結果

問題 No.1115 二つの数列 / Two Sequences
ユーザー yuji9511
提出日時 2020-07-17 21:45:21
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 89 ms / 2,000 ms
コード長 2,330 bytes
コンパイル時間 2,249 ms
コンパイル使用メモリ 200,432 KB
最終ジャッジ日時 2025-01-11 22:32:14
ジャッジサーバーID
(参考情報)
judge1 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 5
other AC * 35
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

/*** author: yuji9511 ***/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using lpair = pair<ll, ll>;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define rep(i,m,n) for(ll i=(m);i<(n);i++)
#define rrep(i,m,n) for(ll i=(m);i>=(n);i--)
#define printa(x,n) for(ll i=0;i<n;i++){cout<<(x[i])<<" \n"[i==n-1];};
void print() {}
template <class H,class... T>
void print(H&& h, T&&... t){cout<<h<<" \n"[sizeof...(t)==0];print(forward<T>(t)...);}
struct AddLazySegmentTree {
private:
ll n;
vector<ll> node, lazy;
public:
AddLazySegmentTree(vector<ll> v) {
ll sz = (ll)v.size();
n = 1; while(n < sz) n *= 2;
node.resize(2*n-1);
lazy.resize(2*n-1, 0);
for(ll i=0; i<sz; i++) node[i+n-1] = v[i];
for(ll i=n-2; i>=0; i--) node[i] = node[i*2+1] + node[i*2+2];
}
void eval(ll k, ll l, ll r) {
if(lazy[k] != 0) {
node[k] += lazy[k] * (r-l);
if(r - l > 1) {
lazy[2*k+1] += lazy[k];
lazy[2*k+2] += lazy[k];
}
lazy[k] = 0;
}
}
void add(ll a, ll b, ll x, ll k=0, ll l=0, ll r=-1) {
if(r < 0) r = n;
eval(k, l, r);
if(b <= l || r <= a) return;
if(a <= l && r <= b) {
lazy[k] += x;
eval(k, l, r);
}
else {
add(a, b, x, 2*k+1, l, (l+r)/2);
add(a, b, x, 2*k+2, (l+r)/2, r);
node[k] = node[2*k+1] + node[2*k+2];
}
}
ll getsum(ll a, ll b, ll k=0, ll l=0, ll r=-1) {
if(r < 0) r = n;
eval(k, l, r);
if(b <= l || r <= a) return 0;
if(a <= l && r <= b) return node[k];
ll vl = getsum(a, b, 2*k+1, l, (l+r)/2);
ll vr = getsum(a, b, 2*k+2, (l+r)/2, r);
return vl + vr;
}
};
void solve(){
ll N;
cin >> N;
ll a[100100], b[100010];
rep(i,0,N) cin >> a[i];
rep(i,0,N) cin >> b[i];
ll ps[100010] = {};
rep(i,0,N) ps[a[i]] = i;
ll ans = 0;
vector<ll> v(N+1,0);
AddLazySegmentTree sg(v);
rep(i,0,N){
ll n = b[i];
ll pos_a = ps[n] + sg.getsum(ps[n], ps[n]+1);
ans += abs(pos_a - i);
sg.add(0, ps[n], 1);
}
print(ans);
}
int main(){
cin.tie(0);
ios::sync_with_stdio(false);
solve();
}
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