Submission #243482

#	Time	Username	Problem	Language	Result	Execution time	Memory
243482		ne4eHbKa	Zoltan (COCI16_zoltan)	C++17	70 / 140	1098 ms	2680 KiB

zoltan

//{ <defines>
#include <bits/stdc++.h>
using namespace std;

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,-O3")

#define fr(i, n) for(int i = 0; i < n; ++i)
#define fo(n) fr(i, n)

#define re return
#define ef else if
#define ifn(x) if(!(x))
#define _  << ' ' <<

#define ft first
#define sd second
#define ve vector
#define pb push_back
#define eb emplace_back

#define sz(x) int((x).size())
#define bnd(x) x.begin(), x.end()
#define clr(x, y) memset((x), (y), sizeof (x))

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef ve<int> vi;

inline ll time() {re chrono :: system_clock().now().time_since_epoch().count();}
mt19937 rnd(time());
mt19937_64 RND(time());

template<typename t> inline void umin(t &a, t b) {a = min(a, b);}
template<typename t> inline void umax(t &a, t b) {a = max(a, b);}

int md = 998244353;

inline int m_add(int&a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_sum(int a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_mul(int&a, int b) {re a = 1ll * a * b % md;}
inline int m_prod(int a, int b) {re 1ll * a * b % md;}

int m_bpow(ll A, ll b) {
    int a = A % md;
    ll ans = 1;
    for(ll p = 1ll << 63 - __builtin_clzll(b); p; p >>= 1) {
        (ans *= ans) %= md;
        if(p & b)
            (ans *= a) %= md;
    }
    re ans;
}

//const ld pi = arg(complex<ld>(-1, 0));
//const ld pi2 = pi + pi;
const int oo = 2e9;
const ll OO = 4e18;
//} </defines>
const int N = 2e5 + 5;

int n, m;
int a[N], p2[N];
pii le[N], gr[N];

void solve() {
    md = 1e9 + 7;
    for(int i = 0; i < N; ++i) p2[i] = !i ? 1 : m_sum(p2[i - 1], p2[i - 1]);
    cin >> n;
    fo(n) cin >> a[i];
//    map<int, int> sh;
//    fo(n) sh[a[i]];
//    m = 0;
//    for(auto &i : sh) i.sd = ++m;
//    fo(n) a[i] = sh[a[i]];
    int M = 1;
    for(int i = n - 1; ~i; --i) {
        le[i] = gr[i] = {1, 1};
        for(int j = i + 1; j < n; ++j) {
            if(a[j] < a[i] && le[j].ft + 1 >= le[i].ft) {
                if(le[j].ft + 1 == le[i].ft) {
                    m_add(le[i].sd, le[j].sd);
                } else {
                    le[i] = le[j];
                    le[i].ft++;
                }
            }
            if(a[j] > a[i] && gr[j].ft + 1 >= gr[i].ft) {
                if(gr[j].ft + 1 == gr[i].ft) {
                    m_add(gr[i].sd, gr[j].sd);
                } else {
                    gr[i] = gr[j];
                    gr[i].ft++;
                }
            }
        }
        umax(M, le[i].ft + gr[i].ft - 1);
    }
    int ans = 0;
    for(int i = 0; i < n; ++i) {
        if(le[i].ft + gr[i].ft < M + 1) continue;
        m_add(ans, m_prod(p2[n - M], m_prod(le[i].sd, gr[i].sd)));
    }
    cout << M _ ans << endl;
}

int main() {
    cout.precision(15); cout << fixed;
    cerr.precision(15); cerr << fixed;
#ifdef _LOCAL
    freopen("in.txt", "r", stdin);
    int tests; cin >> tests;
    for(int test = 1; test <= tests; ++test) {
		cerr << "case #" << test << endl;
        solve();
        cerr << endl;
    }
#else
//    freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
    ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
    solve();
#endif
    return 0;
}

Compilation message (stderr)

zoltan.cpp: In function 'int m_bpow(ll, ll)':
zoltan.cpp:50:26: warning: suggest parentheses around '-' inside '<<' [-Wparentheses]
     for(ll p = 1ll << 63 - __builtin_clzll(b); p; p >>= 1) {
                       ~~~^~~~~~~~~~~~~~~~~~~~

• Subtask 1: 70 / 140