Number sequence

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

题目就是统计序列中Ai < Aj > Ak（i < j < k）的个数，可以从前往后统计每个元素之前小于它的数的个数，在从后往前统计每个元素之后小于它的数的个数。然后乘积加和即可。用树状数组轻松搞定！！

AC代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#define M 50010
using namespace std;

int c[M],num[M];
int l[M],n;

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

void add(int a,int b)
{
    while (a<M)
    {
        c[a]+=b;
        a+=lowbit (a);
    }
}

int sum(int a)
{
    int ans=0;
    while(a>0)
    {
        ans+=c[a];
        a-=lowbit(a);
    }
    return ans;
}

int main ()
{
    int i,j;
    int a,b;
    while(~scanf("%d",&n))
    {
        memset(c,0,sizeof c);
        memset(num,0,sizeof num);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
            l[i]=sum(num[i]-1);
            add(num[i],1);
        }
        memset(c,0,sizeof c);
        long long ans=0;
        for(i=n;i>=1;i--)
        {
            ans=ans+(long long)sum(num[i]-1)*l[i];
            add(num[i],1);
        }
        printf("%lld\n",ans);
    }
    return 0;
}