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(+);
 #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<=;i+=lowbit(i)) tr[i]+=x;
 }
 inline int Query(int x)
 {
     int ret=;
     for(;x;x-=lowbit(x)) ret+=tr[x];
     return ret;
 }

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