結果

問題 No.3525 擬奇平方数
コンテスト
ユーザー 👑 binap
提出日時 2025-04-29 18:57:05
言語 C++17
(gcc 15.2.0 + boost 1.89.0)
コンパイル:
g++-15 -O2 -lm -std=c++17 -Wuninitialized -DONLINE_JUDGE -o a.out _filename_
実行:
./a.out
結果
RE  
(最新)
AC  
(最初)
実行時間 -
コード長 3,823 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 4,922 ms
コンパイル使用メモリ 281,236 KB
実行使用メモリ 30,320 KB
平均クエリ数 1.00
最終ジャッジ日時 2026-05-01 21:03:03
合計ジャッジ時間 31,998 ms
ジャッジサーバーID
（参考情報） 		judge3_1 / judge2_0
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
other RE * 60
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 #
raw source code 

#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);}

const int M = 21;

int main(){
	int len = 1;
	auto plus = [&](int i, int j){
		cout << "? " << i << " + " << j << "\n";
		char c;
		cin >> c;
		if(c == '+') len++;
		return c;
	};
	auto minus = [&](int i, int j){
		cout << "? " << i << " - " << j << "\n";
		char c;
		cin >> c;
		if(c == '+') len++;
		return c;
	};
	auto multiply = [&](int i, int j){
		cout << "? " << i << " * " << j << "\n";
		char c;
		cin >> c;	
		if(c == '+') len++;
		return c;
	};
	auto root = [&](int i, int j){
		cout << "? " << i << " r " << j << "\n";
		char c;
		cin >> c;	
		if(c == '+') len++;
		return c;
	};
	{
		root(1, 1);
	}
	map<long long, int> memo;
	for(int i = 0; i < M; i++){
		plus(i + 2, i + 2);
		memo[1 << i] = i + 2;
	}
	// ok <= n < ng
	int idx_ok;
	long long ok = -1, ng = -1;
	for(int i = 0; i < M; i++){
		auto c = minus(1, i + 2);
		assert(c != 'e');
		if(c == '+' or c == '0'){
			continue;
		}else{
			idx_ok = (i + 2) - 1;
			ok = (1 << (i - 1));
			ng = (1 << i);
			break;
		}
	}
	assert(ok > 0);
	assert(ng > 0);
	while(ng - ok > 1){
		long long mid = (ok + ng) / 2;
		long long delta = mid - ok;
		if(memo.find(delta) == memo.end()){
			assert(false);
		}
		int idx_delta = memo[delta];
		int idx_mid = -1;
		{
			plus(idx_ok, idx_delta);
			idx_mid = len;
		}
		{
			auto c = minus(1, idx_mid);
			if(c == '+' or c == '0'){
				ok = mid;
				idx_ok = idx_mid;
			}else{
				ng = mid;
			}
		}
	}
	long double SMALL = 0.0000001;
	long double r2 = pow(ok, 0.5);
	long double r3 = pow(ok, (long double)1/ 3);
	for(int d = 1; d <= ok; d++){
		if(ok % d == 0){
			if(r2 - r3 - SMALL <= d and d <= r2 + r3 + SMALL){
				cout << "! Yes\n";
				return 0;
			}
		}
	}
	cout << "! No\n";
	return 0;
}
0