Submission #222155

#	Time	Username	Problem	Language	Result	Execution time	Memory
222155		kartel	새로운 문제 (COCI19_akvizna)	C++14	65 / 130	525 ms	225144 KiB

Main

#include <bits/stdc++.h>
#define in(x) freopen(x,"r",stdin)
#define out(x) freopen(x,"w",stdout)
#define F first
#define S second
#define pb push_back
#define M ll(998244353)
#define inf ll(2e9+7)
#define N ll(200500)
#define sz(x) x.size()
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#define el '\n'
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;

//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//typedef tree <ll, null_type, less_equal <ll>, rb_tree_tag, tree_order_statistics_node_update> os;

ld f[3001][3001];
ll n, m, i, j, kol, sz, num[N], pr, K;
ld cht[N], b[3001][3001], k[3001][3001];

void AddInCHT(int i, int j)
{
    ld x;
    while (1)
    {
        pr = num[sz];

        x = (b[pr][j] - b[i][j]) / (k[i][j] - k[pr][j]);
//        cerr << x << el;

        if (x - cht[sz] > 1e-9) break;

        sz--;
    }

    sz++;

    cht[sz] = x;
    num[sz] = i;
}

int main()
{
//    in("input.txt");
//    out("output.txt");

    ios_base::sync_with_stdio(false);
    iostream::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n >> m;

    f[n][0] = 0;
    for (i = 1; i < n; i++) b[i][0] = -1e18, k[i][0] = 0;
    for (i = 1; i <= m; i++) f[n][i] = 0, k[n][i] = 0, b[n][i] = -1e18;

    k[n][0] = ld(1) / ld(n);
    b[n][0] = 0;

    for (j = 1; j <= m; j++)
        {
            sz = 1;

            cht[sz] = -1e18;

            num[sz] = n - j + 1;

            for (i = n - j; i >= 0; i--)
                {
                    int l = 1;
                    int r = sz;

                    while (l < r)
                    {
                        int md = (l + r + 1) >> 1;
                        ld x = cht[md];

                        if (x + ld(i) < 1e-9) l = md;
                                         else r = md - 1;
                    }

                    K = num[l];

                    f[i][j] = b[K][j - 1] + 1 - ld(i) * k[K][j - 1];

                    k[i][j] = ld(1) / ld(i);

                    b[i][j] = f[i][j];

                    if (k[i][j - 1] != 0) AddInCHT(i, j - 1);
                }
        }

    cout << setprecision(9) << fixed << f[0][m];
}

• Subtask 1: 5 / 5
• Subtask 2: 5 / 5
• Subtask 3: 5 / 5
• Subtask 4: 5 / 5
• Subtask 5: 5 / 5
• Subtask 6: 5 / 5
• Subtask 7: 5 / 5
• Subtask 8: 5 / 5
• Subtask 9: 5 / 5
• Subtask 10: 5 / 5
• Subtask 11: 5 / 5
• Subtask 12: 5 / 5
• Subtask 13: 5 / 5
• Subtask 14: 0 / 5
• Subtask 15: 0 / 5
• Subtask 16: 0 / 5
• Subtask 17: 0 / 5
• Subtask 18: 0 / 5
• Subtask 19: 0 / 5
• Subtask 20: 0 / 5
• Subtask 21: 0 / 5
• Subtask 22: 0 / 5
• Subtask 23: 0 / 5
• Subtask 24: 0 / 5
• Subtask 25: 0 / 5
• Subtask 26: 0 / 5