https://codeforces.com/contest/1311/problem/F
The meaning of problems: there are n points on the coordinate axis, the position of each point xi (not overlap), and having a fixed speed vi, ask for all the shortest distance between the point and the.
Solution: Analysis can be properly xi> yi && vi> yi and shortest distance xi-yi, are 0 otherwise.
Sorting by x, statistics have to coordinate and the speed is smaller than xi vi smaller than the number of coordinates and sum1 and sum2.
The point contribution can be counted sum1 * xi - sum2 because of the larger range of v so be discrete.
#include<bits/stdc++.h>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j) for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}} Return only
#define SC scanf
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
using namespace std;
const int maxn = 2e5+9;
const int N = 10000 ;
int s1[maxn] , s2[maxn];
pii a[maxn];
int b[maxn] , len;
int lowerbit(int x){
return x&(-x);
}
void add(int x , int val){
while(x <= len){
s1[x]++;
s2[x] += val;
x += lowerbit(x);
}
}
int getsum(int x , int s[]){
int ans = 0 ;
while(x){
ans += s[x];
x -= lowerbit(x);
}
return ans ;
}
void solve(){
int n ;
cin >> n ;
rep(i , 1 , n){
cin >> a[i].fi ;
}
rep(i , 1 , n){
cin >> a[i].se;
b[i] = a[i].se;
}
int ans = 0 ;
sort(b + 1 , b + 1 + n);
sort(a + 1 , a + 1 + n);
len = unique(b + 1 , b + 1 + n) - b - 1 ;
rep(i , 1 , n){
int pos = lower_bound(b + 1 , b + 1 + len , a[i].se) - b ;
int sum1 = getsum(pos , s1) , sum2 = getsum(pos , s2);
ans += sum1 * a[i].fi - sum2 ;
add(pos , a[i].fi);
}
cout << ans << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
//int t ;
//cin >> t ;
//while(t--){
solve();
//}
}