結果

問題 No.3081 Make Palindromic Multiple
ユーザー 👑 binap
提出日時 2025-02-25 14:04:48
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,413 bytes
コンパイル時間 4,584 ms
コンパイル使用メモリ 262,680 KB
実行使用メモリ 94,680 KB
最終ジャッジ日時 2025-03-27 13:24:00
合計ジャッジ時間 9,495 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge5
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 2 WA * 1
other AC * 4 WA * 5 TLE * 2 -- * 43
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;

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);}

template<typename T>
T pow_mod(T A, T N, T MOD){
	T res = 1 % MOD;
	A %= MOD;
	while(N){
		if(N & 1) res = (res * A) % MOD;
		A = (A * A) % MOD;
		N >>= 1;
	}
	return res;
}

int main(){
	long long n;
	cin >> n;
	
	const int K = 12;
	
	vector<long long> f(K);
	rep(i, K) f[i] += pow_mod<__uint128_t>(10, i, n);
	rep(i, K) f[(K - 1) - i] += pow_mod<__uint128_t>(10, i + K, n);
	
	vector<vector<long long>> g(K, vector<long long>(10)); 
	rep(i, K) rep(num, 10) g[i][num] = (f[i] * num) % n;
	
	int K1 = (K + 1)/ 2;
	int K2 = K - K1;
	
	vector<unordered_set<long long>> se1(K1 + 1);
	se1[0].insert(0);
	rep(i, K1){
		for(auto val : se1[i]){
			if(i > 0){
				se1[i + 1].insert(val);
			}
			for(int num = 1; num <= 9; num++){
				long long val_next = val + g[i][num];
				if(val_next >= n) val_next -= n;
				se1[i + 1].insert(val_next);
			}
		}
	}
	vector<unordered_set<long long>> se2(K2 + 1);
	se2[0].insert(0);
	rep(i, K2){
		for(auto val : se2[i]){
			for(int num = 0; num <= 9; num++){
				long long val_next = val + g[K1 + i][num];
				if(val_next >= n) val_next -= n;
				se2[i + 1].insert(val_next);
			}
		}
	}
	
	auto out = [&](long long val){
		string u1, d1;
		string u2, d2;
		long long val1 = val;
		long long val2 = (n - val) % n;
		
		for(int i = K1; i > 0; i--){
			assert(se1[i].find(val1) != se1[i].end());
			int num;
			for(num = 0; num <= 9; num++){
				if(se1[i - 1].find((val1 - g[i - 1][num] + n) % n) != se1[i - 1].end()) break;
			}
			u1 += '0' + num;
			d1 += '0' + num;
			val1 = (val1 - g[i - 1][num] + n) % n;
		}
		for(int i = K2; i > 0; i--){
			assert(se2[i].find(val2) != se2[i].end());
			int num;
			for(num = 0; num <= 9; num++){
				if(se2[i - 1].find((val2 - g[K1 + (i - 1)][num] + n) % n) != se2[i - 1].end()) break;
			}
			u2 += '0' + num;
			d2 += '0' + num;
			val2 = (val2 - g[K1 + (i - 1)][num] + n) % n;
		}
		reverse(u1.begin(), u1.end());
		reverse(u2.begin(), u2.end());
		cout << "1\n";
		cout << u1 << u2 << d2 << d1 << " 1\n";
		return 0;
	};
	
	for(auto val : se1[K1]){
		if(se2[K2].find((n - val) % n) != se2[K2].end()){
			out(val);
			return 0;
		}
	}
	
	assert(false);
	return 0;
}
0