제출 #1102551

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1102551		ndan	Knapsack (NOI18_knapsack)	C++14	29 / 100	1 ms	424 KiB

knapsack

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

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define file "test"
#define forr(i, a, b) for (int i = (a), _b = (b); i <= _b; ++i)
#define ford(i, b, a) for (int i = (b), _a = (a); i >= _a; --i)
#define forf(i, a, b) for (int i = (a), _b = (b); i < _b; ++i)
#define rep(i, n) for (int i = 0, _n = (n); i < _n; ++i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(x) x.begin(), x.end()
const int maxn = 2e3 + 10;
const int MOD = 1e9 + 7;
const ll inf = 1e18;
const int oo = 1e9 + 7;
const float eps = 1e-6;

ll n, s, v[maxn], w[maxn], k[maxn];

namespace sub1 {
    bool check() {
        return n == 1;
    }
    void solve() {
        cout << v[1] * min(s / w[1], k[1]);
    }
}

namespace sub2 {
    bool check() {
        if (n <= 100) {
            bool kt = 1;
            forr(i, 1, n)
                if (k[i] != 1) {
                    kt = 0;
                    break;
                }
            return kt;
        }
        return 0;
    }
    void solve() {
        ll dp[2005];
        memset(dp, 0, sizeof dp);
        forr(i, 1, n)
            ford(j, s, w[i])
                dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
        cout << dp[s];
    }
}

namespace sub3 {
    bool check() {
        if (n <= 100) {
            bool kt = 1;
            forr(i, 1, n)
                if (k[i] > 10) {
                    kt = 0;
                    break;
                }
            return kt;
        }
        return 0;
    }
    void solve() {

    }
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    //freopen(file".in", "r", stdin);
    //freopen(file".out", "w", stdout);
    cin >> s >> n;
    forr(i, 1, n)
        cin >> v[i] >> w[i] >> k[i];
    if (sub1::check()) return sub1::solve(), 0;
    if (sub2::check()) return sub2::solve(), 0;
    if (sub3::check()) return sub3::solve(), 0;
    return 0;
}

• Subtask 1: 12 / 12
• Subtask 2: 17 / 17
• Subtask 3: 0 / 20
• Subtask 4: 0 / 24
• Subtask 5: 0 / 27