結果

問題 No.3089 Base M Numbers, But Only 0~9
ユーザー 👑 binap
提出日時 2025-04-05 10:13:07
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,138 bytes
コンパイル時間 4,689 ms
コンパイル使用メモリ 254,064 KB
実行使用メモリ 6,528 KB
最終ジャッジ日時 2025-04-05 10:13:15
合計ジャッジ時間 6,967 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
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 * 5
other AC * 15 WA * 4
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 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);}

using mint = modint998244353;

using S = mint;
S op(S a, S b){
	return a * b;
}
S e(){
	return 1;
}

int main(){
	long long m;
	cin >> m;
	string s;
	cin >> s;
	
	int n = s.size();
	
	segtree<S, op, e> seg(n);
	rep(i, n){
		int c = s[i] - '0';
		if(i == 0 and s[i] == '0'){
			long long cnt = (m - c - 1) / 10;
			seg.set(i, cnt);
		}else{
			long long cnt = (m - c - 1) / 10 + 1;
			seg.set(i, cnt);
		}
	}
	mint ans = 0;
	rep(i, n){
		int c = s[i] - '0';
		mint coeff1 = seg.prod(0, i);
		mint coeff2 = seg.prod(i + 1, n);
		mint coeff3;
		if(m % 998244353 == 0){
			if((n - 1) - i == 0) coeff3 = 1;
			else coeff3 = 0;
		}else{
			coeff3 = mint(m).pow((n - 1) - i);
		}
		if(i == 0 and s[i] == '0'){
			long long cnt = (m - c - 1) / 10;
			mint val = cnt * (cnt + 1) / 2 * 10 + c * cnt;
			ans += coeff1 * coeff2 * val * coeff3;
		}else{
			long long cnt = (m - c - 1) / 10 + 1;
			mint val = cnt * (cnt - 1) / 2 * 10 + c * cnt;
			ans += coeff1 * coeff2 * val * coeff3;
		}
	}
	cout << ans << "\n";
	return 0;
}
0