+ Segment tree thinking --cf1311F

/*
1-100t
3+2t
2+3t 
v<0的分一组，v>=0的分一组 
*/ 
#include<bits/stdc++.h>
using namespace std;
#define N 400005
#define ll long long

struct Node{
    ll x,v;
}a[N],b[N];
int n,tot1,tot2;
int cmp1(Node a,Node b){return a.x<b.x;}
ll ans,Sum[N],x[N],m;

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
ll sum[N<<2],cnt[N<<2];
void build(int l,int r,int rt){
    memset(sum,0,sizeof sum);
    memset(cnt,0,sizeof cnt);
} 
void update(int pos,ll v,int l,int r,int rt){
    if(l==r){
        cnt[rt]++;sum[rt]+=v;return;
    }
    int m=l+r>>1;
    if(pos<=m)update(pos,v,lson);
    else update(pos,v,rson);
    cnt[rt]=cnt[rt<<1]+cnt[rt<<1|1];
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
ll query1(int L,int R,int l,int r,int rt){
    if(L<=l && R>=r)return cnt[rt];
    int m=l+r>>1;
    ll res=0;
    if(L<=m)res+=query1(L,R,lson);
    if(R>m)res+=query1(L,R,rson);
    return res;
}
ll query2(int L,int R,int l,int r,int rt){
    if(L<=l && R>=r)return sum[rt];
    int m=l+r>>1;
    ll res=0;
    if(L<=m)res+=query2(L,R,lson);
    if(R>m)res+=query2(L,R,rson);
    return res;
}

ll t[N],tt[N];
int main(){
    cin>>n;
    for(int i=1;i<=n;i++)cin>>t[i];
    for(int i=1;i<=n;i++)cin>>tt[i];
    for(int i=1;i<=n;i++){
        int x,v;x=t[i];v=tt[i];
        if(v<0){
            tot1++;
            a[tot1].x=x;a[tot1].v=v;
        }else {
            tot2++;
            b[tot2].x=x;b[tot2].v=v;
        }
    }
    sort(a+1,a+1+tot1,cmp1);
    sort(b+1,b+1+tot2,cmp1);
    if(tot1 && tot2){//两组间的贡献 
        for(int i=1;i<=tot2;i++)Sum[i]=Sum[i-1]+b[i].x;
        for(int i=1;i<=tot1;i++){
            if(a[i].x<b[1].x){
                ans+=Sum[tot2]-a[i].x*tot2;
                continue;
            }
            int L=1, R & lt = tot2, MID, POS;
             the while (L <= R & lt) { 
                MID = L + R & lt >> . 1 ;
                 IF (B [MID] .x <= A [I] .x) 
                    POS = MID, MID = L + . 1 ;
                 the else R & lt mid- = . 1 ; 
            } 
            ANS + = (the Sum [tot2] -sum [POS]) - a [I] .x * (tot2- POS); 
        } 
    } 
    
    // in a contribution of the same group: in a [i] and the speed of the left side of a large absolute value 
    for ( int I = . 1 ; I <= Tot1; I ++) X [++ m] = a [i] .v; 
    Sort (X + . 1 , X +1+m);
    m=unique(x+1,x+1+m)-x-1;
    build(1,m,1);
    for(int i=1;i<=tot1;i++){
        int pos=lower_bound(x+1,x+1+m,a[i].v)-x;
        ll num=query1(1,pos,1,m,1);
        ll sum=query2(1,pos,1,m,1);
        ans+=num*a[i].x-SUM; 
        Update (POS, A [I] .x, . 1 , m, . 1 ); 
    } 
    
    // in the same group contribution b: the b [i] and the speed of the left small absolute value of 
    m = 0 ; 
     for ( int I = . 1 ; I <= tot2; I ++) X [++ m] = B [I] .v; 
    Sort (X + . 1 , X + . 1 + m); 
    m = UNIQUE (X + . 1 , X + . 1 + m) -X- . 1 ; 
    Build ( . 1 , m, . 1 );
     for ( int I = . 1 ; I <= tot2; I ++ ) {
         int pos=lower_bound(x+1,x+1+m,b[i].v)-x;
        ll num=query1(1,pos,1,m,1);
        ll sum=query2(1,pos,1,m,1);
        ans+=num*b[i].x-sum;
        update(pos,b[i].x,1,m,1); 
    }
    
    cout<<ans<<'\n';
}