Submission #1199074

#	Time	Username	Problem	Language	Result	Execution time	Memory
1199074		thieunguyenhuy	Koala Game (APIO17_koala)	C++17	100 / 100	35 ms	468 KiB

koala

#ifndef hwe
	#include "koala.h"
#endif

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

#define popcount(n) (__builtin_popcountll((n)))
#define clz(n) (__builtin_clzll((n)))
#define ctz(n) (__builtin_ctzll((n)))
#define lg(n) (63 - __builtin_clzll((n)))
#define BIT(n, i) (((n) >> (i)) & 1ll)
#define MASK(i) (1ll << (i))
#define FLIP(n, i) ((n) ^ (1ll << (i)))
#define ON(n, i) ((n) | MASK(i))
#define OFF(n, i) ((n) & ~MASK(i))

#define Int __int128
#define fi first
#define se second

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef pair<long long, int> pli;
typedef pair<int, long long> pil;
typedef vector<pair<int, int>> vii;
typedef vector<pair<long long, long long>> vll;
typedef vector<pair<long long, int>> vli;
typedef vector<pair<int, long long>> vil;

template <class T1, class T2>
bool maximize(T1 &x, T2 y) {
    if (x < y) {
        x = y;
        return true;
    }
    return false;
}
template <class T1, class T2>
bool minimize(T1 &x, T2 y) {
    if (x > y) {
        x = y;
        return true;
    }
    return false;
}

template <class T>
void remove_duplicate(vector<T> &ve) {
    sort (ve.begin(), ve.end());
    ve.resize(unique(ve.begin(), ve.end()) - ve.begin());
}

mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
template <class T> T random(T l, T r) {
    return uniform_int_distribution<T>(l, r)(rng);
}
template <class T> T random(T r) {
    return rng() % r;
}

const int N = 100;
const int MOD = 1e9 + 7;
const int inf = 1e9;
const long long INF = 1e18;

int B[N], R[N];
int ans[N];

#ifdef hwe
namespace Jury {
    int dp[N][205], cnt[N][205];
    int perm[N];
    int W = 100;

    void test() {
        iota(perm, perm + N, 1);
        shuffle(perm, perm + N, rng);
        for (int i = 0; i < N; ++i)
            cerr << perm[i] << ' ';
        cerr << '\n';
    }

    void umax(int i, int j, int u, int v, int val) {
        if (maximize(dp[i][j], dp[u][v] + val))
            cnt[i][j] = cnt[u][v] + 1;
        if (dp[i][j] == dp[u][v] + val)
            maximize(cnt[i][j], cnt[u][v] + 1);
    }

    void playRound(int *B, int *R) {
        for (int j = 0; j <= B[0]; ++j) dp[0][j] = cnt[0][j] = 0;
        for (int j = B[0] + 1; j <= W; ++j)
            dp[0][j] = perm[0], cnt[0][j] = 1;
        for (int i = 1; i < N; ++i) {
            // use B[i] + 1 get perm[i]
            for (int j = 0; j <= W; ++j) {
                dp[i][j] = dp[i - 1][j];
                cnt[i][j] = cnt[i - 1][j];
                if (j > 0) umax(i, j, i, j - 1, 0);
            }
            for (int j = W; j >= B[i] + 1; --j) {
                umax(i, j, i - 1, j - (B[i] + 1), perm[i]);
            }
        }

        for (int i = N - 1, j = W; i >= 0; --i) {
            if (i == 0) {
                if (j > 0) R[i] = B[i] + 1;
                else R[i] = 0;
            }
            else {
                if (dp[i][j] == dp[i - 1][j - (B[i] + 1)] + perm[i] && 
                    cnt[i][j] == cnt[i - 1][j - (B[i] + 1)] + 1) {
                    R[i] = B[i] + 1;
                    j -= B[i] + 1;
                }
                else R[i] = 0;
            }
        }
    }
}

void playRound(int *B, int *R) {
    Jury::playRound(B, R);
}
#endif

vector<int> play(const vector<int> &veB) {
	for (int i = 0; i < veB.size(); ++i) B[i] = veB[i];
	playRound(B, R); vector<int> veR(veB.size());
	for (int i = 0; i < veB.size(); ++i) veR[i] = R[i];
	return veR;
}

int minValue(int n, int W) {
    // TODO: Implement Subtask 1 solution here.
    // You may leave this function unmodified if you are not attempting this
    // subtask.
	vector<int> B(n, 0); B[0] = 1;
	vector<int> R = play(B);

	if (R[0] <= B[0]) return 0;
	for (int i = 1; i < n; ++i) if (R[i] == 0) {
		return i;
	}
	return -1;
}

int maxValue(int n, int W) {
    // TODO: Implement Subtask 2 solution here.
    // You may leave this function unmodified if you are not attempting this
    // subtask.
	
    vector<int> candidates(n);
    iota(candidates.begin(), candidates.end(), 0);

    while (candidates.size() > 1) {
        vector<int> B(n, 0);
        for (auto &x : candidates) B[x] = W / candidates.size();
        vector<int> R = play(B), new_candidates;
        for (auto &x : candidates) if (R[x] > B[x]) {
            new_candidates.emplace_back(x);
        }
        swap(candidates, new_candidates);
    }

    return candidates[0];
}

int greaterValue(int n, int W) {
    // TODO: Implement Subtask 3 solution here.
    // You may leave this function unmodified if you are not attempting this
    // subtask.
    // Nếu B[0] = B[1] = 9 thì không thể lấy được cả 2 thằng này
    int low = 1, high = 9;
    while (low <= high) {
        int mid = (low + high) >> 1;
        vector<int> B(n, 0); B[0] = B[1] = mid;
        vector<int> R = play(B);
        bool isTake0 = R[0] > B[0], isTake1 = R[1] > B[1];
        if (isTake0 != isTake1) return isTake1;
        if (isTake0) low = mid + 1;
        else high = mid - 1;
    }
    return -1;
}

void dnc(int n, int W, int *P, int l, int r, vector<int> &ve) {
    if (l == r) {
        P[ve[0]] = l;
        return;
    }

    int stone = (W == 2 * n ? W / (r - l + 1) : min(int(sqrt(2 * l)), W / (r - l + 1)));

    vector<int> B(n, 0);
    for (auto &x : ve) B[x] = stone;
    vector<int> R = play(B);

    vector<int> less, greater;
    for (auto &x : ve) {
        if (R[x] > B[x]) greater.emplace_back(x);
        else less.emplace_back(x);
    }

    dnc(n, W, P, l, l + less.size() - 1, less);
    dnc(n, W, P, r - greater.size() + 1, r, greater);
}

void allValues(int n, int W, int *P) {
    vector<int> ve(n); iota(ve.begin(), ve.end(), 0);
    dnc(n, W, P, 1, n, ve);
}

#ifdef hwe
signed main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

    Jury::test();
    allValues(N, Jury::W, ans);

    for (int i = 0; i < N; ++i) cerr << ans[i] << ' ';
    cerr << '\n';

    cerr << '\n'; return 0;
}
#endif

• Subtask 1: 4 / 4
• Subtask 2: 15 / 15
• Subtask 3: 18 / 18
• Subtask 4: 10 / 10
• Subtask 5: 53 / 53