답안 #314319

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
314319 2020-10-19T16:54:12 Z Benq Zoltan (COCI16_zoltan) C++14
140 / 140
226 ms 14408 KB
#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
using ld = long double;
using db = double; 
using str = string; // yay python!

using pi = pair<int,int>;
using pl = pair<ll,ll>;
using pd = pair<db,db>;

using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>; 
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>; 
using vpd = vector<pd>;

#define tcT template<class T
#define tcTU tcT, class U
// ^ lol this makes everything look weird but I'll try it
tcT> using V = vector<T>; 
tcT, size_t SZ> using AR = array<T,SZ>; 
tcT> using PR = pair<T,T>;

// pairs
#define mp make_pair
#define f first
#define s second

// vectors
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define sor(x) sort(all(x)) 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back()
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 

// loops
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5;
const ll INF = 1e18; // not too close to LLONG_MAX
const ld PI = acos((ld)-1);
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; // for every grid problem!!
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 
template<class T> using pqg = priority_queue<T,vector<T>,greater<T>>;

// helper funcs
constexpr int pct(int x) { return __builtin_popcount(x); } // # of bits set
constexpr int bits(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down

tcT> bool ckmin(T& a, const T& b) {
	return b < a ? a = b, 1 : 0; } // set a = min(a,b)
tcT> bool ckmax(T& a, const T& b) {
	return a < b ? a = b, 1 : 0; }

tcTU> T fstTrue(T lo, T hi, U f) {
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo)/2;
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}
tcTU> T lstTrue(T lo, T hi, U f) {
	lo --; assert(lo <= hi); // assuming f is decreasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo+1)/2;
		f(mid) ? lo = mid : hi = mid-1;
	} 
	return lo;
}
tcT> void remDup(vector<T>& v) { // sort and remove duplicates
	sort(all(v)); v.erase(unique(all(v)),end(v)); }
tcTU> void erase(T& t, const U& u) { // don't erase
	auto it = t.find(u); assert(it != end(t));
	t.erase(u); } // element that doesn't exist from (multi)set

// INPUT
#define tcTUU tcT, class ...U
tcT> void re(complex<T>& c);
tcTU> void re(pair<T,U>& p);
tcT> void re(vector<T>& v);
tcT, size_t SZ> void re(AR<T,SZ>& a);

tcT> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
tcTUU> void re(T& t, U&... u) { re(t); re(u...); }

tcT> void re(complex<T>& c) { T a,b; re(a,b); c = {a,b}; }
tcTU> void re(pair<T,U>& p) { re(p.f,p.s); }
tcT> void re(vector<T>& x) { trav(a,x) re(a); }
tcT, size_t SZ> void re(AR<T,SZ>& x) { trav(a,x) re(a); }

// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) { 
	#ifdef LOCAL
		return b ? "true" : "false"; 
	#else 
		return ts((int)b);
	#endif
}
tcT> str ts(complex<T> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(vector<bool> v) {
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
tcTU> str ts(pair<T,U> p);
tcT> str ts(T v) { // containers with begin(), end()
	#ifdef LOCAL
		bool fst = 1; str res = "{";
		for (const auto& x: v) {
			if (!fst) res += ", ";
			fst = 0; res += ts(x);
		}
		res += "}"; return res;
	#else
		bool fst = 1; str res = "";
		for (const auto& x: v) {
			if (!fst) res += " ";
			fst = 0; res += ts(x);
		}
		return res;

	#endif
}
tcTU> str ts(pair<T,U> p) {
	#ifdef LOCAL
		return "("+ts(p.f)+", "+ts(p.s)+")"; 
	#else
		return ts(p.f)+" "+ts(p.s);
	#endif
}

// OUTPUT
tcT> void pr(T x) { cout << ts(x); }
tcTUU> void pr(const T& t, const U&... u) { 
	pr(t); pr(u...); }
void ps() { pr("\n"); } // print w/ spaces
tcTUU> void ps(const T& t, const U&... u) { 
	pr(t); if (sizeof...(u)) pr(" "); ps(u...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
tcTUU> void DBG(const T& t, const U&... u) {
	cerr << ts(t); if (sizeof...(u)) cerr << ", ";
	DBG(u...); }
#ifdef LOCAL // compile with -DLOCAL, chk -> fake assert
	#define dbg(...) cerr << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
	#define chk(...) if (!(__VA_ARGS__)) cerr << "Line(" << __LINE__ << ") -> function(" \
		 << __FUNCTION__  << ") -> CHK FAILED: (" << #__VA_ARGS__ << ")" << "\n", exit(0);
#else
	#define dbg(...) 0
	#define chk(...) 0
#endif

// FILE I/O
void setIn(str s) { freopen(s.c_str(),"r",stdin); }
void setOut(str s) { freopen(s.c_str(),"w",stdout); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
void setIO(str s = "") {
	unsyncIO();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

/**
 * Description: modular arithmetic operations 
 * Source: 
	* KACTL
	* https://codeforces.com/blog/entry/63903
	* https://codeforces.com/contest/1261/submission/65632855 (tourist)
	* https://codeforces.com/contest/1264/submission/66344993 (ksun)
	* also see https://github.com/ecnerwala/cp-book/blob/master/src/modnum.hpp (ecnerwal)
 * Verification: 
	* https://open.kattis.com/problems/modulararithmetic
 */

template<int MOD, int RT> struct mint {
	static const int mod = MOD;
	static constexpr mint rt() { return RT; } // primitive root for FFT
	int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
	mint() { v = 0; }
	mint(ll _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
		if (v < 0) v += MOD; }
	friend bool operator==(const mint& a, const mint& b) { 
		return a.v == b.v; }
	friend bool operator!=(const mint& a, const mint& b) { 
		return !(a == b); }
	friend bool operator<(const mint& a, const mint& b) { 
		return a.v < b.v; }
	friend void re(mint& a) { ll x; re(x); a = mint(x); }
	friend str ts(mint a) { return ts(a.v); }
   
	mint& operator+=(const mint& m) { 
		if ((v += m.v) >= MOD) v -= MOD; 
		return *this; }
	mint& operator-=(const mint& m) { 
		if ((v -= m.v) < 0) v += MOD; 
		return *this; }
	mint& operator*=(const mint& m) { 
		v = (ll)v*m.v%MOD; return *this; }
	mint& operator/=(const mint& m) { return (*this) *= inv(m); }
	friend mint pow(mint a, ll p) {
		mint ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans; }
	friend mint inv(const mint& a) { assert(a.v != 0); 
		return pow(a,MOD-2); }
		
	mint operator-() const { return mint(-v); }
	mint& operator++() { return *this += 1; }
	mint& operator--() { return *this -= 1; }
	friend mint operator+(mint a, const mint& b) { return a += b; }
	friend mint operator-(mint a, const mint& b) { return a -= b; }
	friend mint operator*(mint a, const mint& b) { return a *= b; }
	friend mint operator/(mint a, const mint& b) { return a /= b; }
};

typedef mint<MOD,5> mi; // 5 is primitive root for both common mods
typedef vector<mi> vmi;
typedef pair<mi,mi> pmi;
typedef vector<pmi> vpmi;

vector<vmi> scmb; // small combinations
void genComb(int SZ) {
	scmb.assign(SZ,vmi(SZ)); scmb[0][0] = 1;
	FOR(i,1,SZ) F0R(j,i+1) 
		scmb[i][j] = scmb[i-1][j]+(j?scmb[i-1][j-1]:0);
}

/**
 * Description: pre-compute factorial mod inverses,
 	* assumes $MOD$ is prime and $SZ < MOD$.
 * Time: O(SZ)
 * Source: KACTL
 * Verification: https://dmoj.ca/problem/tle17c4p5
 */

vi invs, fac, ifac; // make sure to convert to LL before doing any multiplications ...
void genFac(int SZ) {
	invs.rsz(SZ), fac.rsz(SZ), ifac.rsz(SZ); 
	invs[1] = fac[0] = ifac[0] = 1; 
	FOR(i,2,SZ) invs[i] = int(MOD-(ll)MOD/i*invs[MOD%i]%MOD);
	FOR(i,1,SZ) {
		fac[i] = int((ll)fac[i-1]*i%MOD);
		ifac[i] = int((ll)ifac[i-1]*invs[i]%MOD);
	}
}
/**
ll comb(int a, int b) {
	if (a < b || b < 0) return 0;
	return (ll)fac[a]*ifac[b]%MOD*ifac[a-b]%MOD;
}
*/

/**
 * Description: 1D point update, range query where \texttt{comb} is
 	* any associative operation. If $N=2^p$ then \texttt{seg[1]==query(0,N-1)}.
 * Time: O(\log N)
 * Source: 
	* http://codeforces.com/blog/entry/18051
	* KACTL
 * Verification: SPOJ Fenwick
 */

template<class T> struct Seg { // comb(ID,b) = b
	const T ID = {0,0}; 
	T comb(T a, T b) { 
		if (a.f > b.f) return a;
		if (a.f < b.f) return b;
		return {a.f,a.s+b.s};
	} 
	int n; vector<T> seg;
	void init(int _n) { n = _n; seg.assign(2*n,ID); }
	void pull(int p) { seg[p] = comb(seg[2*p],seg[2*p+1]); }
	void upd(int p, T val) { // set val at position p
		p += n;
		seg[p] = comb(seg[p],val); for (p /= 2; p; p /= 2) pull(p); }
	T query(int l, int r) {	// sum on interval [l, r]
		T res = {0,1};
		for (l += n, r += n+1; l < r; l /= 2, r /= 2) {
			if (l&1) res = comb(res,seg[l++]);
			if (r&1) res = comb(seg[--r],res);
		}
		return res;
	}
};

int N;


V<pair<int,mi>> solve(vi A) {
	Seg<pair<int,mi>> S; S.init(N);
	V<pair<int,mi>> res;
	trav(t,A) {
		res.pb(S.query(t+1,N-1)); res.bk.f ++;
		S.upd(t,res.bk);
	}
	return res;
}

int main() {
	setIO(); re(N); 
	genFac(N+1);
	vi A(N); re(A);
	reverse(all(A));
	{
		vi dis = A; remDup(dis);
		trav(t,A) t = lb(all(dis),t)-begin(dis);
	}
	V<pair<int,mi>> x = solve(A);
	trav(t,A) t = N-1-t;
	V<pair<int,mi>> y = solve(A);
	int ans = 0;
	F0R(i,N) ckmax(ans,x[i].f+y[i].f-1);
	mi tot = 0;
	F0R(i,N) if (x[i].f+y[i].f-1 == ans) {
		// if includes 1 -> N-1-(ans-1) remaining
		// otherwise -> 1, N-1-(ans-1)
		// if (i == N-1) tot += pow(mi(2),N-ans)*x[i].s*y[i].s;
		// else 
		tot += pow(mi(2),N-ans)*x[i].s*y[i].s;
	}
	ps(ans,tot);
	// ps(ans,tot*fac[N]*ifac[ans]*ifac[N-ans]);
	// you should actually read the stuff at the bottom
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
*/

Compilation message

zoltan.cpp: In function 'void setIn(str)':
zoltan.cpp:184:28: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
  184 | void setIn(str s) { freopen(s.c_str(),"r",stdin); }
      |                     ~~~~~~~^~~~~~~~~~~~~~~~~~~~~
zoltan.cpp: In function 'void setOut(str)':
zoltan.cpp:185:29: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
  185 | void setOut(str s) { freopen(s.c_str(),"w",stdout); }
      |                      ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~
# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 384 KB Output is correct
2 Correct 1 ms 384 KB Output is correct
3 Correct 1 ms 384 KB Output is correct
4 Correct 1 ms 384 KB Output is correct
5 Correct 1 ms 384 KB Output is correct
6 Correct 0 ms 384 KB Output is correct
7 Correct 1 ms 384 KB Output is correct
8 Correct 1 ms 384 KB Output is correct
9 Correct 1 ms 384 KB Output is correct
10 Correct 1 ms 384 KB Output is correct
11 Correct 131 ms 11740 KB Output is correct
12 Correct 113 ms 10492 KB Output is correct
13 Correct 108 ms 8980 KB Output is correct
14 Correct 148 ms 10732 KB Output is correct
15 Correct 190 ms 12756 KB Output is correct
16 Correct 226 ms 14408 KB Output is correct
17 Correct 162 ms 13864 KB Output is correct
18 Correct 162 ms 13860 KB Output is correct
19 Correct 161 ms 13880 KB Output is correct
20 Correct 168 ms 13900 KB Output is correct