题意

给一个数列\(a\),定义\(f(l,r)\)为\(b_1, b_2, \dots, b_{r - l + 1}\),\(b_i = a_{l - 1 + i}\),将\(b\)排序,\(f(l,r)\)=\(\sum\limits_{i = 1}^{r - l + 1}{b_i \cdot i}\)

计算\(\left(\sum\limits_{1 \le l \le r \le n}{f(l, r)}\right) \mod (10^9+7)\)

分析

考虑每个数字对答案的贡献，首先每个\(a_i\)的区间贡献为\((n-i+1)\times i\times a_i\)

在\(a_i\)左边的比它小的数\(a_j\)的贡献为\((n-i+1)\times j\times a_i\)

在\(a_i\)右边的比它小的数\(a_j\)的贡献为\(i\times (n-j+1)\times a_i\)

将所有贡献加起来即为答案

按排序后的下标建树状数组，维护原下标前缀和,边更新边加答案

Code

#include<bits/stdc++.h>
#define fi first
#define se second
#define bug cout<<"--------------"<<endl
using namespace std;
typedef long long ll;
const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=1e9;
const ll llf=1e18;
const int mod=1e9+7;
const int maxn=5e5+10;
int n;
ll a[maxn],c[maxn],tr[maxn];
void add(int x,ll k){
	while(x<=n){
		tr[x]=(tr[x]+k)%mod;
		x+=x&-x;
	}
}
ll dor(int x){
	ll ret=0;
	while(x){
		ret=(ret+tr[x])%mod;
		x-=x&-x;
	}
	return ret;
}
ll ans=0;
void solve(){
	memset(tr,0,sizeof(tr));
	for(int i=1;i<=n;i++){
		ll t=dor(a[i])*(n-i+1)%mod;
		ans+=c[a[i]]*t%mod;
		ans%=mod;
		add(a[i],i);
	}
}
int main(){
	ios::sync_with_stdio(false);
	//freopen("in","r",stdin);
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		ans+=a[i]*(n-i+1)%mod*i%mod;
		ans%=mod;
		c[i]=a[i];
	}
	sort(c+1,c+n+1);
	for(int i=1;i<=n;i++){
		a[i]=lower_bound(c+1,c+n+1,a[i])-c;
	}
	solve();
	reverse(a+1,a+n+1);
	solve();
	cout<<ans<<endl;
	return 0;
}