CF1167F. Scalar Queries

Solution

拆贡献，自身的贡献为 $b_i*(i-1)*(n-i)$ ，每一个左边比他小的数 $b_j$ 会产生 $b_i*j*(n-i)$ 的贡献，需要维护 $∑j\sum_{j}$ ，右边比它小的数会产生 $b_i*(i-1)*(n-j)$ 的贡献，需要维护 $∑n−j\sum_{n-j}$ 。

把 $b_i$ 从小到大排序，分别求一下左边的贡献和右边的贡献，树状数组维护即可。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=1e9+7;
const int MAXN=500005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
int ans=0,a[MAXN],b[MAXN],c[MAXN],s[MAXN],n;
int upd(int x,int y) { return x+y>=mods?x+y-mods:x+y; }
void add(int x,int y) { for (;x<=n;x+=x&(-x)) s[x]=upd(s[x],upd(y,mods)); }
int query(int x) { int ans=0; for (;x;x-=x&(-x)) ans=upd(ans,s[x]); return ans; }
void solve()
{for (int i=1;i<=n;i++){ans=upd(ans,1ll*query(a[i])*(n-i+1)%mods*c[i]%mods);add(a[i],i);}for (int i=1;i<=n;i++) add(a[i],-i);
}
signed main()
{n=read();for (int i=1;i<=n;i++) a[i]=b[i]=c[i]=read(),ans=upd(ans,1ll*a[i]*(n-i+1)%mods*i%mods);sort(b+1,b+n+1);for (int i=1;i<=n;i++) a[i]=lower_bound(b+1,b+n+1,a[i])-b;solve();reverse(a+1,a+n+1);reverse(c+1,c+n+1);solve();printf("%d\n",ans);return 0;
}