HDU 5147 Sequence II 树状数组

Sequence II

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 375    Accepted Submission(s): 173

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

/*HDU 5147 树状数组*/#include<iostream>#include<stdio.h>using namespace std;#define N 50010int c[N],f[N],g[N],a[N];int n;//所有大于x的加上d void add(int x,int d){while(x<=n){c[x]+=d;x+=x&(-x);}}//统计小于x的个数 int sum(int x){int ans=0;while(x>0){ans+=c[x];x-=x&(-x);}return ans;}int main(){int t,i;__int64 ans,s;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=1;i<=n;i++)scanf("%d",&a[i]);memset(c,0,sizeof(c));for(i=1;i<=n;i++){add(a[i],1);f[i]=sum(a[i])-1;//左边小于a[i]的个数 }memset(c,0,sizeof(c));for(i=n;i>=1;i--){add(a[i],1);g[i]=n-i+1-sum(a[i]);//右边大于a[i]的个数 }ans=0;s=0;for(i=1;i<=n;i++){ans+=s*g[i];//与左边的相乘 s+=f[i];//统计右边的和 } printf("%I64d\n",ans); }return 0;}