結果

問題 No.808 Kaiten Sushi?
ユーザー ミドリムシ
提出日時 2019-03-22 22:27:07
言語 C++11（廃止可能性あり）
(gcc 13.3.0)
結果
WA  
実行時間 -
コード長 4,439 bytes
コンパイル時間 1,980 ms
コンパイル使用メモリ 122,504 KB
実行使用メモリ 30,080 KB
最終ジャッジ日時 2024-09-19 05:57:05
合計ジャッジ時間 12,089 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
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 * 2 WA * 1
other AC * 19 WA * 37
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 # 

#include <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
#include <queue>
#include <cmath>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
using namespace std;
using lint = long long;
const long mod = 1e9 + 7;
#define all(x) (x).begin(), (x).end()
#define bitcount(n) __builtin_popcountl(long(n))
#define fcout cout << fixed << setprecision(15)
#define highest(x) (63 - __builtin_clzl(x))
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T> inline void POSS(T condition){ if(condition) cout << "POSSIBLE" << endl; else cout << "IMPOSSIBLE" << endl; }
template<class T> inline void Poss(T condition){ if(condition) cout << "Possible" << endl; else cout << "Impossible" << endl; }
template<class T> inline void First(T condition){ if(condition) cout << "First" << endl; else cout << "Second" << endl; }
template<class T = string, class U = char>int character_count(T text, U character){ int ans = 0; for(U i: text){ ans += (i == character); } return ans; }
long power(long base, long exponent, long module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ long root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
struct position{ int y, x; }; position mv[4] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // double euclidean(position first, position second){ return sqrt((second.x - first.x) * (second.x - first.x) + (second.y - first.y) * (second.y - first.y)); }
template<class T, class U> string to_string(pair<T, U> x){ return to_string(x.first) + "," + to_string(x.second); } string to_string(string x){ return x; }
template<class itr> void array_output(itr start, itr goal){ string ans; for(auto i = start; i != goal; i++) ans += to_string(*i) + " "; if(!ans.empty()) ans.pop_back(); cout << ans << endl; }
template<class itr> void cins(itr start, itr goal){ for(auto i = start; i != goal; i++){ cin >> (*i); } }
template<class T> T gcd(T a, T b){ if(a && b){ return gcd(min(a, b), max(a, b) % min(a, b)); }else{ return a; }} template<class T> T lcm(T a, T b){ return a / gcd(a, b) * b; }
struct combination{ vector<long> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2, mod); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } long C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };

using namespace __gnu_pbds;
using Node = long;
using fast_set = tree<Node, null_type, less<Node>, rb_tree_tag, tree_order_statistics_node_update>;

int main(){
    int N;
    long L;
    cin >> N >> L;
    long x[N], y[N];
    fast_set x_set, y_set;
    for(int i = 0; i < N; i++){
    	cin >> x[i];
        x_set.insert(x[i]);
        x_set.insert(x[i] + L);
    }
    for(int i = 0; i < N; i++){
        cin >> y[i];
        y_set.insert(y[i]);
        y_set.insert(y[i] + L);
    }
    long ans = 0, now = 0;
    for(int i = 0; i < N; i++){
    	auto next_itr = x_set.find_by_order(x_set.order_of_key(now));
    	long next = (*next_itr);
    	ans += next - now;
    	next %= L;
    	x_set.erase(next);
    	x_set.erase(next + L);
    	now = next;

    	next_itr = y_set.find_by_order(y_set.order_of_key(now));
    	next = (*next_itr);
    	ans += next - now;
    	next %= L;
    	y_set.erase(next);
    	y_set.erase(next + L);
    	now = next;
    }
    if(N == 1){
    	cout << ans << endl;
    	return 0;
    }
    int last = int(lower_bound(y, y + N, now) - y);
    int last_keep = last;
    int x_last = int(lower_bound(x, x + N, now) - x);
    long last_now = y[last];
    long x_last_back;
    if(x_last == 0){
    	x_last_back = x[N - 1] - L;
    }else{
    	x_last_back = x[x_last - 1];
    }
    while(last_now > x_last_back){
	    if(last == 0){
	    	last_now += y[N - 1] - y[0] - L;
	    	last = N - 1;
	    }else{
	    	last_now += y[last - 1] - y[last];
	    	last--;
	    }
    }
    if(last == N - 1){
    	last_now += y[0] - y[N - 1] + L;
    }else{
    	last_now += y[last + 1] - y[last];
    }
    cout << ans - y[last_keep] + last_now << endl;
}
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