uva 1428 - Ping pong（树状数组，4级）

N (3N20000)

Input

The first line of the input contains an integer T (1T20) , indicating the number of test cases, followed by T

Every test case consists of N + 1 integers. The first integer is N , the number of players. Then N distinct integers a₁, a₂...a_N follow, indicating the skill rank of each player, in the order of west to east ( 1a_i100000 , i = 1...N

Output

For each test case, output a single line contains an integer, the total number of different games.

Sample Input

1
3 1 2 3

Sample Output

1

思路：第i个为裁判，则有ci为左边比i小的，di为右边比i大的。ci*(n-i-di)+di*(i-1-ci)

          结果就是它们的和。

         ci.di咋求呢？第i个的值当位置来加就好了。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int mn=1e5+9;
const int mm=2e4+9;
long long sum[mn];
long long hh[mm],tt[mm];
long long f[mm];
int lowbit(int x)
{
  return x&(-x);
}
void update(int x)
{
  for(int i=x;i<mn;i+=lowbit(i))
  {
    sum[i]++;
  }
}
int query(int x)
{ int ret=0;
  for(int i=x;i>0;i-=lowbit(i))
  {
    ret+=sum[i];
  }
  return ret;
}
int main()
{
  int cas,n;
  while(~scanf("%d",&cas))
  {
    while(cas--)
    { memset(sum,0,sizeof(sum));
      scanf("%d",&n);
      int a;
      for(int i=0;i<n;++i)
      {
        scanf("%d",&f[i]);
      }
      long long ans=0;
      for(int i=0;i<n;++i)
      {
        hh[i+1]=query(f[i]);
        update(f[i]);
      }
      memset(sum,0,sizeof(sum));
      for(int i=n-1;i>=0;--i)
      {
        tt[i+1]=query(f[i]);
        update(f[i]);
      }
      for(int i=1;i<=n;++i)
        ans+=hh[i]*((long long)(n-i-tt[i]))+(i-hh[i]-1)*(tt[i]);
      printf("%lld\n",ans);
    }
  }
  return 0;
}