#1081240 (C++17) No.3525 擬奇平方数

• 問題: No.3525 擬奇平方数
• ユーザー: 👑 binap
• 提出日時: 2025-05-01 11:24:06
• 言語: C++17 (gcc 15.2.0 + boost 1.89.0)
• 結果: RE (最新)
  AC
  (最初)
• 実行時間: -
• コード長: 3,887 bytes
• コンパイル時間: 4,898 ms
• コンパイル使用メモリ: 280,416 KB
• 実行使用メモリ: 30,320 KB
• 平均クエリ数: 42.00
• 最終ジャッジ日時: 2026-05-01 21:03:47
• 合計ジャッジ時間: 28,409 ms
• ジャッジサーバーID （参考情報）: judge3_0 / judge1_1

ファイルパターン	結果
other	RE * 60

ソースコード

#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;
	memo[1] = 2;
	for(int i = 0; i < (M - 1); i++){
		plus(i + 2, i + 2);
		memo[1 << (i + 1)] = i + 3;
	}
	{
		root(1, 1);
	}
	for(int i = 0; i < (M - 1); i++){
		plus((M + 2), i + (M + 2));
		memo[i + 2] = i + (M + 3);
	}
	flush(cout);
	
	// n < 2 ^ k
	auto assume = [&](auto self, int idx, int k) -> int{
		for(int i = k - 1; i > 0; i--){
			root(memo[i], idx);
			int idx_r = len;
			auto c = minus(idx_r, memo[1]);
			if(c == '0'){
				// n < 2 ^ i
				continue;
			}else{
				// 2 ^ i <= n < 2 ^ (i + 1)
				assert(c == '+');
				auto d = minus(idx, memo[1 << i]);
				if(d == '0') return 1 << i;
				int idx_new = len;
				return self(self, idx_new, i) + (1 << i);
			}
		}
		return 1;
	};
	
	int n = assume(assume, 1, M);
	
//	cout << n << "\n";
	
	auto is_pseudo = [&](int n){
		long double SMALL = 0.0000001;
		long double r2 = pow(n, 0.5);
		long double r3 = pow(n, (long double)1/ 3);
		for(int d = 1; d <= n; d++){
			if(n % d == 0){
				if(r2 - r3 - SMALL <= d and d <= r2 + r3 + SMALL){
					return true;
				}
			}
		}
		return false;
	};
	if(is_pseudo(n)) cout << "! Yes\n";
	else cout << "! No\n";
	return 0;
}