結果
問題 |
No.3089 Base M Numbers, But Only 0~9
|
ユーザー |
![]() |
提出日時 | 2025-04-05 01:29:50 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 31 ms / 2,000 ms |
コード長 | 4,479 bytes |
コンパイル時間 | 3,598 ms |
コンパイル使用メモリ | 183,204 KB |
実行使用メモリ | 8,192 KB |
最終ジャッジ日時 | 2025-04-05 01:29:56 |
合計ジャッジ時間 | 4,976 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 5 |
other | AC * 19 |
ソースコード
#include <algorithm> #include <atcoder/all> #include <bitset> #include <cassert> #include <cmath> #include <ctime> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <random> #include <set> #include <stack> #include <string> #include <tuple> #include <unordered_map> #include <unordered_set> #include <utility> #include <vector> using namespace std; using namespace atcoder; typedef long long ll; #define rep(i, n) for (int i = 0; i < (int)(n); i++) #define repr(i, n) for (int i = (int)(n) - 1; i >= 0; i--) #define repk(i, k, n) for (int i = k; i < (int)(n); i++) #define all(v) v.begin(), v.end() #define mod1 1000000007 #define mod2 998244353 #define mod3 100000007 #define vi vector<int> #define vs vector<string> #define vc vector<char> #define vl vector<ll> #define vb vector<bool> #define vvi vector<vector<int>> #define vvc vector<vector<char>> #define vvl vector<vector<ll>> #define vvb vector<vector<bool>> #define vvvi vector<vector<vector<int>>> #define vvvl vector<vector<vector<ll>>> #define pii pair<int, int> #define pil pair<int, ll> #define pli pair<ll, int> #define pll pair<ll, ll> #define vpii vector<pair<int, int>> #define vpll vector<pair<ll, ll>> #define vvpii vector<vector<pair<int, int>>> #define vvpll vector<vector<pair<ll, ll>>> // using mint = modint998244353; template <typename T> void debug(T e) { cerr << e << endl; } template <typename T> void debug(vector<T> &v) { rep(i, v.size()) { cerr << v[i] << " "; } cerr << endl; } template <typename T> void debug(vector<vector<T>> &v) { rep(i, v.size()) { rep(j, v[i].size()) { cerr << v[i][j] << " "; } cerr << endl; } } template <typename T> void debug(vector<pair<T, T>> &v) { rep(i, v.size()) { cerr << v[i].first << " " << v[i].second << endl; } } template <typename T> void debug(set<T> &st) { for (auto itr = st.begin(); itr != st.end(); itr++) { cerr << *itr << " "; } cerr << endl; } template <typename T> void debug(multiset<T> &ms) { for (auto itr = ms.begin(); itr != ms.end(); itr++) { cerr << *itr << " "; } cerr << endl; } template <typename T> void debug(map<T, T> &mp) { for (auto itr = mp.begin(); itr != mp.end(); itr++) { cerr << itr->first << " " << itr->second << endl; } } void debug_out() { cerr << endl; } template <typename Head, typename... Tail> void debug_out(Head H, Tail... T) { cerr << H << " "; debug_out(T...); } using mint = modint998244353; void debug_mint1(vector<mint> &vec) { for (int i = 0; i < vec.size(); i++) { cerr << vec[i].val() << " "; } cerr << endl; } void debug_mint2(vector<vector<mint>> &vec) { for (int i = 0; i < vec.size(); i++) { for (int j = 0; j < vec[i].size(); j++) { cerr << vec[i][j].val() << " "; } cerr << endl; } } mint inv2 = mint(499122177); mint prod_sum(ll start, ll M, ll num) { // 等差数列の和の公式なのは恐らく正解 // debug_out(start, M, num); mint mstart = mint(start); mint mgoal = mint(start) + mint(num - 1) * mint(10); mint ret = (mstart + mgoal) * mint(num) * inv2; return ret; } int main() { ll M; cin >> M; string S; cin >> S; ll N = S.size(); ll rem = M % 10; vector<ll> num(N); vector<ll> rems(N); for (ll i = 0; i < N; i++) { rems[i] = (ll)(S[i] - '0'); if (rems[i] < rem) { num[i] = M / 10 + 1; } else { num[i] = M / 10; } if (i == 0 && rems[i] == 0) { num[i]--; } } vector<mint> left_prod(N + 2, mint(1)); vector<mint> right_prod(N + 2, mint(1)); for (ll i = 1; i <= N; i++) { left_prod[i] = left_prod[i - 1] * mint(num[i - 1]); } for (ll i = N; i >= 1; i--) { right_prod[i] = right_prod[i + 1] * mint(num[i - 1]); } // debug_mint1(left_prod); // debug_mint1(right_prod); mint ans = mint::raw(0); for (ll i = 0; i < N; i++) { mint pnum = mint(M).pow(N - i - 1); ll start = rems[i]; ll last = rems[i] + (M / 10) * 10; if (rems[i] >= rem) { last -= 10; } if (i == 0 && rems[i] == 0) { start += 10; } ans += pnum * prod_sum(start, M % mod2, num[i]) * left_prod[i] * right_prod[i + 2]; // debug(ans.val()); } cout << ans.val() << endl; }