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| #include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
struct PairHash {
inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }
};
#define M_PI 3.14159265358979323846
#define MAX_N 202020
#define INF 987654321
#define EPS 1e-9
#define ll long long
#define pll pair<ll, ll>
#define vpll vector<pll>
#define vall3 vector<array<ll,3>>
#define all3 array<ll,3>
#define all5 array<ll,5>
#define vall5 vector<all5>
#define vll vector<ll>
#define vi vector<int>
#define vs vector<string>
#define usll unordered_set<ll>
#define uspll unordered_set<pll, PairHash>
#define umll unordered_map<ll,ll>
#define vvs vector<vs>
#define vvll vector<vll>
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
using namespace std;
struct Seg {
ll mi, ma, c;
Seg *le, *ri;
Seg(vll& A, ll l, ll r) : mi(A[l]), ma(A[r]), c(0), le(nullptr), ri(nullptr) {
if(l == r) return;
ll m = l + (r - l) / 2;
le = new Seg(A, l, m);
ri = new Seg(A, m + 1, r);
}
ll query(ll l, ll r) {
if(l > ma or r < mi) return 0;
if(l <= mi and ma <= r) return c;
return (le ? le->query(l, r) : 0) + (ri ? ri->query(l, r) : 0);
}
ll update(ll n, ll v) {
if(mi <= n and n <= ma) {
c += v;
if(le) le->update(n, v);
if(ri) ri->update(n, v);
}
}
};
ll solve(vll& A, vll& B) {
ll n = A.size(), res = 0;
vll D(n);
for(ll i = 0; i < n; i++) {
D[i] = B[i] - A[i];
}
sort(all(D));
D.erase(unique(all(D)), end(D));
Seg* seg = new Seg(D, 0, D.size() - 1);
for(ll i = 0; i < n; i++) {
seg->update(B[i] - A[i], 1);
}
for(ll i = 0; i < n; i++) {
seg->update(B[i] - A[i], -1);
res += seg->query(LLONG_MIN, A[i] - B[i] - 1);
}
return res;
}
int main() {
ios_base::sync_with_stdio(0);
ios::sync_with_stdio(0);
cin.tie(0);
cout.precision(12);
ll tc = 1;
for (ll i = 1; i <= tc; i++) {
ll n;
cin>>n;
vll A(n), B(n);
for(ll j = 0; j < n; j++) cin>>A[j];
for(ll j = 0; j < n; j++) cin>>B[j];
cout<<solve(A,B)<<endl;
}
return 0;
}
|
