World is Exploding

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Given a sequence A with length n,count how many quadruple (a,b,c,d) satisfies:

Input

The input consists of multiple test cases.
Each test case begin with an integer n in a single line.

The next line contains n integers

Output

For each test case,output a line contains an integer.

Sample Input

2 4 1 3

1 2 3 4

Sample Output

题意：

问符合题目给的四元组有多少个;

思路：

容斥,先算出a,b,c,d满足Aa<Ab&&Ac<Ad的个数,再减去a==c，a==d,b==c,b==d的个数,就是答案了,因为不可能有两个相等的出现;然后就是用树状数组

求pres[i],preb[i],nexs[i],nexb[i];分别表示第i个数前边比它小,比它大,后面比它小比它大的个数具体的看代码吧;

AC代码：

/************************************************
┆  ┏┓　　　┏┓ ┆   
┆┏┛┻━━━┛┻┓ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　━　　　┃ ┆
┆┃　┳┛　┗┳　┃ ┆
┆┃　　　　　　　┃ ┆ 
┆┃　　　┻　　　┃ ┆
┆┗━┓　  　┏━┛ ┆
┆　　┃　  　┃　　┆　　　　　　
┆　　┃　  　┗━━━┓ ┆
┆　　┃　 AC代马 　　┣┓┆
┆　　┃　        　　┏┛┆
┆　　┗┓┓┏━┳┓┏┛ ┆
┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆      
************************************************ */ 
 
 
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
 
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=5e4+10;
const int maxn=1e3+14;
const double eps=1e-8;

int n,fa[N],pres[N],preb[N],nexs[N],nexb[N],sum[N];
struct node
{
    int a,id;
}po[N];
int cmp(node x,node y)
{
    if(x.a==y.a)return x.id<y.id;
    return x.a<y.a;
}
int cmp1(node x,node y)
{
    if(x.a==y.a)return x.id<y.id;
    return x.a>y.a;
}

int lowbit(int x){return x&(-x);}

inline void update(int x)
{
    while(x<=n)
    {
        sum[x]++;
        x+=lowbit(x);
    }
}
int query(int x)
{
    int s=0;
    while(x)
    {
        s+=sum[x];
        x-=lowbit(x);
    }
    return s;
}
int main()
{      
        while(scanf("%d",&n)!=EOF)
        {
            For(i,1,n)read(po[i].a),po[i].id=i;
            sort(po+1,po+n+1,cmp);
            po[0].a=-1;
            mst(sum,0);
            For(i,1,n)
            {
                if(po[i].a==po[i-1].a)fa[i]=fa[i-1];
                else fa[i]=i;
                pres[po[i].id]=query(po[i].id)-(i-fa[i]);
                nexs[po[i].id]=fa[i]-1-pres[po[i].id];
                update(po[i].id);
            }
            mst(sum,0);
            sort(po+1,po+n+1,cmp1);
            For(i,1,n)
            {
                if(po[i].a==po[i-1].a)fa[i]=fa[i-1];
                else fa[i]=i;
                preb[po[i].id]=query(po[i].id)-(i-fa[i]);
                nexb[po[i].id]=fa[i]-1-preb[po[i].id];
                update(po[i].id);
            }
            sort(po+1,po+n+1,cmp1);
            LL ans1=0,ans2=0,ans;
            For(i,1,n)
            {
                ans1=ans1+pres[i];
                ans2=ans2+preb[i];
            }
            ans=ans1*ans2;
            For(i,1,n)
            {
                ans=ans-nexs[i]*nexb[i];//a==c
                ans=ans-preb[i]*nexb[i];//a==d
                ans=ans-pres[i]*nexs[i];//b==c
                ans=ans-pres[i]*preb[i];//b==d
            }         
            cout<<ans<<endl;
        }
        return 0;
}