HDU2492 Ping pong

题解:

每个数，求出左右两边比这个数大的和比这个数小的，然后以每个数作为裁判，求和就行了

代码:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define LL long long
#define CLR(x) memset(x,0,sizeof x)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef pair<int,int> P;
const double eps=1e-9;
const int maxn=20100;
const int N=100010;
const int mod=1e9+7;
const int INF=1e9;

int a[maxn],l_num[N],r_num[N];
LL L[N],R[N];
int n;

int lowbit(int x){return x&-x;}

void add(int x,int v,int*c){
    while(x<N){
        c[x]+=v;
        x+=lowbit(x);
    }
}

LL sum(int x,int*c){
    LL cnt=0;
    while(x){
        cnt+=c[x];
        x-=lowbit(x);
    }
    return cnt;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        CLR(l_num);CLR(r_num);
        for(int i=1;i<=n;i++){
            L[i]=sum(a[i]-1,l_num);//左边比a[i]小
            R[n+1-i]=sum(a[n+1-i]-1,r_num);//右边比a[n+1-i]小
            add(a[i],1,l_num);
            add(a[n+1-i],1,r_num);
        }
        LL ans=0;
        for(int i=1;i<=n;i++){
            //cout<<L[i]<<" "<<R[i]<<endl;
            ans+=L[i]*(n-i-R[i])+(i-1-L[i])*R[i];
        }
        printf("%lld\n",ans);
    }
    return 0;
}