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 first line of the input contains an integer T (1$ \le$T$ \le$20) , indicating the number of test cases, followed by T lines each of which describes a test case.
Every test case consists of N + 1 integers. The first integer is N, the number of players. Then N distinct integers a1, a2...aNfollow, indicating the skill rank of each player, in the order of west to east ( 1$ \le$ai$ \le$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

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define maxn 20000+20#define maxm 100000+20using namespace std;#define mem0(a) memset(a,0,sizeof(a))int a[maxn],c[maxn],d[maxn],e[maxm];void add(int x){    while( x <= maxm) {        e[x]+= 1;        x+= x&(-x);// lowbit(x)=x&(-x)    }}int sum(int x){    int ans = 0 ;    while(x){        ans += e[x];        x -= x&(-x);    }    return ans ;}int main(){    int t;    scanf("%d",&t);    while(t--){ //有T组测试数据        int n ;        scanf("%d",&n);//每组数据有ｎ个数        mem0(a);mem0(c);//初始化标记数组c        mem0(d);mem0(e);        for(int i = 1 ; i <= n ; i++){             scanf("%d",&a[i]); //使每一个比a[i]大的数加1，保存在相应的e[i]中            add(a[i]);            c[i] = sum(a[i]-1);//sum函数是求比a[i]小的数之和        } //用数组c[i]保存比a[i]小的数之和，                mem0(e);/*注：只要比a[i]大的数，都加1；反之c[i]=n就表示在i号数之前共有n个数比它小但在求和时要减一，不能包括它本身*/           //从后向前考虑每一个比t[i]大的数        for(int i = n ; i >= 1 ; i--){            add(a[i]);            d[i]=sum(a[i]-1);        }x        long long ans = 0 ;        for(int i = 1 ; i <= n ; i++){            ans += (long long)(c[i])*(n-i-d[i]) + (long long )d[i]*(i-1-c[i]);        }        printf("%lld\n",ans);    }    return 0;<span style="font-family: Arial, Helvetica, sans-serif;">}</span>