psum[N],ssum[N];
int len[N];
PII q[N];
LL all[N];
int f[N][25];struct Node1 {void rmq_init(){memset(f,0,sizeof(f));for(int i=1;i<=n;i++) f[i][0]=i;int t=log(n)/log(2)+1;for(int j=1;j<t;j++)for(int i=1;i<=n-(1<<j)+1;i++) {if(a[f[i][j-1]]<a[f[i+(1ll<<(j-1))][j-1]]) f[i][j]=f[i][j-1];else if(a[f[i][j-1]]>=a[f[i+(1ll<<(j-1))][j-1]]) f[i][j]=f[i+(1ll<<(j-1))][j-1];}}int query(int l,int r) {int t=len[r-l+1];if(a[f[l][t]]<a[f[r-(1<<t)+1][t]]) return f[l][t];else  return f[r-(1<<t)+1][t];}void init() {top=0;suf[n+1]=0; ssum[n+1]=0;}void get() {init(); rmq_init();memset(suf,0,sizeof(suf));memset(pre,0,sizeof(pre));memset(psum,0,sizeof(psum));memset(ssum,0,sizeof(ssum));for(int i=1;i<=n;i++) {while(top&&a[stk[top]]>a[i]) suf[stk[top--]]=i;pre[i]=stk[top]; stk[++top]=i;}while(top) pre[stk[top]]=stk[top-1],suf[stk[top--]]=n+1;for(int i=1;i<=n;i++) psum[i]=(1ll*(i-pre[i])*a[i]%mod+psum[pre[i]])%mod;for(int i=n;i>=1;i--) ssum[i]=(1ll*(suf[i]-i)*a[i]%mod+ssum[suf[i]])%mod;pre[0]=pre[n+1]=suf[0]=suf[n+1]=0;for(int i=1;i<=n;i++) pre[i]=(pre[i-1]+psum[i])%mod;for(int i=n;i>=1;i--) suf[i]=(suf[i+1]+ssum[i])%mod;for(int i=1;i<=m;i++) {int l=q[i].X,r=q[i].Y;int pos=query(l,r);LL ans=1ll*(1ll*pos-l+1)*(r-pos+1)%mod*a[pos]%mod;ans+=pre[r]-pre[pos]-psum[pos]*(r-pos)%mod;ans+=suf[l]-suf[pos]-ssum[pos]*(pos-l)%mod;all[i]+=ans%mod; all[i]%=mod;all[i]+=mod; all[i]%=mod;}}}x;struct Node2 {void rmq_init(){memset(f,0,sizeof(f));for(int i=1;i<=n;i++) f[i][0]=i;int t=log(n)/log(2)+1;for(int j=1;j<t;j++)for(int i=1;i<=n-(1<<j)+1;i++) {if(a[f[i][j-1]]>a[f[i+(1ll<<(j-1))][j-1]]) f[i][j]=f[i][j-1];else f[i][j]=f[i+(1ll<<(j-1))][j-1];}}int query(int l,int r) {int t=len[r-l+1];if(a[f[l][t]]>a[f[r-(1<<t)+1][t]]) return f[l][t];else return f[r-(1<<t)+1][t];}void init() {top=0;suf[n+1]=0; ssum[n+1]=0;}void get() {init(); rmq_init();memset(suf,0,sizeof(suf));memset(pre,0,sizeof(pre));memset(psum,0,sizeof(psum));memset(ssum,0,sizeof(ssum));for(int i=1;i<=n;i++) {while(top&&a[stk[top]]<a[i]) suf[stk[top--]]=i;pre[i]=stk[top]; stk[++top]=i;}while(top) pre[stk[top]]=stk[top-1],suf[stk[top--]]=n+1;for(int i=1;i<=n;i++) psum[i]=(1ll*(i-pre[i])*a[i]%mod+psum[pre[i]])%mod;for(int i=n;i>=1;i--) ssum[i]=(1ll*(suf[i]-i)*a[i]%mod+ssum[suf[i]])%mod;pre[0]=pre[n+1]=suf[0]=suf[n+1]=0;for(int i=1;i<=n;i++) pre[i]=(pre[i-1]+psum[i])%mod;for(int i=n;i>=1;i--) suf[i]=(suf[i+1]+ssum[i])%mod;for(int i=1;i<=m;i++) {int l=q[i].X,r=q[i].Y;int pos=query(l,r);LL ans=1ll*(1ll*pos-l+1)*(r-pos+1)%mod*a[pos]%mod;ans+=pre[r]-pre[pos]-psum[pos]*(r-pos)%mod;ans+=suf[l]-suf[pos]-ssum[pos]*(pos-l)%mod;all[i]+=ans%mod; all[i]%=mod;all[i]+=mod; all[i]%=mod;}}}y;LL qmi(LL a,LL b) {LL ans=1; a%=mod;while(b) {if(b&1) ans=ans*a%mod;a=a*a%mod;b>>=1;}return ans%mod;
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);int _; scanf("%d",&_);while(_--) {memset(all,0,sizeof(all));scanf("%d%d",&n,&m);for(int i=1;i<=n;i++) len[i]=(int)(log(i)/log(2));for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int i=1;i<=m;i++) {int l,r; scanf("%d%d",&l,&r);q[i]={l,r};}for(int i=1;i<=m;i++) all[i]=0;x.get(); y.get();for(int i=1;i<=m;i++) printf("%lld\n",all[i]%mod*qmi(2,mod-2)%mod*qmi(1ll*(1ll*q[i].Y-q[i].X+1)*(q[i].Y-q[i].X+1+1)/2,mod-2)%mod);}return 0;
}
/*
*/