Submission #222182

#	Time	Username	Problem	Language	Result	Execution time	Memory
222182		ne4eHbKa	새로운 문제 (COCI19_akvizna)	C++17	0 / 130	126 ms	8208 KiB

Main

//{ <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 ip2(x) (1 << (x))
#define lp2(x) (1ll << (x))
#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 = lp2(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 = 1e5 + 5;

int n, k, pr[N], nx[N], a[N], p[N];
ll u[N], U[N];

struct cmp {
    bool operator() (const int &a, const int &b) const {
        ll fi = u[a] * U[b];
        ll se = u[b] * U[a];
        re fi == se ? a > b : fi < se;
    }
};

set<int, cmp> s;

void add(int i) {
    int j = nx[i];
    u[i] = a[i] * a[j];
    U[i] = p[i] * p[j];
    s.insert(i);
}

void del(int i) {
    s.erase(i);
}

void cut(int i) {
    del(i);
    int j = nx[i];
    int p = pr[i];
    if(j < n) del(j);
    if(p) del(p);
    nx[p] = j;
    pr[j] = p;
    u[i] = 0;
    a[j] += a[i];
    a[i] = 0;
    if(p) add(p);
    if(j < n) add(j);
}

void solve() {
    cin >> n >> k;

        fo(n) pr[i + 1] = i, nx[i + 1] = i + 2, a[i + 1] = 1, p[i + 1] = i + 1;
        s.clear();
        fo(n - 1) add(i + 1);
        fr(it, n - k) cut(*s.begin());
        ld ans = 0;
        fo(n) ans += ld(a[i + 1]) / p[i + 1];

    ld v = ans;

        ld z = pow(n, ld(1) / (k - 1));
        ans = 1;
        int f = 1;
        fo(k - 1) {
            int h = round(pow(z, i + 1));
            ans += ld(h - f) / h;
            f = h;
        }

    ld w = ans;

    cout << max(v, w) << endl;
}

int main() {
    cerr.precision(3); cerr << fixed;
    cout.precision(9); cout << fixed;
#ifdef _LOCAL
    freopen("in.txt", "r", stdin);
    int tests; cin >> tests;
    for(int test = 1; test <= tests; ++test) {
        cerr << test << " {\n";
        solve();
        cerr << "}\n\n";
    }
#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;
}

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