HDOJ 5147 Sequence II 树阵

时间：2023-03-08 17:07:53

树阵:

每个号码的前面维修比其数数少,和大量的这后一种数比他的数字

再枚举每一个位置组合一下

Sequence II

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 121    Accepted Submission(s): 58

Problem Description
Long long ago, there is a sequence A with length n. All numbers in this sequence is no smaller than 1 and no bigger than n, and all numbers are different in this sequence.

Please calculate how many quad (a,b,c,d) satisfy:

1. 1≤a<b<c<d≤n

2. Aa<Ab

3. Ac<Ad
Input
The first line contains a single integer T, indicating the number of test cases.

Each test case begins with a line contains an integer n.

The next line follows n integers A1,A2,…,An.



[Technical Specification]

1 <= T <= 100

1 <= n <= 50000

1 <= Ai <=
n
Output
For each case output one line contains a integer,the number of quad.
Sample Input
1

5

1 3 2 4 5
Sample Output
4
Source

/* ***********************************************

Author        :CKboss

Created Time  :2014年12月20日 星期六 21时38分00秒

File Name     :HDOJ5147.cpp

************************************************ */

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <string>

#include <cmath>

#include <cstdlib>

#include <vector>

#include <queue>

#include <set>

#include <map>

using namespace std;

typedef long long int LL;

const int maxn=55000;

int a[maxn];

int n;

LL sum1[maxn],sum2[maxn];

int t1[maxn],t2[maxn];

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

/// 1 找比当前数小的 2 找比当前数大的

void init()

{

	memset(sum1,0,sizeof(sum1));

	memset(sum2,0,sizeof(sum2));

	memset(t1,0,sizeof(t1));

	memset(t2,0,sizeof(t2));

}

void add(int kind,int p)

{

	if(kind==1) for(int i=p;i<maxn;i+=lowbit(i)) t1[i]+=1;

	else if(kind==2) for(int i=p;i;i-=lowbit(i)) t2[i]+=1;

}

int sum(int kind,int p)

{

	int ret=0;

	if(kind==1) for(int i=p;i;i-=lowbit(i)) ret+=t1[i];

	else if(kind==2) for(int i=p;i<maxn;i+=lowbit(i)) ret+=t2[i];

	return ret;

}

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

	int T_T;

	scanf("%d",&T_T);

	while(T_T--)

	{

		scanf("%d",&n);

		for(int i=1;i<=n;i++) scanf("%d",a+i);

		init();

		/// from left to right

		for(int i=1;i<=n;i++)

		{

			int ss=sum(1,a[i]);

			sum1[i]=ss;

			add(1,a[i]);

		}

		/// from right to left

		for(int i=n;i>=1;i--)

		{

			int ss=sum(2,a[i]);

			sum2[i]=sum2[i+1]+ss;

			add(2,a[i]);

		}

		LL ans=0;

		for(int i=2;i<=n-1;i++)

		{

			/// X...i i+1...X

			ans+=sum1[i]*sum2[i+1];

		}

		cout<<ans<<endl;

	}

    return 0;

}

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