結果
問題 | No.284 門松と魔法(2) |
ユーザー |
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提出日時 | 2020-08-21 15:33:40 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 5,120 bytes |
コンパイル時間 | 1,339 ms |
コンパイル使用メモリ | 126,260 KB |
実行使用メモリ | 280,320 KB |
最終ジャッジ日時 | 2024-10-15 01:26:57 |
合計ジャッジ時間 | 24,621 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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ファイルパターン | 結果 |
---|---|
other | AC * 27 WA * 2 TLE * 1 -- * 10 |
ソースコード
#include<iostream>#include<string>#include<cstdio>#include<vector>#include<cmath>#include<algorithm>#include<functional>#include<iomanip>#include<queue>#include<ciso646>#include<random>#include<map>#include<set>#include<bitset>#include<stack>#include<unordered_map>#include<utility>#include<cassert>#include<complex>#include<numeric>using namespace std;//#define int long longtypedef long long ll;typedef unsigned long long ul;typedef unsigned int ui;constexpr ll mod = 1000000007;const ll INF = mod * mod;typedef pair<int, int>P;#define stop char nyaa;cin>>nyaa;#define rep(i,n) for(int i=0;i<n;i++)#define per(i,n) for(int i=n-1;i>=0;i--)#define Rep(i,sta,n) for(int i=sta;i<n;i++)#define rep1(i,n) for(int i=1;i<=n;i++)#define per1(i,n) for(int i=n;i>=1;i--)#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)#define all(v) (v).begin(),(v).end()typedef pair<ll, ll> LP;typedef double ld;typedef pair<ld, ld> LDP;const ld eps = 1e-12;const ld pi = acos(-1.0);ll mod_pow(ll x, ll n, ll m) {ll res = 1;while (n) {if (n & 1)res = res * x % m;x = x * x % m; n >>= 1;}return res;}struct modint {ll n;modint() :n(0) { ; }modint(ll m) :n(m) {if (n >= mod)n %= mod;else if (n < 0)n = (n % mod + mod) % mod;}operator int() { return n; }};bool operator==(modint a, modint b) { return a.n == b.n; }modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; }modint operator+(modint a, modint b) { return a += b; }modint operator-(modint a, modint b) { return a -= b; }modint operator*(modint a, modint b) { return a *= b; }modint operator^(modint a, ll n) {if (n == 0)return modint(1);modint res = (a * a) ^ (n / 2);if (n % 2)res = res * a;return res;}ll inv(ll a, ll p) {return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);}modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }const int max_n = 1 << 2;modint fact[max_n], factinv[max_n];void init_f() {fact[0] = modint(1);for (int i = 0; i < max_n - 1; i++) {fact[i + 1] = fact[i] * modint(i + 1);}factinv[max_n - 1] = modint(1) / fact[max_n - 1];for (int i = max_n - 2; i >= 0; i--) {factinv[i] = factinv[i + 1] * modint(i + 1);}}modint comb(int a, int b) {if (a < 0 || b < 0 || a < b)return 0;return fact[a] * factinv[b] * factinv[a - b];}ll gcd(ll a, ll b) {if (a < b)swap(a, b);while (b) {ll r = a % b; a = b; b = r;}return a;}const int b = 4;struct ste {int val, pre, id;};bool operator==(ste const& a, ste const& b){return a.id == b.id && a.val == b.val && a.pre == b.pre;}struct SegT {private:int sz; vector<vector<ste>> node;const vector<ste> init_c = { ste{0,-1,-1} };public:SegT(int n) {sz = 1;while (sz < n)sz *= 2;node.resize(2 * sz - 1, init_c);}vector<ste> f(vector<ste> a, vector<ste>b) {vector<ste> res;int id1 = 0, id2 = 0;rep(i, a.size()+b.size()) {if (id1 == a.size()) {res.push_back(b[id2]); id2++;}else if (id2 == b.size()) {res.push_back(a[id1]); id1++;}else {if (a[id1].val > b[id2].val) {res.push_back(a[id1]); id1++;}else {res.push_back(b[id2]); id2++;}}}vector<ste> tres;for (ste s : res) {bool valid = true;for (ste p : tres) {if (p.pre == s.pre && p.id == s.id) {valid = false; break;}}if (valid)tres.push_back(s);}if (tres.size() > 4)tres.resize(4);return tres;}void update(int k, int v,int p,int id) {k += sz - 1;vector<ste> val = { {v,p,id} };vector<ste> cop = node[k];node[k] = f(node[k],val);if (cop == node[k])return;while (k > 0) {k = (k - 1) / 2;node[k] = f(node[k * 2 + 1], node[k * 2 + 2]);}}vector<ste> query(int a, int b, int k = 0, int l = 0, int r = -1) {if (r < 0)r = sz;if (r <= a || b <= l)return init_c;else if (a <= l && r <= b)return node[k];else {vector<ste> vl = query(a, b, k * 2 + 1, l, (l + r) / 2);vector<ste> vr = query(a, b, k * 2 + 2, (l + r) / 2, r);return f(vl, vr);}}};void solve() {int n; cin >> n;int sz = 1 << 20;SegT up(sz), dw(sz);vector<int> a(n);rep(i, n)cin >> a[i];rep(i, n) {vector<ste> x = dw.query(a[i] + 1, sz);for (ste s : x) {if (s.pre == a[i])continue;int v = s.val + 1;int p = s.id;int id = a[i];up.update(a[i], v, p, id);}x = up.query(0, a[i]);for (ste s : x) {if (s.pre == a[i])continue;int v = s.val + 1;int p = s.id;int id = a[i];dw.update(a[i], v, p, id);}}int ans = 0;vector<ste> s = up.query(0, sz);ans = max(ans, s[0].val);s = dw.query(0, sz);ans = max(ans, s[0].val);if (ans < 3)ans = 0;cout << ans << "\n";}signed main() {ios::sync_with_stdio(false);cin.tie(0);//cout << fixed << setprecision(10);//init_f();//expr();//int t; cin >> t; rep(i, t)//while (cin >> n >> m>>s1>>s2>>t, n)solve();return 0;}