結果

問題 No.3068 Speedrun (Hard)
ユーザー 👑 binap
提出日時 2025-03-21 22:30:40
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,564 bytes
コンパイル時間 4,409 ms
コンパイル使用メモリ 251,712 KB
実行使用メモリ 7,328 KB
最終ジャッジ日時 2025-03-21 22:30:50
合計ジャッジ時間 10,454 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge7
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 13 WA * 19
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef pair<int, int> P;
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return 
    os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); 
    return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return 
    true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder
    ::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 
    1 ? "\n" : " "); return os;}
template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}
int main(){
    vector<int> a(4);
    cin >> a;
    int n;
    cin >> n;
    vector<int> p(4);
    cin >> p;
    int t;
    cin >> t;
    
    bool FLIP0 = false;
    bool FLIP1 = false;
    if(p[0] < p[1]){
        FLIP0 = true;
        swap(a[0], a[1]);
        swap(p[0], p[1]);
    }
    if(p[2] < p[3]){
        FLIP1 = true;
        swap(a[2], a[3]);
        swap(p[2], p[3]);
    }
    
    vector<vector<int>> alpha(n + 1);
    vector<vector<int>> beta(n + 1);
    
    for(int s0 = 0; s0 <= min(n, a[0] + a[1]); s0++){
        int s1 = n - s0;
        if(s1 > a[2] + a[3]) continue;
        vector<int> f0;
        for(int k0 = 0; k0 <= s0; k0++){
            int k1 = s0 - k0;
            if(k1 > a[1]) continue;
            f0.push_back(p[0] * k0 + p[1] * k1);
        }
        vector<int> f1;
        for(int k2 = 0; k2 <= s1; k2++){
            int k3 = s1 - k2;
            if(k3 > a[3]) continue;
            f1.push_back(p[2] * k2 + p[3] * k3);
        }
        int i1 = f1.size() - 1;
        rep(i0, f0.size()){
            while(i1 >= 0 and f0[i0] + f1[i1] > t) i1--;
            if(i1 < 0) break;
            if(f0[i0] + f1[i1] == t){
                for(int k0 = 0; k0 <= s0; k0++){
                    int k1 = s0 - k0;
                    if(p[0] * k0 + p[1] * k1 == f0[i0]){
                        if(FLIP0) cout << k1 << ' ' << k0 << ' ';
                        else cout << k0 << ' ' << k1 << ' ';
                        break;
                    }
                }
                for(int k2 = 0; k2 <= s1; k2++){
                    int k3 = s1 - k2;
                    if(k3 > a[3]) continue;
                    if(p[2] * k2 + p[3] * k3 == f1[i1]){
                        if(FLIP1) cout << k3 << ' ' << k2 << ' ';
                        else cout << k2 << ' ' << k3 << ' ';
                        break;
                    }
                }
                return 0;
            }
        }
    }
    assert(false);
    return 0;
}
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