P1637 Ternary Ascending Subsequence

Topic description

Erwin has recently become interested in something called a "thair". . .

In the sequence a1,a2...an containing n integers,

Three numbers are called "thair" if and only if i<j<k and ai<aj<ak

Find the number of "thair" in a sequence.

Input and output format

Input format:

start with a positive integer n,

Next n numbers a1~an.

Output format:

number of "thair"

Input and output example

Input Example #1: Copy

4
2 1 3 4

Output Sample #1: Copy

Input example #2:

5
1 2 2 3 4

Sample output #2:

illustrate

Explanation for Example 2:

The seven "thair" are

1 2 3 1 2 4 1 2 3 1 2 4 1 3 4 2 3 4 2 3 4 30% of the data n<=100

60% of the data n<=2000

100% of the data n<=30000

Big data is randomly generated

0<=a[i]<=maxlongint

Solution：

　　This topic is actually quite watery. . .

　　First discretize, use the tree array to find the number $a_i$ of the number in front of each number that is smaller than it, and then find the number $b_i$ behind each number that is larger than it, then $ans=\sum{ a_i*b_i},i\in[1,n]$.

　　The output of $ans$ is $ok$ (note that the answer may explode $long\;long$).

Code:

 1 // luogu-judger-enable-o2
 2 #include<bits/stdc++.h>
 3 #define ll long long
 4 #define il inline
 5 #define lson l,m,rt<<1
 6 #define rson m+1,r,rt<<1|1
 7 using namespace std;
 8 const int N=100005;
 9 ll n,m,a[N],tr[N],tot[N];
10 ll ans;
11 struct node{
12     int v,id;
13     bool operator < (const node a)const{return v<a.v;}
14 }q[N];
15 il int gi()
16 {
17     int a=0;char x=getchar();bool f=0;
18     while((x<'0'||x>'9')&&x!='-')x=getchar();
19     if(x=='-')x=getchar(),f=1;
20     while(x>='0'&&x<='9')a=a*10+x-48,x=getchar();
21     return f?-a:a;
22 }
23 il void update(int k){while(k<=n){tr[k]++;k+=k&-k;}}
24 il ll query(int k){ll sum=0;while(k){sum+=tr[k];k-=k&-k;}return sum;}
25 int main(){
26     n=gi();
27     for(int i=1;i<=n;i++)q[i].v=gi(),q[i].id=i;
28     sort(q+1,q+n+1);q[0].v=-10000000;
29     for(int i=1;i<=n;i++)if(q[i].v!=q[i-1].v)a[q[i].id]=++m;else a[q[i].id]=m;
30     for(int i=1;i<=n;i++)update(a[i]),tot[i]=query(a[i]-1);
31     memset(tr,0,sizeof(tr));
32     for(int i=n;i>=1;i--){update(a[i]);if(tot[i])ans+=tot[i]*(query(n)-query(a[i]));}
33     cout<<ans;
34     return 0;
35 }