#684450 (C++14) No.1631 Sorting Integers (Multiple of K) Easy

• 問題: No.1631 Sorting Integers (Multiple of K) Easy
• ユーザー: nawawan
• 提出日時: 2021-07-30 21:48:53
• 言語: C++14 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 2,461 ms / 3,000 ms
• コード長: 3,210 bytes
• コンパイル時間: 1,677 ms
• コンパイル使用メモリ: 175,136 KB
• 実行使用メモリ: 131,328 KB
• 最終ジャッジ日時: 2024-09-15 23:45:06
• 合計ジャッジ時間: 37,777 ms
• ジャッジサーバーID （参考情報）: judge2 / judge4

ファイルパターン	結果
sample	AC * 4
other	AC * 28

ソースコード

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#include <bits/stdc++.h>
using namespace std;
struct modint{
    long long val;
    modint(): val(0) {}
    modint(long long x){
        if(x < 0) val = x % mod() + mod();
        else val = x % mod();
    }
    modint(const modint &t){
        val = t.val;
    }
    static int &mod(){
        static int m = 0;
        return m;
    }
    static void set_mod(int md){
        mod() = md;
    }
    modint& operator =(const modint m){
        val = m.val;
        return *this;
    }
    modint operator -(){
        return modint(-val);
    }
    modint& operator-=(const modint &m){
        val -= m.val;
        if(val < 0) val += mod();
        return *this;
    }
    modint& operator+=(const modint &m){
        val += m.val;
        if(val >= mod()) val -= mod();
        return *this;
    }
    modint& operator*=(const modint &m){
        val *= m.val;
        val %= mod();
        return *this;
    }
    modint& operator/=(modint m){
        *this *= m.inv();
        return *this;
    }
    modint inv(){
        long long x = 1, y = 0;
        long long a = val, b = mod();
        while(b != 0){
            long long t = a / b;
            a -= t * b;
            x -= t * y;
            swap(a, b);
            swap(x, y);
        }
        x %= mod();
        if(x < 0) x += mod();
        return modint(x);
    }
    modint pow(long long k){
        long long res = 1;
        long long v = val;
        while(k > 0){
            if(k & 1) res = res * v % mod();
            v = v * v % mod();
            k >>= 1;
        }
        return modint(res);
    }
    bool operator==(const modint &m){
        return val == m.val;
    }
    modint operator+(const modint &m){
        return modint(*this) += m;
    }
    modint operator-(const modint &m){
        return modint(*this) -= m;
    }
    modint operator*(const modint &m){
        return modint(*this) *= m;
    }
    modint operator/(const modint &m){
        return modint(*this) /= m;
    }
    bool operator!=(const modint &m){
        return modint(*this).val != m.val;
    }
    bool operator!=(const int &m){
        return modint(*this).val != m;
    }
};
using mint = modint;
int main(){
    int N, K;
    cin >> N >> K;
    modint::set_mod(K); 
    vector<int> c(9);
    for(int i = 0; i < 9; i++) cin >> c[i];
    vector<mint> ten(N + 1);
    ten[0] = 1;
    for(int i = 0; i < N; i++) ten[i + 1] = ten[i] * 10;
    vector<vector<long long>> dp((1 << N), vector<long long>(K));
    dp[0][0] = 1;
    vector<int> num(N);
    int id = 0;
    for(int i = 0; i < 9; i++){
        for(int j = 0; j < c[i]; j++){
            num[id] = i + 1;
            id++;
        }
    }
    for(int bit = 0; bit < (1 << N); bit++){
        int cnt = __builtin_popcount(bit);
        for(int j = 0; j < N; j++){
            if((bit & (1 << j)) == 0){
                for(int k = 0; k < K; k++){
                    mint te = ten[N - 1 - cnt] * num[j] + k;
                    dp[bit ^ (1 << j)][te.val] += dp[bit][k];
                }
            }
        }
    }
    for(int i = 0; i < 9; i++){
        for(int j = 0; j < c[i]; j++){
            dp[(1 << N) - 1][0] /= (j + 1);
        }
    }
    cout << dp[(1 << N) - 1][0] << endl;
}
 
 
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