Submission #222152

#	Time	Username	Problem	Language	Result	Execution time	Memory
222152		ne4eHbKa	새로운 문제 (COCI19_akvizna)	C++17	20 / 130	8 ms	1792 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 = 505;

int n, k;

int f[N][N];
ld e[N][N];

ld solve(int n, int k) {
    if(n < 2) {
        f[n][k] = n;
        re e[n][k] = n;
    }
    if(k == 1) {
        f[n][k] = n;
        re e[n][k] = 1;
    }
    if(~f[n][k]) re e[n][k];
    ld ans = 0;
    int j = -1;
    for(int i = 1; i <= n; ++i) {
        ld t = solve(n - i, k - 1) + i / ld(n);
        if(t > ans) {
            ans = t;
            j = i;
        }
    }
    f[n][k] = j;
    re e[n][k] = ans;
}

void solve() {
    clr(f, -1);
    cin >> n >> k;
    if(n > 100) cout << "wat\n";
    else cout << solve(n, k) << 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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