Educational Codeforces Round 50 (Rated for Div. 2) C. Classy Numbers

問題ページ

解法

桁DPをする。dp[i][j][k] = (i桁目まででA未満が確定しているか(=j)、0以外の数がk個のときの数の個数）とする。jの更新はj or d < (Aのi桁目)とするいつもの、kの更新は0以外なら1足すとすればよい。桁DPの計算量がO(logR)なので全体でO(TlogR)になる。

#include <bits/stdc++.h>
 
using namespace std;
using ll = long long;
// #define int ll
using PII = pair<int, int>;
 
#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
 
template<typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template<typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
 
template<typename T> vector<T> make_v(size_t a) { return vector<T>(a); }
template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts) { 
  return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}
template<typename T,typename V> typename enable_if<is_class<T>::value==0>::type
fill_v(T &t, const V &v) { t=v; }
template<typename T,typename V> typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t, const V &v ) { for(auto &e:t) fill_v(e,v); }
 
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')'; return out;
}
template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
  for(T& x: vec) {is >> x;} return is;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'['; for(T i: a) {out<<i<<',';} out<<']'; return out;
}
 
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0}; // DRUL
const int INF = 1<<30;
const ll LLINF = 1LL<<60;
const int MOD = 1000000007;

signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  auto func = [&](string s) {
    int n = s.size();
    auto dp = make_v<ll>(n+1, 2, 4);
    dp[0][0][0] = 1;
    REP(i, n) REP(j, 2) REP(k, 4) {
      int lim = j?9:s[i]-'0';
      REP(d, lim+1) {
        int nk = k + (d!=0?1:0);
        if(nk > 3) continue;
        (dp[i+1][j || d<lim][nk] += dp[i][j][k]);
      }
    }

    ll ans = 0;
    REP(i, 2) REP(j, 4) ans += dp[n][i][j];
    return ans;
  };

  ll test;
  cin >> test;
  REP(tes, test) {
    ll l, r;
    cin >> l >> r;
    string s = to_string(r), t = to_string(l-1);
    cout << func(s) - func(t) << endl;
  }

  return 0;
}