HDOJ 5147 Sequence II 树状数组

树状数组:

维护每一个数前面比它小的数的个数,和这个数后面比他大的数的个数

再枚举每个位置组合一下 

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

151 3 2 4 5

 

Sample Output

4

 

Source

BestCoder Round #23

 

/* ***********************************************Author        :CKbossCreated 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 rightfor(int i=1;i<=n;i++){int ss=sum(1,a[i]);sum1[i]=ss;add(1,a[i]);}/// from right to leftfor(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...Xans+=sum1[i]*sum2[i+1];}cout<<ans<<endl;}        return 0;}