#1082530 (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: 2025-05-06 21:37:04
• 言語: C++17 (gcc 15.2.0 + boost 1.89.0)
• 結果: WA
• 実行時間: -
• コード長: 4,385 bytes
• コンパイル時間: 4,477 ms
• コンパイル使用メモリ: 275,400 KB
• 実行使用メモリ: 30,320 KB
• 平均クエリ数: 488.00
• 最終ジャッジ日時: 2026-05-01 21:04:19
• 合計ジャッジ時間: 13,936 ms
• ジャッジサーバーID （参考情報）: judge1_1 / judge2_1

ファイルパターン	結果
other	AC * 39 WA * 21

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 = 25;

int main(){
	int len = 1;
	auto plus = [&](int i, int j){
		cout << "a " << i << " + " << j << "\n";
		len++;
		return len;
	};
	auto minus = [&](int i, int j){
		cout << "a " << i << " - " << j << "\n";
		len++;
		return len;
	};
	auto multiply = [&](int i, int j){
		cout << "a " << i << " * " << j << "\n";
		len++;
		return len;
	};
	auto root = [&](int i, int j){
		cout << "a " << i << " r " << j << "\n";
		len++;
		return len;
	};

	auto less = [&](int i, int j){
		cout << "? " << i << " < " << j << "\n";
		char c;
		cin >> c;
		if(c == 'T') return true;
		return false;
	};
	auto equal = [&](int i, int j){
		cout << "? " << i << " < " << j << "\n";
		char c1;
		cin >> c1;
		cout << "? " << j << " < " << i << "\n";
		char c2;
		cin >> c2;
		if(c1 == 'T' and c2 == 'T') return true;
		else return false;
	};
	
	auto answer = [&](bool success){
		if(success) cout << "! Yes\n";
		else cout << "! No\n";
	};
	
	int n = -1;
	int one = -1;
	int two = -1;
	int three = -1;
	
	// a = floor(sqrt(N)) + x
	vector<int> a(M, -1);
	// a^2
	vector<int> a2(M, -1);
	// a^2 - N
	vector<int> a2mN(M, -1);
	// b = floor(sqrt(a^2 - N))
	vector<int> b(M, -1);
	// b^2
	vector<int> b2(M, -1);
	// d = a - b
	vector<int> d(M, -1);
	
	// (N + 3d^2)^2 N
	vector<int> LHS(M, -1);
	// ((3d + 1)N + d^3)^2
	vector<int> RHS(M, -1);
	
	
	n = 1;
	one = root(n, n);
	two = plus(one, one);
	three = plus(one, two);
	
	{
		a[0] = root(two, n);
		a2[0] = multiply(a[0], a[0]);
	}
	for(int x = 1; x < M; x++){
		a[x] = plus(a[x - 1], one);
		a2[x] = multiply(a[x], a[x]);
		a2mN[x] = minus(a2[x], n);
		b[x] = root(two, a2mN[x]);
		b2[x] = multiply(b[x], b[x]);
		
		d[x] = minus(a[x], b[x]);
		
		// N + 3d^2
		int _alpha = plus(n, multiply(three, multiply(d[x], d[x])));
		// (3d + 1)N + d^3
		int _beta = plus(multiply(plus(multiply(three, d[x]), one), n), multiply(d[x], multiply(d[x], d[x])));
		
		LHS[x] = multiply(multiply(_alpha, _alpha), n);
		RHS[x] = multiply(_beta, _beta);
	}
	{
		if(equal(a2[0], n)){
			answer(true);
			return 0;
		}
	}
	{
		for(int x = 1; x < M; x++){
			if(equal(b2[x], a2mN[x])){
				if(less(LHS[x], RHS[x]) or equal(LHS[x], RHS[x])) answer(true);
				else answer(false);
				return 0;
			}
		}
	}
	
	answer(false);
	return 0;
}