제출 #106377

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
106377		polyfish	코알라 (APIO17_koala)	C++14	19 / 100	18 ms	512 KiB

koala

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

#define debug(x) cerr << #x << " = " << x << '\n';
#define BP() cerr << "OK!\n";
#define PR(A, n) {cerr << #A << " = "; for (int _=1; _<=n; ++_) cerr << A[_] << ' '; cerr << '\n';}
#define PR0(A, n) {cerr << #A << " = "; for (int _=0; _<n; ++_) cerr << A[_] << ' '; cerr << '\n';}
#define FILE_NAME "data"

const int MAX_N = 202;

int b[MAX_N], r[MAX_N], a[MAX_N];
pair<int, int> f[MAX_N][MAX_N];

int minValue(int N, int W) {
    // TODO: Implement Subtask 1 solution here.

    b[0] = 1;
    playRound(b, r);

    for (int i=0; i<N; ++i) {
        if (r[i]==0)
            return i;
    }

    return 0;
}

void trace(int n, int w, int& cnt) {
    if (w==0)
        return;

    if (f[n][w]==f[n-1][w])
        return trace(n-1, w, cnt);

    cnt += (a[n]==*max_element(a+1, a+100+1));
    trace(n-1, w-a[n]-1, cnt);
}

void test() {
    int n = 100, w = 100;
    // TODO: Implement Subtask 2 solution here.

    for (int i=1; i<=50; ++i)
        a[i] = 0;
    for (int i=51; i<=75; ++i)
        a[i] = 0;
    for (int i=76; i<=94; ++i)
        a[i] = 0;
    for (int i=95; i<=100; ++i)
        a[i] = 16;

    for (int i=1; i<=n; ++i) {
        for (int j=0; j<=w; ++j) {
            f[i][j] = f[i-1][j];
            if (j>a[i])
                f[i][j] = max(f[i][j], {f[i-1][j-a[i]-1].first + i, f[i-1][j-a[i]-1].second+1});
        }
    }
    debug(f[n][w].first);
    debug(f[n][w].second);
    int cnt = 0;
    trace(n, w, cnt);
    debug(cnt);
}

int maxValue(int n, int w) {
    // TODO: Implement Subtask 2 solution here.

//    test();
    vector<int> b0, b1, b2, b3;

    ///Round 1
    for (int i=0; i<n; ++i)
        b[i] = 1;
    playRound(b, r);
    for (int i=0; i<n; ++i) {
        if (r[i])
            b1.push_back(i);
        else
            b0.push_back(i);
    }

    ///Round 2
    for (auto x : b0)
        b[x] = 0;
    for (auto x : b1)
        b[x] = 2;
    memset(r, 0, sizeof(r));
    playRound(b, r);
    for (auto x : b1) {
        if (r[x]>=3)
            b2.push_back(x);
    }
    for (auto x : b2)
        b1.erase(find(b1.begin(), b1.end(), x));

    ///Round 3
    for (auto x : b0)
        b[x] = 0;
    for (auto x : b1)
        b[x] = 1;
    for (auto x : b2)
        b[x] = 3;
    memset(r, 0, sizeof(r));
    playRound(b, r);
    for (auto x : b2) {
        if (r[x]>=4)
            b3.push_back(x);
    }
    for (auto x : b3)
        b2.erase(find(b2.begin(), b2.end(), x));


    ///Round 4
    memset(b, 0, sizeof(b));
    for (auto x : b3)
        b[x] = 16;
    memset(r, 0, sizeof(r));
    playRound(b, r);
    for (auto x : b3) {
        if (r[x]>=17)
            return x;
    }

    return 0;
}

int greaterValue(int N, int W) {
    // TODO: Implement Subtask 3 solution here.


    return 0;
}

void allValues(int N, int W, int *P) {
    if (W == 2*N) {
        // TODO: Implement Subtask 4 solution here.
        // You may leave this block unmodified if you are not attempting this
        // subtask.
    } else {
        // TODO: Implement Subtask 5 solution here.
        // You may leave this block unmodified if you are not attempting this
        // subtask.
    }
}

• Subtask 1: 4 / 4
• Subtask 2: 15 / 15
• Subtask 3: 0 / 18
• Subtask 4: 0 / 10
• Subtask 5: 0 / 53