Scalar Queries CodeForces - 1167F (计数,树状数组)

You are given an array $a_1,a_2,…,a_n$. All $a_i$ are pairwise distinct.

Let's define function $f(l,r)$ as follows:

• let's define array $b_1,b_2,…,b_{r-l+1}$, where $b_i=a_{l-1+i}$;
• sort array $b$ in increasing order;
• result of the function $f(l,r)$ is $\sum\limits_{i=1}^{r-l+1}b_i\cdot i$.

Calculate $\Bigg(\sum\limits_{1\le l\le r\le n}f(l,r)\Bigg )mod(10^9+7)$, i.e. total sum of $f$ for all subsegments of $a$ modulo $10^9+7$.

可以得到$a_x$的贡献为

$\sum\limits_{\substack{a_i<a_x\\ i<x}} i\cdot (n-x+1)+\sum\limits_{\substack{a_i<a_x\\ i>x}}x\cdot (n-i+1)+x\cdot (n-x+1)$

#include <iostream>
#include <cstdio>
#include <algorithm>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
using namespace std;
typedef long long ll;

const int N = 1e6+10, P = 1e9+7;
int n, a[N], b[N];
ll c[N], cnt[N];
void add(int x, int v) {
	for (; x<=n; x+=x&-x) c[x]+=v;
}
ll query(int x) {
	ll r = 0;
	for (; x; x^=x&-x) r+=c[x];
	return r%P;
}

int main() {
	scanf("%d", &n);
	REP(i,1,n) scanf("%d",a+i),b[i]=a[i];
	sort(b+1,b+1+n);
	REP(i,1,n) a[i]=lower_bound(b+1,b+1+n,a[i])-b;
	REP(i,1,n) {
		cnt[i] += query(a[i])*(n-i+1)%P;
		add(a[i], i);
	}
	REP(i,1,n) cnt[i] += (ll)i*(n-i+1)%P, c[i] = 0;
	PER(i,1,n) {
		cnt[i] += query(a[i])*i%P;
		add(a[i], (n-i+1));
	}
	ll ans = 0;
	REP(i,1,n) ans+=cnt[i]*b[a[i]]%P;
	printf("%lld\n", ans%P);
}