結果
問題 | No.1097 Remainder Operation |
ユーザー |
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提出日時 | 2020-08-16 15:28:13 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 288 ms / 2,000 ms |
コード長 | 8,711 bytes |
コンパイル時間 | 2,609 ms |
コンパイル使用メモリ | 205,620 KB |
最終ジャッジ日時 | 2025-01-13 01:57:11 |
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 21 |
ソースコード
/* #region Head */#include <bits/stdc++.h>using namespace std;using ll = long long;using ull = unsigned long long;using ld = long double;using pll = pair<ll, ll>;template <class T> using vc = vector<T>;template <class T> using vvc = vc<vc<T>>;using vll = vc<ll>;using vvll = vvc<ll>;using vld = vc<ld>;using vvld = vvc<ld>;using vs = vc<string>;using vvs = vvc<string>;template <class T, class U> using um = unordered_map<T, U>;template <class T> using pq = priority_queue<T>;template <class T> using pqa = priority_queue<T, vc<T>, greater<T>>;template <class T> using us = unordered_set<T>;#define REP(i, m, n) for (ll i = (m), i##_len = (ll)(n); i < i##_len; ++(i))#define REPM(i, m, n) for (ll i = (m), i##_max = (ll)(n); i <= i##_max; ++(i))#define REPR(i, m, n) for (ll i = (m), i##_min = (ll)(n); i >= i##_min; --(i))#define REPD(i, m, n, d) for (ll i = (m), i##_len = (ll)(n); i < i##_len; i += (d))#define REPMD(i, m, n, d) for (ll i = (m), i##_max = (ll)(n); i <= i##_max; i += (d))#define REPI(itr, ds) for (auto itr = ds.begin(); itr != ds.end(); itr++)#define ALL(x) begin(x), end(x)#define SIZE(x) ((ll)(x).size())#define PERM(c) \sort(ALL(c)); \for (bool c##p = 1; c##p; c##p = next_permutation(ALL(c)))#define UNIQ(v) v.erase(unique(ALL(v)), v.end());#define endl '\n'#define sqrt sqrtl#define floor floorl#define log2 log2lconstexpr ll INF = 1'010'000'000'000'000'017LL;constexpr ll MOD = 1'000'000'007LL; // 1e9 + 7constexpr ld EPS = 1e-12;constexpr ld PI = 3.14159265358979323846;template <typename T> istream &operator>>(istream &is, vc<T> &vec) { // vector 入力for (T &x : vec) is >> x;return is;}template <typename T> ostream &operator<<(ostream &os, vc<T> &vec) { // vector 出力 (for dump)os << "{";REP(i, 0, SIZE(vec)) os << vec[i] << (i == i_len - 1 ? "" : ", ");os << "}";return os;}template <typename T> ostream &operator>>(ostream &os, vc<T> &vec) { // vector 出力 (inline)REP(i, 0, SIZE(vec)) os << vec[i] << (i == i_len - 1 ? "\n" : " ");return os;}template <typename T, typename U> istream &operator>>(istream &is, pair<T, U> &pair_var) { // pair 入力is >> pair_var.first >> pair_var.second;return is;}template <typename T, typename U> ostream &operator<<(ostream &os, pair<T, U> &pair_var) { // pair 出力os << "(" << pair_var.first << ", " << pair_var.second << ")";return os;}// map, um, set, us 出力template <class T> ostream &out_iter(ostream &os, T &map_var) {os << "{";REPI(itr, map_var) {os << *itr;auto itrcp = itr;if (++itrcp != map_var.end()) os << ", ";}return os << "}";}template <typename T, typename U> ostream &operator<<(ostream &os, map<T, U> &map_var) { return out_iter(os, map_var); }template <typename T, typename U> ostream &operator<<(ostream &os, um<T, U> &map_var) {os << "{";REPI(itr, map_var) {auto [key, value] = *itr;os << "(" << key << ", " << value << ")";auto itrcp = itr;if (++itrcp != map_var.end()) os << ", ";}os << "}";return os;}template <typename T> ostream &operator<<(ostream &os, set<T> &set_var) { return out_iter(os, set_var); }template <typename T> ostream &operator<<(ostream &os, us<T> &set_var) { return out_iter(os, set_var); }template <typename T> ostream &operator<<(ostream &os, pq<T> &pq_var) {pq<T> pq_cp(pq_var);os << "{";if (!pq_cp.empty()) {os << pq_cp.top(), pq_cp.pop();while (!pq_cp.empty()) os << ", " << pq_cp.top(), pq_cp.pop();}return os << "}";}// dump#define DUMPOUT cerrvoid dump_func() { DUMPOUT << endl; }template <class Head, class... Tail> void dump_func(Head &&head, Tail &&... tail) {DUMPOUT << head;if (sizeof...(Tail) > 0) DUMPOUT << ", ";dump_func(move(tail)...);}// chmax (更新「される」かもしれない値が前)template <typename T, typename U, typename Comp = less<>> bool chmax(T &xmax, const U &x, Comp comp = {}) {if (comp(xmax, x)) {xmax = x;return true;}return false;}// chmin (更新「される」かもしれない値が前)template <typename T, typename U, typename Comp = less<>> bool chmin(T &xmin, const U &x, Comp comp = {}) {if (comp(x, xmin)) {xmin = x;return true;}return false;}// ローカル用#define DEBUG_#ifdef DEBUG_#define DEB#define dump(...) \DUMPOUT << " " << string(#__VA_ARGS__) << ": " \<< "[" << to_string(__LINE__) << ":" << __FUNCTION__ << "]" << endl \<< " ", \dump_func(__VA_ARGS__)#else#define DEB if (false)#define dump(...)#endifstruct AtCoderInitialize {static constexpr int IOS_PREC = 15;static constexpr bool AUTOFLUSH = false;AtCoderInitialize() {ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);cout << fixed << setprecision(IOS_PREC);if (AUTOFLUSH) cout << unitbuf;}} ATCODER_INITIALIZE;void Yn(bool p) { cout << (p ? "Yes" : "No") << endl; }void YN(bool p) { cout << (p ? "YES" : "NO") << endl; }/* #endregion */// ダブリングで,k ステップ先の要素を log(k) で得るためのクラスtemplate <typename T> class DoublingStepVal {using Data = pair<ll, T>;using FNext = function<Data(ll)>; // (i) -> (次の要素の番号, 追加データ)using FMerge = function<T(T, T)>;ll n;ll log_n;T ti;FNext fNext;FMerge fMerge;vc<vc<Data>> next; // ある回数次のインデックス,超越した回数public:// コンストラクタ(要素数を指定)DoublingStepVal(const ll n, const T ti, FNext fNext, FMerge fMerge): n(n), log_n((ll)floorl(log2l(n))), ti(ti), fNext(fNext), fMerge(fMerge),next(log_n + 1, vc<pll>(n, {0, ti})) {init();}// コンストラクタ(要素数,最大距離(n より大きい時は必ずそれを指定))DoublingStepVal(const ll n, const T ti, FNext fNext, FMerge fMerge, const ll max_step): n(n), log_n((ll)floorl(log2l(max_step))), ti(ti), fNext(fNext), fMerge(fMerge),next(log_n + 1, vc<pll>(n, {0, ti})) {init();}// O(n * log_n)void init() {REP(i, 0, n) next[0][i] = fNext(i); // i の次の要素のインデックスを返す関数,次がない場合は -1REP(e, 0, log_n) REP(i, 0, n) {if (next[e][i].first == -1)next[e + 1][i] = {-1, ti}; // 2^e 個次の要素がない -> 2^(e+1) 個次の要素もないelse {auto [nxt, ov] = next[e][i];next[e + 1][i] = {next[e][nxt].first,fMerge(next[e][nxt].second, ov)}; // 「2^e 個次の要素」の 2^e 個次の要素 = 2^(e+1) 個次の要素}}}// cur の k 個先の要素を返す, O(log k)pll step(ll _cur, const ll k) {pll cur = {_cur, ti};REPR(e, log_n, 0) {if (cur.first == -1)break; // cur が既に存在しない要素を指している -> それ以降で存在する要素を指すことはないif ((k >> e) & 1)// k を二進展開した際、e 番目のビットが立っていたら、cur の位置を 2^e だけ次にずらすcur = {next[e][cur.first].first, fMerge(next[e][cur.first].second, cur.second)};}return cur;}};// Problemvoid solve() {ll n;cin >> n;vll a(n);cin >> a;ll q;cin >> q;vll k(q);cin >> k;using T = ll;auto fNext = [&a, &n](ll i) -> pair<ll, T> {ll x = i + a[i];T ov = x / n;x -= ov * n;return {x, ov};};auto fMerge = [](T data1, T data2) -> T {return data1 + data2; //};DoublingStepVal<T> ds(n, 0, fNext, fMerge, *max_element(ALL(k)));REP(i, 0, q) {auto [nxt, ov] = ds.step(0, k[i]);cout << (nxt + ov * n) << endl;}}// entry pointint main() {solve();return 0;}