結果

問題 No.2828 Remainder Game
ユーザー 👑 binapbinap
提出日時 2024-08-02 21:36:26
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
RE  
実行時間 -
コード長 2,873 bytes
コンパイル時間 4,692 ms
コンパイル使用メモリ 262,496 KB
実行使用メモリ 25,532 KB
平均クエリ数 47.90
最終ジャッジ日時 2024-08-02 21:36:36
合計ジャッジ時間 9,292 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge4
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 RE -
testcase_01 RE -
testcase_02 RE -
testcase_03 RE -
testcase_04 RE -
testcase_05 RE -
testcase_06 RE -
testcase_07 RE -
testcase_08 RE -
testcase_09 RE -
testcase_10 RE -
testcase_11 WA -
testcase_12 AC 72 ms
24,824 KB
testcase_13 AC 69 ms
24,952 KB
testcase_14 RE -
testcase_15 AC 68 ms
24,568 KB
testcase_16 AC 65 ms
25,208 KB
testcase_17 AC 68 ms
25,196 KB
testcase_18 AC 65 ms
24,568 KB
testcase_19 AC 64 ms
24,824 KB
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 vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;

ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}
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<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(){
	int n;
	cin >> n;
	vector<int> ans;
	auto solve = [&](auto solve, int left, int right, int rest){
		if(right - left == 1){
			rep(time, rest) ans.push_back(left);
			return;
		}
		int mid = (left + right) / 2;
		int k = mid - left;
		cout << n << ' ' << k << "\n";
		for(int x = left; x < mid; x++) cout << x << ' ';
		cout << "\n";
		int rest_left;
		cin >> rest_left;
		int rest_right = rest - rest_left;
		if(rest != rest_right){
			solve(solve, left, mid, rest_left);
		}
		if(rest != rest_left){
			solve(solve, mid, right, rest_right);
		}
	};
	solve(solve, 0, n, 5);
	int sum = 0;
	assert(int(ans.size()) == 5);
	rep(i, 5) if(ans[i] == 0) ans[i] = n;
	rep(i, 5) sum += ans[i];
	cout << "0 1\n";
	cout << sum << "\n";
	return 0;
}
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