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HDOJ 5147 Sequence II 樹狀數組

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HDOJ 5147 Sequence II 樹狀數組

樹狀數組:

維護每一個數前面比它小的數的個數,和這個數後面比他大的數的個數

再枚舉每個位置組合一下

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
2. Aa
3. Ac
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

Source BestCoder Round #23

/* ***********************************************
Author        :CKboss
Created Time  :2014年12月20日 星期六 21時38分00秒
File Name     :HDOJ5147.cpp
************************************************ */

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

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 找比當前數小的 2 找比當前數大的

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=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<