結果

問題 No.992 最長増加部分列の数え上げ
ユーザー akua
提出日時 2023-04-20 22:29:08
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 832 ms / 2,000 ms
コード長 6,211 bytes
コンパイル時間 22,258 ms
コンパイル使用メモリ 272,920 KB
最終ジャッジ日時 2025-02-12 10:36:06
ジャッジサーバーID
(参考情報)
judge5 / judge5
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 3
other AC * 42
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

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <atcoder/all>
#include <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include <math.h>
#include <iomanip>
#include <functional>
using namespace std;
using namespace atcoder;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repi(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
typedef long long ll;
typedef unsigned long long ull;
const ll inf=1e18;
using graph = vector<vector<int> > ;
using P= pair<ll,ll>;
using vi=vector<int>;
using vvi=vector<vi>;
using vll=vector<ll>;
using vvll=vector<vll>;
using vp=vector<P>;
using vvp=vector<vp>;
using vd=vector<double>;
using vvd =vector<vd>;
//string T="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
//string S="abcdefghijklmnopqrstuvwxyz";
//g++ main.cpp -std=c++17 -I .
//cout <<setprecision(20);
//cout << fixed << setprecision(10);
//cin.tie(0); ios::sync_with_stdio(false);
const double PI = acos(-1);
int vx[]={0,1,0,-1,-1,1,1,-1},vy[]={1,0,-1,0,1,1,-1,-1};
void putsYes(bool f){cout << (f?"Yes":"No") << endl;}
void putsYES(bool f){cout << (f?"YES":"NO") << endl;}
void putsFirst(bool f){cout << (f?"First":"Second") << endl;}
void debug(int test){cout << "TEST" << " " << test << endl;}
ll pow_pow(ll x,ll n,ll mod){
if(n==0) return 1;
x%=mod;
ll res=pow_pow(x*x%mod,n/2,mod);
if(n&1)res=res*x%mod;
return res;
}
ll gcd(ll x,ll y){
if(y==0)return x;
return gcd(y,x%y);
}
ll lcm(ll x,ll y){
return ll(x/gcd(x,y))*y;
}
template<class T> bool chmin(T& a, T b) {
if (a > b) {
a = b;
return true;
}
else return false;
}
template<class T> bool chmax(T& a, T b) {
if (a < b) {
a = b;
return true;
}
else return false;
}
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
//ll mod =998244353;
ll mod =1e9+7;
struct mint {
ll x; // typedef long long ll;
mint(ll x=0):x((x%mod+mod)%mod){}
mint operator-() const { return mint(-x);}
mint& operator+=(const mint a) {
if ((x += a.x) >= mod) x -= mod;
return *this;
}
mint& operator-=(const mint a) {
if ((x += mod-a.x) >= mod) x -= mod;
return *this;
}
mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
mint operator+(const mint a) const { return mint(*this) += a;}
mint operator-(const mint a) const { return mint(*this) -= a;}
mint operator*(const mint a) const { return mint(*this) *= a;}
mint pow(ll t) const {
if (!t) return 1;
mint a = pow(t>>1);
a *= a;
if (t&1) a *= *this;
return a;
}
// for prime mod
mint inv() const { return pow(mod-2);}
mint& operator/=(const mint a) { return *this *= a.inv();}
mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
vector<mint> fact, ifact;
combination(int n):fact(n+1),ifact(n+1) {
//assert(n < mod);
fact[0] = 1;
for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
ifact[n] = fact[n].inv();
for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
}
mint operator()(int n, int k) {
if (k < 0 || k > n) return 0;
if (n<0) return 0;
return fact[n]*ifact[k]*ifact[n-k];
} mint p(int n, int k) { return fact[n]*ifact[n-k]; } } c(2000000);
using vm=vector<mint> ;
using vvm=vector<vm> ;
struct edge{
int to; ll cost;
edge(int to,ll cost) : to(to),cost(cost){}
};
using ve=vector<edge>;
using vve=vector<ve>;
struct Compress{
vll a;
map<ll,ll> d,d2;
int cnt;
Compress() :cnt(0) {}
void add(ll x){a.push_back(x);}
void init(){
set<ll>s(a.begin(),a.end());
for(auto y:s)d[y]=cnt++;
for(auto&y:a)y=d[y];
for(auto u:d)d2[u.second]=u.first;
}
ll to(ll x){return d[x];} //
ll from(ll x){return d2[x];}//
int size(){return cnt;}
};
vll anss;
using S=ll;
S op(S a,S b){return max(a,b);}
S e(){return -1;}
void solve(int test){
int n; cin >> n;
vll a(n+1,-inf);rep(i,n)cin >> a[i+1];
Compress comp;
rep(i,n+1)comp.add(a[i]);
comp.init();
rep(i,n+1)a[i]=comp.to(a[i]);
int m=comp.size();
vll dp(n+1,-inf);
dp[0]=0;
segtree<S,op,e> seg(dp);
for(int i=1; i<=n; i++){
dp[i]=seg.prod(0,a[i])+1;
seg.set(a[i],max(seg.get(a[i]),dp[i]));
}
ll M=seg.all_prod();
vll maxv(M+1);
vi is(n+1);
for(int i=n; i>=0; i--){
if(dp[i]==M){
chmax(maxv[dp[i]],a[i]+0ll);
is[i]=1;
}
else {
if(maxv[dp[i]+1]>a[i]){
is[i]=1;
chmax(maxv[dp[i]],a[i]);
}
}
}
vi ids;
for(int i=0; i<=n; i++)if(is[i])ids.push_back(i);
fenwick_tree<mint> fw(m);
vm dp2(n+1);
dp2[0]=1;
vvi g(M+1);
for(auto id: ids){
g[dp[id]].push_back(id);
}
rep(i,M-1+1){
vi vs;
for(auto u:g[i])vs.push_back(u);
for(auto u:g[i+1])vs.push_back(u);
sort(vs.begin(),vs.end());
for(auto id:vs){
if(dp[id]==i){
fw.add(a[id],dp2[id]);
}
else {
dp2[id]=fw.sum(0,a[id]);
}
}
for(auto id:vs){
if(dp[id]==i){
fw.add(a[id],-dp2[id]);
}
}
}
mint ans=0;
repi(i,1,n+1)if(dp[i]==M){
ans+=dp2[i];
}
cout << ans << endl;
}
//g++ main.cpp -std=c++17 -I .
int main(){cin.tie(0);ios::sync_with_stdio(false);
int t=1;//cin >> t;
rep(test,t)solve(test);
for(auto u:anss)cout << u << endl;
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0