# [CH4201] Loulan Totem

I use tree array to solve this problem

After we read in the data, we analyze the problem and solve two problems: the number of '^' and the number of 'v'. Let's first consider^

Suppose we make the vertex whose i is ^, then the number of ^ whose i is the vertex is the product of the number of heights whose left side height is less than i and the number whose right side height is less than i. We just need to enumerate the positions of i and accumulate the answers.

How can we efficiently maintain the number of sides less than i height? We build a tree array (of course, the line tree can also) to store the number of times each height appears.

We are enumerating I in positive order. For each I, we look for the number of 0~a[i]-1 in the tree array, which is the number of heights on the left side of I that are less than I. after the search is completed, we put the height of a[i] in the tree array + 1, indicating that it has occurred once.

We can do another reverse enumeration to find out the number that the height on the right side of i is less than i. Then we can work out the number of ^

Similarly, we can also solve for the number of V.

``` 1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6 typedef long long ll;
7 const int N=200010;
8 int n,a[N],sum[N];
9 ll ans,l[N],r[N];
11     int ret=0;
12     int op=1;
13     char c=getchar();
14     while(c<'0'||c>'9') {if(c=='-') op=-1; c=getchar();}
15     while(c<='9'&&c>='0') ret=ret*10+c-'0',c=getchar();
16     return ret*op;
17 }
18 inline int lowbit(int x) {
19     return -x&x;
20 }
21 inline int query(int x) {
22     int ret=0;
23     while(x) {
24         ret+=sum[x];
25         x-=lowbit(x);
26     }
27     return ret;
28 }
29 inline void add(int x) {
30     while(x<=n) {
31         sum[x]++;
32         x+=lowbit(x);
33     }
34 }
35 int main() {
38     for(int i=1;i<=n;i++) {
39         l[i]=query(n)-query(a[i]);
41     }
42     memset(sum,0,sizeof(sum));
43     for(int i=n;i>=1;i--) {
44         r[i]=query(n)-query(a[i]);
46     }
47     for(int i=1;i<=n;i++)
48         ans+=(long long)l[i]*r[i];
49     printf("%lld ",ans);
50     ans=0;
51     memset(sum,0,sizeof(sum));
52     memset(l,0,sizeof(l));
53     memset(r,0,sizeof(r));
54     for(int i=1;i<=n;i++) {
55         l[i]=query(a[i]-1);
57     }
58     memset(sum,0,sizeof(sum));
59     for(int i=n;i>=1;i--) {
60         r[i]=query(a[i]-1);