結果

問題 No.3068 Speedrun (Hard)
ユーザー 👑 binap
提出日時 2025-03-21 22:47:21
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 405 ms / 2,000 ms
コード長 3,759 bytes
コンパイル時間 4,708 ms
コンパイル使用メモリ 252,688 KB
実行使用メモリ 7,324 KB
最終ジャッジ日時 2025-03-21 22:47:40
合計ジャッジ時間 6,967 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
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 * 32
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 # 

#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 <= min(s0, a[0]); 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 <= min(s1, a[2]); 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){
				vector<int> ans(4, -1);
				for(int k0 = 0; k0 <= min(s0, a[0]); k0++){
					int k1 = s0 - k0;
					if(k1 > a[1]) continue;
					if(p[0] * k0 + p[1] * k1 == f0[i0]){
						if(FLIP0){
							ans[0] = k1;
							ans[1] = k0;
						}else{
							ans[0] = k0;
							ans[1] = k1;
						}
						break;
					}
				}
				for(int k2 = 0; k2 <= min(s1, a[2]); k2++){
					int k3 = s1 - k2;
					if(k3 > a[3]) continue;
					if(p[2] * k2 + p[3] * k3 == f1[i1]){
						if(FLIP1){
							ans[2] = k3;
							ans[3] = k2;
						}else{
							ans[2] = k2;
							ans[3] = k3;
						}
						break;
					}
				}
				rep(i, 4) assert(ans[i] != -1);
				cout << ans;
				return 0;
			}
		}
	}
	return 0;
}
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