結果

問題 No.1097 Remainder Operation
ユーザー iiljjiiljj
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
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
other AC * 21
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

/* #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 log2l
constexpr ll INF = 1'010'000'000'000'000'017LL;
constexpr ll MOD = 1'000'000'007LL; // 1e9 + 7
constexpr ld EPS = 1e-12;
constexpr ld PI = 3.14159265358979323846;
template <typename T> istream &operator>>(istream &is, vc<T> &vec) { // vector
for (T &x : vec) is >> x;
return is;
}
template <typename T> ostream &operator<<(ostream &os, vc<T> &vec) { // vector (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 (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
is >> pair_var.first >> pair_var.second;
return is;
}
template <typename T, typename U> ostream &operator<<(ostream &os, pair<T, U> &pair_var) { // pair
os << "(" << pair_var.first << ", " << pair_var.second << ")";
return os;
}
// map, um, set, us
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 cerr
void 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 ()
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 ()
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;
}
//
#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(...)
#endif
struct 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 */
// k log(k)
template <typename T> class DoublingStepVal {
using Data = pair<ll, T>;
using FNext = function<Data(ll)>; // (i) -> (, )
using FMerge = function<T(T, T)>;
ll n;
ll log_n;
T ti;
FNext fNext;
FMerge fMerge;
vc<vc<Data>> next; //
public:
//
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();
}
// n
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 -1
REP(e, 0, log_n) REP(i, 0, n) {
if (next[e][i].first == -1)
next[e + 1][i] = {-1, ti}; // 2^e -> 2^(e+1)
else {
auto [nxt, ov] = next[e][i];
next[e + 1][i] = {
next[e][nxt].first,
fMerge(next[e][nxt].second, ov)}; // 2^e 2^e = 2^(e+1)
}
}
}
// cur k , 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 ->
if ((k >> e) & 1)
// k e cur 2^e
cur = {next[e][cur.first].first, fMerge(next[e][cur.first].second, cur.second)};
}
return cur;
}
};
// Problem
void 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 point
int main() {
solve();
return 0;
}
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