CF1167F. Scalar Queries

Solution

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

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; }

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