UVA 1428 Ping pong

树状数组

枚举裁判位置，设裁判为第i 个人，左边有l[i]个比他小的选手，右边有r[i]个比他小的选手；

令c[i]表示技能值为i 的人是否存在，计算l[i] 即c[1]~c[i-1]的和，计算l[i]后使c[a[i]]=1；同理求r[i]；

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 100005;

int n;
int a[maxn];
long long c[maxn];
long long l[maxn],r[maxn];

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

void add (int x,long long d){
    while (x<=maxn-5){　　//这里是我坑了很久的地方。。。这里的边界是技能值边界。。。
        c[x]+=d;
        x+=lowbit(x);
    }
}

long long sum (int x){
    long long ans=0;
    while (x>0){
        ans+=c[x];
        x-=lowbit(x);
    }
    return ans;
}

int main (){
    int t;    //while (cin>>t) cout<<lowbit(t)<<endl;
    cin>>t;
    while (t--){
        cin>>n;
        for (int i=1;i<=n;i++){
            cin>>a[i];
        }
        memset (c,0,sizeof c);
        for (int i=1;i<=n;i++){
            l[i]=sum(a[i]);
            add (a[i],1);
        }
        memset (c,0,sizeof c);
        for (int i=n;i>=1;i--){
            r[i]=sum(a[i]);
            add (a[i],1);
        }
        long long ans=0;
        for (int i=1;i<=n;i++)
            ans+=l[i]*(n-i-r[i])+(i-l[i]-1)*r[i];//cout<<l[i]<<" "<<r[i]<<endl;
        cout<<ans<<endl;
    }
    return 0;
}