結果

問題 No.3089 Base M Numbers, But Only 0~9
ユーザー 👑 amentorimaru
提出日時 2025-02-19 23:32:54
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
(最新)
AC  
(最初)
実行時間 -
コード長 5,796 bytes
コンパイル時間 4,597 ms
コンパイル使用メモリ 268,556 KB
実行使用メモリ 7,328 KB
最終ジャッジ日時 2025-04-04 12:33:11
合計ジャッジ時間 6,057 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
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ファイルパターン 結果
sample AC * 5
other AC * 13 WA * 6
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 # 

#define ATCODER
#define _USE_MATH_DEFINES
#include<stdio.h>
#include <bit>
#include <cstdint>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<vector>
#include<string>
#include <cassert>
#include <numeric>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <math.h>
#include <climits>
#include <set>
#include <map>
#include <list>
#include <random>
#include <iterator>
#include <bitset>
#include <chrono>
#include <type_traits>
using namespace std;

using ll = long long;
using ld = long double;
using pll = pair<ll, ll>;
using pdd = pair<ld, ld>;

//template<claLR T> using pq = priority_queue<T, vector<T>, greater<T>>;

#define FOR(i, a, b) for(ll i=(a); i<(b);i++)
#define REP(i, n) for(ll i=0; i<(n);i++)
#define ROF(i, a, b) for(ll i=(b-1); i>=(a);i--)
#define PER(i, n) for(ll i=n-1; i>=0;i--)
#define REPREP(i,j,a,b) for(ll i=0;i<a;i++)for(ll j=0;j<b;j++)
#define VV(type) vector< vector<type> >
#define VV2(type,n,m,val) vector< vector<type> > val;val.resize(n);for(ll i;i<n;i++)val[i].resize(m)
#define vec(type) vector<type>
#define VEC(type,n,val) vector<type> val;val.resize(n)
#define VL vector<ll>
#define VVL vector< vector<ll> >
#define VP vector< pair<ll,ll> >
#define SZ size()
#define all(i) begin(i),end(i)
#define SORT(i) sort(all(i))
#define BITI(i) (1<<i)
#define BITSET(x,i) x | (ll(1)<<i)
#define BITCUT(x,i) x & ~(ll(1)<<i)
#define EXISTBIT(x,i) (((x>>i) & 1) != 0)
#define ALLBIT(n) (ll(1)<<n-1)
#define CHMAX(n,v) n=n<v?v:n
#define CHMIN(n,v) n=n>v?v:n
#define MP(a,b) make_pair(a,b)
#define DET2(x1,y1,x2,y2) (x1)*(y2)-(x2)*(y1)
#define DET3(x1,y1,z1,x2,y2,z2,x3,y3,z3) (x1)*(y2)*(z3)+(x2)*(y3)*(z1)+(x3)*(y1)*(z2)-(z1)*(y2)*(x3)-(z2)*(y3)*(x1)-(z3)*(y1)*(x2)
#define INC(a) for(auto& v:a)v++;
#define DEC(a) for(auto& v:a)v--;
#define SQU(x) (x)*(x)
#define L0 ll(0)
#ifdef ATCODER
#include <atcoder/all>
using namespace atcoder;
//using mint = modint1000000007;
using mint2 = modint998244353;
#endif
template<typename T = ll>
vector<T> read(size_t n) {
  vector<T> ts(n);
  for (size_t i = 0; i < n; i++) cin >> ts[i];
  return ts;
}

template<typename TV, const ll N> void read_tuple_impl(TV&) {}
template<typename TV, const ll N, typename Head, typename... Tail>
void read_tuple_impl(TV& ts) {
  get<N>(ts).emplace_back(*(istream_iterator<Head>(cin)));
  read_tuple_impl<TV, N + 1, Tail...>(ts);
}
template<typename... Ts> decltype(auto) read_tuple(size_t n) {
  tuple<vector<Ts>...> ts;
  for (size_t i = 0; i < n; i++) read_tuple_impl<decltype(ts), 0, Ts...>(ts);
  return ts;
}

using val = ll;
using func = ll;

val op(val a, val b) {
  return a + b;
}


val e() { return 0; }

val mp(func f, val a)
{
  return a + f;
}
func comp(func f, func g) {
  return f + g;
}

func id() {
  return 0;
}

ll g;
bool f(val a) {
  return a < g;
}

ll di[4] = { 1,0,-1,0 };
ll dj[4] = { 0,1,0,-1 };
ll si[4] = { 0,3,3,0 };
ll sj[4] = { 0,0,3,3 };
//ll di[4] = { -1,-1,1,1 };
//ll dj[4] = { -1,1,-1,1 };
//ll di8[8] = { 0,-1,-1,-1,0,1,1,1 };
//ll dj8[8] = { -1,-1,0,1,1,1,0,-1 };

class Rational {
public:
  Rational(ll nn = 0, ll dd = 1) {
    n = nn; d = dd; Reduction();
  }
  Rational operator+() const { return *this; }

  Rational& operator=(ll nn) { n = nn; d = 1; return *this; }
  Rational& operator = (const Rational& r) { n = r.n; d = r.d; return *this; }
  Rational& operator += (const Rational& r) { n = n * r.d + r.n * d; d *= r.d; Reduction(); return *this; }
  Rational& operator -= (const Rational& r) { n = n * r.d - r.n * d; d *= r.d; Reduction(); return *this; }
  Rational& operator *= (const Rational& r) { n *= r.n; d *= r.d; Reduction();  return *this; }
  Rational& operator /= (const Rational& r) { n *= r.d; d *= r.n; Reduction();  return *this; }

  bool operator ==(const Rational& r) { return n * r.d == r.n * d; }
  bool operator !=(const Rational& r) { return !(*this == r); }
  bool operator <(const Rational& r) {
    if (d == 0 && r.d == 0) return n < 0 && 0 < r.n;
    if (r.d == 0) return r.n > 0;
    if (d == 0) return n < 0;
    return n * r.d < r.n * d;
  }
  bool operator >(const Rational& r) {
    if (d == 0 && r.d == 0) return r.n < 0 && 0 < n;
    if (r.d == 0) return r.n < 0;
    if (d == 0) return n > 0;
    return n * r.d < r.n * d;
  }

  bool operator <=(const Rational& r) { return !(*this > r); }
  bool operator >=(const Rational& r) { return !(*this < r); }

  void Reduction() {
    ll g = gcd(n, d);
    if (g == 0)return;
    if (d < 0)g *= -1;
    n /= g; d /= g;
  }
  // d is not negative
  ll n, d;
};

static Rational operator + (const Rational& l, const Rational& r) { return Rational(l) += r; }
static Rational operator - (const Rational& l, const Rational& r) { return Rational(l) -= r; }
static Rational operator * (const Rational& l, const Rational& r) { return Rational(l) *= r; }
static Rational operator / (const Rational& l, const Rational& r) { return Rational(l) /= r; }
static bool operator < (const Rational& l, const Rational& r) { return Rational(l) < r; }
static bool operator > (const Rational& l, const Rational& r) { return Rational(l) > r; }
static bool operator <= (const Rational& l, const Rational& r) { return Rational(l) <= r; }
static bool operator >= (const Rational& l, const Rational& r) { return Rational(l) >= r; }


void solve() {
  ll m;
  string n;
  cin >> m;
  cin >> n;
  mint2 v = 1;
  mint2 sum = 0;
  ll mm = n.size();
  REP(i, mm) {
    v *= (m - (n[i] - '0') - 1) / 10 + 1;
    sum *= m;
    sum += n[i] - '0';
  }
  mint2 ans = sum * v;
  mint2 rv = 1;
  mint2 half = mint2(2).inv();
  PER(i, mm) {
    ll c = (m - (n[i] - '0') - 1) / 10 + 1;
    ans += 10 * v * (c - 1) * half * rv;
    rv *= m;
  }
  cout << ans.val();
  return;
}

int main() {
  ll t = 1;
  //cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
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