hoj 2275 Number sequence

Number sequence

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

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 50010using 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;}