Submission #316298

#	Time	Username	Problem	Language	Result	Execution time	Memory
316298		lauran	Jelly Flavours (IOI20_jelly)	C++14	59 / 100	363 ms	193272 KiB

jelly

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

const int INF = 1e9;

int dp[205][505][505];
int dp2[2005][10005];

bool subtask4(vector <int> b) {
    for (auto x: b)
        if (x != b[0])
            return 0;
    return 1;
}

int find_maximum_unique(int x, int y, vector <int> a, vector <int> b) {
    int n = a.size();
    if (x <= 500 && y <= 500 && n <= 200) {
        // subtask 1, 2
        //int dp[n + 10][x + 10][y + 10] = {0};
        for (int i = 1; i <= n; i++) {
            for (int xx = 0; xx <= x; xx++) {
                for (int yy = 0; yy <= y; yy++) {
                    dp[i][xx][yy] = dp[i - 1][xx][yy];
                    if (xx >= a[i - 1])
                        dp[i][xx][yy] = max(dp[i][xx][yy], dp[i - 1][xx - a[i - 1]][yy] + 1);
                    if (yy >= b[i - 1])
                        dp[i][xx][yy] = max(dp[i][xx][yy], dp[i - 1][xx][yy - b[i - 1]] + 1);
                }
            }
        }
        int ans = dp[n][0][0];
        for (int xx = 0; xx <= x; xx++)
            for (int yy = 0; yy <= y; yy++)
                ans = max(ans, dp[n][xx][yy]);
        return ans;
    }
    else if (subtask4(b)) {
        // iau cele mai ieftine din a, dupa care cat pot din b
        sort(a.begin(), a.end());
        int poz = 0, sum = 0;
        while (poz <= n - 1 && sum + a[poz] <= x)
            sum += a[poz], poz++;
        sum = 0;
        while (poz <= n - 1 && sum + b[poz] <= y)
            sum += b[poz], poz++;
        return poz;
    }
    else {
        // subtaskul 5
        sort(a.begin(), a.end());
        sort(b.begin(), b.end());

        for (int i = 1; i <= n; i++) {
            for (int xx = 0; xx <= x; xx++) {
                // iau primele i, consum xx in A; cat consum minim in B?
                dp2[i][xx] = INF;
                if (xx >= a[i - 1])
                    dp2[i][xx] = min(dp2[i][xx], dp2[i - 1][xx - a[i - 1]]);
                dp2[i][xx] = min(dp2[i][xx], dp2[i - 1][xx] + b[i - 1]);
            }
        }

        int ret = 0;
        while (ret < n) {
            // caut dp2[ret + 1][xx] <= y
            bool ok = 0;
            for (int xx = 0; xx <= x; xx++)
                if (dp2[ret + 1][xx] <= y)
                    ok = 1;
            if (ok)
                ret++;
            else
                break;
        }
        return ret;
    }
}

• Subtask 1: 11 / 11
• Subtask 2: 24 / 24
• Subtask 3: 0 / 9
• Subtask 4: 10 / 10
• Subtask 5: 14 / 14
• Subtask 6: 0 / 32