Submission #20736872


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#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
//#pragma GCC optimize ("-O3")
using namespace std;
void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
//---------------------------------------------------------------------------------------------------
template<int MOD> struct ModInt {
    static const int Mod = MOD; unsigned x; ModInt() : x(0) { }
    ModInt(signed sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    ModInt(signed long long sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    int get() const { return (int)x; }
    ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
    ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
    ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
    ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
    ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
    ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
    ModInt inverse() const { long long a = x, b = MOD, u = 1, v = 0;
        while (b) { long long t = a / b; a -= t * b; std::swap(a, b); u -= t * v; std::swap(u, v); }
        return ModInt(u); }
    bool operator==(ModInt that) const { return x == that.x; }
    bool operator!=(ModInt that) const { return x != that.x; }
    ModInt operator-() const { ModInt t; t.x = x == 0 ? 0 : Mod - x; return t; }
};
template<int MOD> ostream& operator<<(ostream& st, const ModInt<MOD> a) { st << a.get(); return st; };
template<int MOD> ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
    ModInt<MOD> r = 1; while (k) { if (k & 1) r *= a; a *= a; k >>= 1; } return r; }
typedef ModInt<1000000007> mint;
/*---------------------------------------------------------------------------------------------------
            ∧_∧
      ∧_∧  (´<_` )  Welcome to My Coding Space!
     ( ´_ゝ`) /  ⌒i     @hamayanhamayan0
    /   \     | |
    /   / ̄ ̄ ̄ ̄/  |
  __(__ニつ/     _/ .| .|____
     \/____/ (u ⊃
---------------------------------------------------------------------------------------------------*/














string N; int K;
map<char, int> dic;
mint dp[201010][18][2][2]; // dp[digit][kind][isless][leadingzero]
//---------------------------------------------------------------------------------------------------
mint solve() {
    bitset<16> used;
    int len = N.length();

    dp[0][0][0][1] = 1;
    rep(digit, 0, len) {
        int x = dic[N[digit]];

        rep(kind, 0, 17) rep(isless, 0, 2) rep(leadingzero, 0, 2) {
            mint& cur = dp[digit][kind][isless][leadingzero];

            if (isless) {
                // 既にNより小さい場合において、既に選択された数字を選ぶ場合の遷移
                dp[digit + 1][kind][isless][leadingzero] += cur * kind;

                if (!leadingzero) {
                    // 既にNより小さい場合、かつ、leadingzeroじゃない場合において、新しい数を選ぶ場合
                    dp[digit + 1][kind + 1][isless][leadingzero] += cur * (16 - kind);
                }
                else {
                    // leadingzeroの場合は種類数をカウントしない(leadingzeroは種類にカウントしない)ためkind=0であるはず

                    // 既にNより小さい場合、かつ、leadingzeroである場合において、0を選択する場合(leadingzeroを継続する場合)
                    dp[digit + 1][kind][isless][1] += cur;

                    // 既にNより小さい場合、かつ、leadingzeroである場合において、0以外を選択する場合
                    dp[digit + 1][kind + 1][isless][0] += cur * (16 - kind - 1);
                }
            }
            else {
                // まだNより小さくない(Nと等しい)場合は、小さくなるように数を選択する
                rep(nxt, 0, x) {
                    // 先頭で0を選ぶ場合のみleading-zeroになるので、そちらに遷移をさせる
                    if (digit == 0 && nxt == 0) dp[digit + 1][0][1][1] += cur;
                    else {
                        // それ以外の場合は、種類数が増えるかを確認して、遷移を行う
                        bitset<16> used2 = used;
                        used2.set(nxt);
                        dp[digit + 1][used2.count()][1][0] += cur;
                    }
                }
            }
        }

        // こちらの遷移は先頭digit桁がNと同じになるように遷移させるもの
        int pre = used.count();
        used.set(x);
        int post = used.count();
        if (digit == 0) dp[digit + 1][post][0][0] += dp[digit][pre][0][1];
        else dp[digit + 1][post][0][0] += dp[digit][pre][0][0];
    }

    mint ans = dp[len][K][0][0]; // Nと全く同じパターン。K種類であればここで計算される
    ans += dp[len][K][1][0];     // N未満でK種類のもの
    return ans;
}
//---------------------------------------------------------------------------------------------------
void _main() {
    rep(i, 0, 10) dic[char('0' + i)] = i;
    rep(i, 0, 6) dic[char('A' + i)] = 10 + i;

    cin >> N >> K;
    cout << solve() << endl;
}




Submission Info

Submission Time
Task F - Digits Paradise in Hexadecimal
User hamayanhamayan
Language C++ (GCC 9.2.1)
Score 600
Code Size 6151 Byte
Status AC
Exec Time 321 ms
Memory 60324 KB

Judge Result

Set Name Sample All
Score / Max Score 0 / 0 600 / 600
Status
AC × 5
AC × 33
Set Name Test Cases
Sample sample_01.txt, sample_02.txt, sample_03.txt, sample_04.txt, sample_05.txt
All handmade_00.txt, handmade_01.txt, handmade_02.txt, handmade_03.txt, random_00.txt, random_01.txt, random_02.txt, random_03.txt, random_04.txt, random_05.txt, random_06.txt, random_07.txt, random_08.txt, random_09.txt, random_10.txt, random_11.txt, random_12.txt, random_13.txt, random_14.txt, random_15.txt, random_16.txt, random_17.txt, random_18.txt, random_19.txt, random_20.txt, random_21.txt, random_22.txt, random_23.txt, sample_01.txt, sample_02.txt, sample_03.txt, sample_04.txt, sample_05.txt
Case Name Status Exec Time Memory
handmade_00.txt AC 50 ms 60152 KB
handmade_01.txt AC 46 ms 60120 KB
handmade_02.txt AC 39 ms 60116 KB
handmade_03.txt AC 40 ms 60188 KB
random_00.txt AC 321 ms 60324 KB
random_01.txt AC 321 ms 60120 KB
random_02.txt AC 317 ms 60324 KB
random_03.txt AC 320 ms 60284 KB
random_04.txt AC 320 ms 60268 KB
random_05.txt AC 318 ms 60176 KB
random_06.txt AC 320 ms 60216 KB
random_07.txt AC 318 ms 60324 KB
random_08.txt AC 318 ms 60280 KB
random_09.txt AC 320 ms 60312 KB
random_10.txt AC 318 ms 60268 KB
random_11.txt AC 317 ms 60312 KB
random_12.txt AC 94 ms 60248 KB
random_13.txt AC 167 ms 60196 KB
random_14.txt AC 239 ms 60232 KB
random_15.txt AC 303 ms 60172 KB
random_16.txt AC 103 ms 60256 KB
random_17.txt AC 129 ms 60164 KB
random_18.txt AC 241 ms 60232 KB
random_19.txt AC 268 ms 60160 KB
random_20.txt AC 51 ms 60124 KB
random_21.txt AC 171 ms 60308 KB
random_22.txt AC 194 ms 60216 KB
random_23.txt AC 313 ms 60272 KB
sample_01.txt AC 48 ms 60044 KB
sample_02.txt AC 40 ms 60104 KB
sample_03.txt AC 42 ms 60100 KB
sample_04.txt AC 45 ms 60104 KB
sample_05.txt AC 39 ms 60180 KB