LA4329 Ping pong(树状数组基础)

LA4329 Ping pong

N (3  N  20000) ping pong players live along a west-east street(consider the street as a line segment).
Each player has a unique skill rank. To improve their skill rank, they often compete with each other. If
two players want to compete, they must choose a referee among other ping pong players and hold the
game in the referee’s house. For some reason, the contestants can’t choose a referee whose skill rank is
higher or lower than both of theirs. The contestants have to walk to the referee’s house, and because
they are lazy, they want to make their total walking distance no more than the distance between their
houses. Of course all players live in different houses and the position of their houses are all different. If
the referee or any of the two contestants is different, we call two games different. Now is the problem:
how many different games can be held in this ping pong street?
Input
The rst line of the input contains an integer T (1  T  20), indicating the number of test cases,
followed by T lines each of which describes a test case.
Every test case consists of N + 1 integers. The rst integer is N, the number of players. Then N
distinct integers a1; a2 : : : aN follow, indicating the skill rank of each player, in the order of west to east
(1  ai  100000, 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

树状数组与组合原理基础题，注意结果要用long long int.

#include
#include
#include
#include
using namespace std;
int a[20005];
//b[i]表示i左边比i小的个数，d[i]表示i右边比i小的个数
int b[20005];
int d[20005];
int C[100005];
int T,n;
int max_a;
int Sum(int x)
{
      int ret=0;
      while(x>0)
      {
            ret+=C[x];
            x-=x&(-x);
      }
      return ret;
}
int Add(int x,int m)
{
      while(x<=100000)
      {
            C[x]+=m;
            x+=(x&(-x));
      }
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        max_a=0;
        memset(C,0,sizeof(C));
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
           scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
        {
           Add(a[i],1);
           b[i]=Sum(a[i]-1);
        }
        memset(C,0,sizeof(C));
        for(int i=n;i>=1;i--)
        {
           Add(a[i],1);
           d[i]=Sum(a[i]-1);
        }
        long long  int ans=0;
        for(int i=2;i1-b[i])*d[i];
        }
        printf("%lld\n",ans);
    }
}