HDU 2838 Cow Sorting

HDU 2838 Cow Sorting

用三个树状数组维护一下：比a[i]小的数有多少、前i个数之和是多少、比a[i]小的数之和是多少。
以下是我的代码：

/*
 * Author:  lee1r
 * Created Time:  2011/8/2 9:33:01
 * File Name: hdu2838.cpp
  */
#include < iostream >
#include < sstream >
#include < fstream >
#include < vector >
#include < list >
#include < deque >
#include < queue >
#include < stack >
#include < map >
#include < set >
#include < bitset >
#include < algorithm >
#include < cstdio >
#include < cstdlib >
#include < cstring >
#include < cctype >
#include < cmath >
#include < ctime >
#define  lowbit(x) ((x)&(-(x)))
using   namespace  std;
typedef  long   long  int64;
typedef unsigned  long   long  uint64;
const   int  kMaxn( 100007 );

int  n;
int64 ans,bit[kMaxn],bit2[kMaxn],bit3[kMaxn];

void  Add(int64  * a, int  x, int  delta)
{
     for ( int  i = x;i <= n;i += lowbit(i))
        a[i] += delta;
}

int64 Sum(int64  * a, int  x)
{
    int64 re( 0 );
     for ( int  i = x;i > 0 ;i -= lowbit(i))
        re += a[i];
     return  re;
}

int  main()
{
     // freopen("data.in","r",stdin);

     while (scanf( " %d " , & n) == 1 )
    {
        ans = 0 ;
        memset(bit, 0 , sizeof (bit));
        memset(bit2, 0 , sizeof (bit2));
        memset(bit3, 0 , sizeof (bit3));

         for ( int  i = 1 ;i <= n;i ++ )
        {
             int  t;
            scanf( " %d " , & t);
            Add(bit,t, 1 );
            Add(bit2,i,t);
            Add(bit3,t,t);
            ans += (i - Sum(bit,t - 1 ) - 1 ) * t;
            ans += Sum(bit2,i - 1 );
            ans -= Sum(bit3,t - 1 );
        }

        cout << ans << endl;
    }

     return   0 ;
}