Number sequence

Submitted : 1598, Accepted : 432

Given a number sequence which has N element(s), please calculate the number of different collocation for three number Ai, Aj, Ak, which satisfy that Ai < Aj > Ak and i < j < k.

Input

The first line is an integer N (N <= 50000). The second line contains N integer(s): A1, A2, ..., An(0 <= Ai <= 32768).

Output

There is only one number, which is the the number of different collocation.

Sample Input

5
1 2 3 4 1

Sample Output

6

分别统计每个数左边小的，然后在逆序统计一下乘起来就好啊

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<algorithm>
using namespace std;
#define N 50010
int n,a[N],c[N],lef[N];
long long ans;
int lowbit(int x)
{
      return x&(-x);
}
void update(int pos,int m)
{
      while(pos<N)
      {
            c[pos]+=m;
            pos+=lowbit(pos);
      }
}
int sum(int x)
{
      int sum=0;
      while(x>0)
      {
            sum+=c[x];
            x-=lowbit(x);
      }
      return sum;
}
int main()
{
      while(scanf("%d",&n)!=EOF)
      {
            for(int i=1;i<=n;i++)
                  scanf("%d",&a[i]);
            ans=0;
            memset(c,0,sizeof(c));
            memset(lef,0,sizeof(lef));
            for(int i=1;i<=n;i++)
            {
                 lef[i]=sum(a[i]-1);
                 update(a[i],1);
            }
            memset(c,0,sizeof(c));
            for(int i=n;i>=1;i--)
            {
                  ans+=lef[i]*sum(a[i]-1);
                  update(a[i],1);
            }
            printf("%lld\n",ans);
      }
      return 0;
}