HOJ——T 2275 Number sequence

http://acm.hit.edu.cn/hoj/problem/view?id=2275

Source : SCU Programming Contest 2006 Final
	Time limit : 1 sec		Memory limit : 64 M

Submitted : 1864, Accepted : 498

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

对于每个数求出两侧的小于当前数的数量，乘法原理求和
（woc开longlong mmp多组数据 ）

#include <algorithm>
#include <cstring>
#include <cstdio>

using namespace std;

const int N(50000+5);
#define LL long long
int an1[N],an2[N];
int n,a[N],tr[N];

#define lowbit(x) (x&((~x)+1))
inline void Update(int i,int x)
{
    for(;i<=32770;i+=lowbit(i)) tr[i]+=x;
}
inline int Query(int x)
{
    int ret=0;
    for(;x;x-=lowbit(x)) ret+=tr[x];
    return ret;
}

int main()
{
    for(LL ans=0;~scanf("%d",&n);ans=0)
    {
        for(int i=1;i<=n;i++) scanf("%d",a+i);
        memset(tr,0,sizeof(tr));
        for(int i=1;i<=n;i++)
            an1[i]=Query(a[i]-1),Update(a[i],1);
        memset(tr,0,sizeof(tr));
        for(int i=n;i>=1;i--)
            an2[i]=Query(a[i]-1),Update(a[i],1);
        for(int i=1;i<=n;i++) ans=ans+(LL)an1[i]*an2[i];
        printf("%lld
",ans);
    }
    return 0;
}