Archie

显然是要求逆序对的，这东西当然可以用树状数组求就可以了

然后呢，对于每一个点i，它为后面的点的话，那$j>i$的时候就会有贡献

扫就行了

#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
int n;
struct hair{
    int id;
    ll key;
}h[100005];
ll tr[100005];
ll ans[100005];
int lowbit(int x){
    return x&-x;
}
void add(int x,int k){
    for(int i=x;i<=n;i+=lowbit(i)){
        tr[i]+=k;
    }
} 
ll suf[100005];
ll query(int x){
    ll ans=0;
    while(x){
        ans+=tr[x];
        x-=lowbit(x);
    }
    return ans;
}
int cnt;
bool cmp(hair x,hair y){
    if(x.key==y.key){
        return x.id>y.id;
    }else{
        return x.key>y.key;
    }
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;++i){
        scanf("%lld",&h[i].key);
        h[i].id=i;
    }
    sort(h+1,h+n+1,cmp);
    for(int i=1;i<=n;++i){
        add(h[i].id,1);
        ans[h[i].id]=query(h[i].id-1);
    }
    cnt=n;
    ll anss=0;
    for(int i=0;i<n;++i){
        while(cnt>=1&&i>h[cnt].key){
            anss+=ans[h[cnt].id];
            cnt--;
        }
        cout<<anss<<endl;
    }
    return 0;
}