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• 题意：
• 思路：

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; } /* */

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