結果
| 問題 | No.3081 Make Palindromic Multiple |
| ユーザー |
👑  binap
|
| 提出日時 | 2025-03-26 22:46:57 |
| 言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
RE
|
| 実行時間 | - |
| コード長 | 7,090 bytes |
| コンパイル時間 | 6,828 ms |
| コンパイル使用メモリ | 336,088 KB |
| 実行使用メモリ | 33,600 KB |
| 最終ジャッジ日時 | 2025-03-27 13:27:22 |
| 合計ジャッジ時間 | 15,206 ms |
|
ジャッジサーバーID （参考情報） |
judge1 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 51 RE * 3 |
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 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" : " "); returnos;}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){returntrue;}); 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);}// Sieve of Eratosthenes// https://youtu.be/UTVg7wzMWQc?t=2774struct Sieve {int n;vector<int> f, primes;Sieve(int n=1):n(n), f(n+1) {f[0] = f[1] = -1;for (long long i = 2; i <= n; ++i) {if (f[i]) continue;primes.push_back(i);f[i] = i;for (long long j = i*i; j <= n; j += i) {if (!f[j]) f[j] = i;}}}bool isPrime(int x) { return f[x] == x;}vector<int> factorList(int x) {vector<int> res;while (x != 1) {res.push_back(f[x]);x /= f[x];}return res;}vector<pair<ll,int>> factor(ll x) {vector<pair<ll,int>> res;for (int p : primes) {int y = 0;while (x%p == 0) x /= p, ++y;if (y != 0) res.emplace_back(p,y);}if (x != 1) res.emplace_back(x,1);return res;}};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;Sieve sieve(1000005);auto factor = sieve.factor(n);int two = 0;int five = 0;for(auto [key, val] : factor){if(key == 2) two = val;if(key == 5) five = val;}assert(two == 0 or five == 0);if(two <= 3 and five <= 1){int c = 1;long long phi = n;rep(_, two) c *= 2, phi /= 2;rep(_, five) c *= 5, phi /= 5;for(auto [p, e] : factor){if(p == 2) continue;if(p == 5) continue;phi /= p;phi *= p - 1;}cout << "1\n";cout << c << ' ' << 9 * phi << "\n";return 0;}const int K = 12;const int L = 40;// n = 2^e2 * 5^e5 * mlong long m = n;string s;if(two > 0){long long x = 1;rep(i, two) x *= 2;m /= x;s = to_string(x);}else if(five > 0){long long x = 1;rep(i, five) x *= 5;m /= x;s = to_string(x);}else{assert(false);}while(s.size() < L){s.insert(0, "0");}string s_rev = s;reverse(s_rev.begin(), s_rev.end());long long reminder = 0;rep(i, L) reminder += (s[i] - '0') * pow_mod<__uint128_t>(10, (L - 1) - i, m);rep(i, L) reminder += (s_rev[i] - '0') * pow_mod<__uint128_t>(10, (2 * L + 2 * K - 1) - i, m);reminder %= m;vector<long long> f(K);rep(i, K) f[i] += pow_mod<__uint128_t>(10, L + i, m);rep(i, K) f[(K - 1) - i] += pow_mod<__uint128_t>(10, L + i + K, m);vector<vector<long long>> g(K, vector<long long>(10));rep(i, K) rep(num, 10) g[i][num] = (f[i] * num) % m;int K1 = (K + 1)/ 2;int K2 = K - K1;vector<vector<long long>> vec1(K1 + 1);vec1[0] = {0LL};rep(i, K1){for(auto val : vec1[i]){if(i > 0){vec1[i + 1].push_back(val);}for(int num = 1; num <= 9; num++){long long val_next = val + g[i][num];if(val_next >= m) val_next -= m;vec1[i + 1].push_back(val_next);}}}vector<vector<long long>> vec2(K2 + 1);vec2[0] = {reminder % m};rep(i, K2){for(auto val : vec2[i]){for(int num = 0; num <= 9; num++){long long val_next = val + g[K1 + i][num];if(val_next >= m) val_next -= m;vec2[i + 1].push_back(val_next);}}}auto out = [&](long long val){string u1, d1;string u2, d2;long long val1 = val;long long val2 = (2 * n - val) % m;for(int i = K1; i > 0; i--){int num;for(num = 0; num <= 9; num++){if(i == 1 and num == 0) continue;long long val = (val1 - g[i - 1][num] + m) % m;int idx = lower_bound(vec1[i - 1].begin(), vec1[i - 1].end(), val) - vec1[i - 1].begin();if(idx < int(vec1[i - 1].size()) and vec1[i - 1][idx] == val) break;}{long long val = (val1 - g[i - 1][num] + m) % m;u1 += '0' + num;d1 += '0' + num;val1 = val;}}for(int i = K2; i > 0; i--){int num;for(num = 0; num <= 9; num++){long long val = (val2 - g[K1 + (i - 1)][num] + m) % m;int idx = lower_bound(vec2[i - 1].begin(), vec2[i - 1].end(), val) - vec2[i - 1].begin();if(idx < int(vec2[i - 1].size()) and vec2[i - 1][idx] == val) break;}{long long val = (val2 - g[K1 + (i - 1)][num] + m) % m;u2 += '0' + num;d2 += '0' + num;val2 = val;}}reverse(u1.begin(), u1.end());reverse(u2.begin(), u2.end());cout << "3\n";cout << s_rev << ' ' << "1\n";cout << u1 << u2 << d2 << d1 << " 1\n";cout << s << ' ' << "1\n";return 0;};rep(i, K1 + 1) sort(vec1[i].begin(), vec1[i].end());rep(i, K2 + 1) sort(vec2[i].begin(), vec2[i].end());{if(vec1[K1][0] == 0 and vec2[K2][0] == 0){out(0LL);return 0;}}int idx = int(vec2[K2].size()) - 1;for(auto val1 : vec1[K1]){if(val1 == 0) continue;long long val2 = (m - val1) % m;while(idx > 0){if(vec2[K2][idx - 1] >= val2) idx--;else break;}if(vec2[K2][idx] == val2){out(val1);return 0;}}assert(false);return 0;}