結果

問題 No.2204 Palindrome Splitting (No Rearrangement ver.)
ユーザー stoqstoq
提出日時 2023-02-03 22:08:28
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 139 ms / 2,000 ms
コード長 7,349 bytes
コンパイル時間 4,963 ms
コンパイル使用メモリ 274,808 KB
実行使用メモリ 7,040 KB
最終ジャッジ日時 2024-07-02 20:07:50
合計ジャッジ時間 9,410 ms
ジャッジサーバーID
(参考情報)
judge2 / judge1
このコードへのチャレンジ
(要ログイン)

テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 11 ms
6,784 KB
testcase_01 AC 10 ms
6,784 KB
testcase_02 AC 10 ms
6,784 KB
testcase_03 AC 83 ms
7,040 KB
testcase_04 AC 27 ms
6,784 KB
testcase_05 AC 19 ms
6,784 KB
testcase_06 AC 135 ms
6,784 KB
testcase_07 AC 122 ms
7,040 KB
testcase_08 AC 111 ms
6,912 KB
testcase_09 AC 117 ms
6,912 KB
testcase_10 AC 135 ms
6,912 KB
testcase_11 AC 111 ms
6,912 KB
testcase_12 AC 134 ms
6,912 KB
testcase_13 AC 136 ms
6,912 KB
testcase_14 AC 134 ms
6,912 KB
testcase_15 AC 124 ms
6,912 KB
testcase_16 AC 58 ms
7,040 KB
testcase_17 AC 73 ms
6,912 KB
testcase_18 AC 135 ms
6,912 KB
testcase_19 AC 135 ms
7,040 KB
testcase_20 AC 135 ms
7,040 KB
testcase_21 AC 135 ms
6,912 KB
testcase_22 AC 135 ms
6,912 KB
testcase_23 AC 134 ms
6,912 KB
testcase_24 AC 135 ms
6,784 KB
testcase_25 AC 135 ms
6,912 KB
testcase_26 AC 135 ms
6,784 KB
testcase_27 AC 92 ms
6,912 KB
testcase_28 AC 136 ms
6,912 KB
testcase_29 AC 139 ms
6,912 KB
testcase_30 AC 11 ms
6,784 KB
testcase_31 AC 10 ms
6,784 KB
testcase_32 AC 115 ms
6,912 KB
testcase_33 AC 135 ms
6,912 KB
testcase_34 AC 10 ms
6,784 KB
testcase_35 AC 114 ms
6,912 KB
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

#define MOD_TYPE 2
#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
// #include <atcoder/lazysegtree>
// #include <atcoder/modint>
// #include <atcoder/segtree>
using namespace atcoder;
#if 0
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
using Int = boost::multiprecision::cpp_int;
using lld = boost::multiprecision::cpp_dec_float_100;
#endif
#if 0
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
template <typename T>
using extset =
tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#endif
#if 1
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif
#pragma region Macros
using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
template <typename Q_type>
using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>;
#if MOD_TYPE == 1
constexpr ll MOD = ll(1e9 + 7);
#else
#if MOD_TYPE == 2
constexpr ll MOD = 998244353;
#else
constexpr ll MOD = 1000003;
#endif
#endif
using mint = static_modint<MOD>;
constexpr int INF = (int)1e9 + 10;
constexpr ll LINF = (ll)4e18;
const double PI = acos(-1.0);
constexpr ld EPS = 1e-10;
constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0};
#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define REPI(i, m, n) for (int i = m; i < (int)(n); ++i)
#define repi(i, n) REPI(i, 0, n)
#define YES(n) cout << ((n) ? "YES" : "NO") << "\n"
#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << "\n";
#define UNIQUE(v) v.erase(unique(all(v)), v.end())
struct io_init {
io_init() {
cin.tie(nullptr);
ios::sync_with_stdio(false);
cout << setprecision(30) << setiosflags(ios::fixed);
};
} io_init;
template <typename T>
inline bool chmin(T &a, T b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template <typename T>
inline bool chmax(T &a, T b) {
if (a < b) {
a = b;
return true;
}
return false;
}
inline ll floor(ll a, ll b) {
if (b < 0) a *= -1, b *= -1;
if (a >= 0) return a / b;
return -((-a + b - 1) / b);
}
inline ll ceil(ll a, ll b) { return floor(a + b - 1, b); }
template <typename A, size_t N, typename T>
inline void Fill(A (&array)[N], const T &val) {
fill((T *)array, (T *)(array + N), val);
}
template <typename T>
vector<T> compress(vector<T> &v) {
vector<T> val = v;
sort(all(val)), val.erase(unique(all(val)), val.end());
for (auto &&vi : v) vi = lower_bound(all(val), vi) - val.begin();
return val;
}
template <typename T, typename U>
constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept {
is >> p.first >> p.second;
return is;
}
template <typename T, typename U>
constexpr ostream &operator<<(ostream &os, pair<T, U> p) noexcept {
os << p.first << " " << p.second;
return os;
}
ostream &operator<<(ostream &os, mint m) {
os << m.val();
return os;
}
ostream &operator<<(ostream &os, modint m) {
os << m.val();
return os;
}
template <typename T>
constexpr istream &operator>>(istream &is, vector<T> &v) noexcept {
for (int i = 0; i < v.size(); i++) is >> v[i];
return is;
}
template <typename T>
constexpr ostream &operator<<(ostream &os, vector<T> &v) noexcept {
for (int i = 0; i < v.size(); i++)
os << v[i] << (i + 1 == v.size() ? "" : " ");
return os;
}
template <typename T>
constexpr void operator--(vector<T> &v, int) noexcept {
for (int i = 0; i < v.size(); i++) v[i]--;
}
random_device seed_gen;
mt19937_64 engine(seed_gen());
inline ll randInt(ll l, ll r) { return engine() % (r - l + 1) + l; }
struct BiCoef {
vector<mint> fact_, inv_, finv_;
BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
for (int i = 2; i < n; i++) {
fact_[i] = fact_[i - 1] * i;
inv_[i] = -inv_[MOD % i] * (MOD / i);
finv_[i] = finv_[i - 1] * inv_[i];
}
}
mint C(ll n, ll k) const noexcept {
if (n < k || n < 0 || k < 0) return 0;
return fact_[n] * finv_[k] * finv_[n - k];
}
mint P(ll n, ll k) const noexcept { return C(n, k) * fact_[k]; }
mint H(ll n, ll k) const noexcept { return C(n + k - 1, k); }
mint Ch1(ll n, ll k) const noexcept {
if (n < 0 || k < 0) return 0;
mint res = 0;
for (int i = 0; i < n; i++)
res += C(n, i) * mint(n - i).pow(k) * (i & 1 ? -1 : 1);
return res;
}
mint fact(ll n) const noexcept {
if (n < 0) return 0;
return fact_[n];
}
mint inv(ll n) const noexcept {
if (n < 0) return 0;
return inv_[n];
}
mint finv(ll n) const noexcept {
if (n < 0) return 0;
return finv_[n];
}
};
BiCoef bc(300010);
#pragma endregion
// -------------------------------
using M1 = modint1000000007;
using M2 = modint998244353;
using hash_t = pair<M1, M2>;
namespace atcoder {
bool operator<(const M1 &t1, const M1 &t2) { return t1.val() < t2.val(); }
bool operator<(const M2 &t1, const M2 &t2) { return t1.val() < t2.val(); }
} // namespace atcoder
template <typename T>
struct Hash {
int n;
T s;
static long long B;
bool calc_rev;
vector<M1> sum1, sum1_rev;
vector<M2> sum2, sum2_rev;
vector<M1> Bp1;
vector<M2> Bp2;
Hash() {}
Hash(T &s, bool calc_rev = true)
: s(s), n(s.size()), calc_rev(calc_rev), Bp1(n + 1), Bp2(n + 1) {
Bp1[0] = 1, Bp2[0] = 1;
for (int i = 0; i < n; i++) {
Bp1[i + 1] = Bp1[i] * B, Bp2[i + 1] = Bp2[i] * B;
}
sum1.assign(n + 1, 0), sum2.assign(n + 1, 0);
for (int i = n - 1; i >= 0; i--) {
sum1[i] = sum1[i + 1] * B + s[i];
sum2[i] = sum2[i + 1] * B + s[i];
}
if (not calc_rev) return;
sum1_rev.assign(n + 1, 0), sum2_rev.assign(n + 1, 0);
for (int i = 0; i < n; i++) {
sum1_rev[i + 1] = sum1_rev[i] * B + s[i];
sum2_rev[i + 1] = sum2_rev[i] * B + s[i];
}
}
hash_t query(int l, int r, bool rev = false) {
assert(l <= r);
if (rev) {
assert(calc_rev);
hash_t res = {sum1_rev[r] - sum1_rev[l] * Bp1[r - l],
sum2_rev[r] - sum2_rev[l] * Bp2[r - l]};
return res;
}
hash_t res = {sum1[l] - sum1[r] * Bp1[r - l],
sum2[l] - sum2[r] * Bp2[r - l]};
return res;
}
bool is_palindrome(int l, int r) {
assert(calc_rev);
return query(l, (l + r) / 2, false) == query((l + r + 1) / 2, r, true);
}
hash_t calc_hash(T &t) {
M1 t1 = 0;
M2 t2 = 0;
for (int i = int(t.size()) - 1; i >= 0; i--) {
t1 = t1 * B + t[i];
t2 = t2 * B + t[i];
}
return {t1, t2};
}
};
random_device B_seed;
mt19937 B_engine(B_seed());
template <typename T>
long long Hash<T>::B = (long long)(B_engine()) + 500010;
void solve() {
string s;
cin >> s;
int n = s.size();
vector<int> dp(n + 1, -INF);
dp[0] = INF;
Hash hash(s);
repi(i, n) {
REPI(j, i + 1, n + 1) {
if (not hash.is_palindrome(i, j)) continue;
chmax(dp[j], min(dp[i], j - i));
}
}
cout << dp[n] << "\n";
}
int main() { solve(); }
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