#1192400 (C++17) No.3762 Glowing Utility Pole

• 問題: No.3762 Glowing Utility Pole
• ユーザー: 👑 potato167
• 提出日時: 2026-10-09 22:28:03
• 言語: C++17 (gcc 15.3.0 + boost 1.92.0 + ACL)
• 結果: AC
• 実行時間: 1,981 ms / 2,000 ms
• コード長: 8,328 bytes
• コンパイル時間: 3,111 ms
• コンパイル使用メモリ: 284,748 KB
• 実行使用メモリ: 34,160 KB
• 最終ジャッジ日時: 2026-10-09 22:28:31
• 合計ジャッジ時間: 16,540 ms
• ジャッジサーバーID （参考情報）: judge5_1 / judge1_1

ファイルパターン	結果
sample	AC * 3
other	AC * 47

ソースコード

#line 1 "b.cpp"
# pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
const ll ILL=2167167167167167167;
const int INF=2100000000;
#define rep(i,a,b) for (int i=(int)(a);i<(int)(b);i++)
#define all(p) p.begin(),p.end()
template<class T> using pq_ = priority_queue<T, vector<T>, greater<T>>;
template<class T> int LB(vector<T> &v,T a){return lower_bound(v.begin(),v.end(),a)-v.begin();}
template<class T> int UB(vector<T> &v,T a){return upper_bound(v.begin(),v.end(),a)-v.begin();}
template<class T> bool chmin(T &a,T b){if(b<a){a=b;return 1;}else return 0;}
template<class T> bool chmax(T &a,T b){if(a<b){a=b;return 1;}else return 0;}
template<class T> void So(vector<T> &v) {sort(v.begin(),v.end());}
template<class T> void Sore(vector<T> &v) {sort(v.begin(),v.end(),[](T x,T y){return x>y;});}
bool yneos(bool a,bool upp=false){if(a){cout<<(upp?"YES\n":"Yes\n");}else{cout<<(upp?"NO\n":"No\n");}return a;}
template<class T> void vec_out(vector<T> &p,int ty=0){
    if(ty==2){cout<<'{';for(int i=0;i<(int)p.size();i++){if(i){cout<<",";}cout<<'"'<<p[i]<<'"';}cout<<"}\n";}
    else{if(ty==1){cout<<p.size()<<"\n";}for(int i=0;i<(int)(p.size());i++){if(i) cout<<" ";cout<<p[i];}cout<<"\n";}}
template<class T> T vec_min(vector<T> &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmin(ans,x);return ans;}
template<class T> T vec_max(vector<T> &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmax(ans,x);return ans;}
template<class T> T vec_sum(vector<T> &a){T ans=T(0);for(auto &x:a) ans+=x;return ans;}
int pop_count(long long a){int res=0;while(a){res+=(int)(a&1),a>>=1;}return res;}
template<class T> T square(T a){return a * a;}

#include <atcoder/convolution>
using mint = atcoder::modint998244353;
#line 2 "/Users/Shared/po167_library/math/Binomial.hpp"

#line 5 "/Users/Shared/po167_library/math/Binomial.hpp"

namespace po167{
template<class T>
struct Binomial{
    std::vector<T> fact_vec, fact_inv_vec;
    void extend(int m = -1){
        int n = fact_vec.size();
        if (m == -1) m = n * 2;
        if (n >= m) return;
        fact_vec.resize(m);
        fact_inv_vec.resize(m);
        for (int i = n; i < m; i++){
            fact_vec[i] = fact_vec[i - 1] * T(i);
        }
        fact_inv_vec[m - 1] = T(1) / fact_vec[m - 1];
        for (int i = m - 1; i > n; i--){
            fact_inv_vec[i - 1] = fact_inv_vec[i] * T(i);
        }
    }
    Binomial(int MAX = 0){
        fact_vec.resize(1, T(1));
        fact_inv_vec.resize(1, T(1));
        extend(MAX + 1);
    }

    T fact(int i){
        if (i < 0) return 0;
        while (int(fact_vec.size()) <= i) extend();
        return fact_vec[i];
    }
    T invfact(int i){
        if (i < 0) return 0;
        while (int(fact_inv_vec.size()) <= i) extend();
        return fact_inv_vec[i];
    }
    T C(int a, int b){
        if (a < b || b < 0) return 0;
        return fact(a) * invfact(b) * invfact(a - b);
    }
    T invC(int a, int b){
        if (a < b || b < 0) return 0;
        return fact(b) * fact(a - b) *invfact(a);
    }
    T P(int a, int b){
        if (a < b || b < 0) return 0;
        return fact(a) * invfact(a - b);
    }
    T inv(int a){
        if (a < 0) return inv(-a) * T(-1);
        if (a == 0) return 1;
        return fact(a - 1) * invfact(a);
    }
    T Catalan(int n){
        if (n < 0) return 0;
        return fact(2 * n) * invfact(n + 1) * invfact(n);
    }
    T narayana(int n, int k){
        if (n <= 0 || n < k || k < 1) return 0;
        return C(n, k) *  C(n, k - 1) * inv(n);
    }
    T Catalan_pow(int n,int d){
        if (n < 0 || d < 0) return 0;
        if (d == 0){
            if (n == 0) return 1;
            return 0;
        }
        return T(d) * inv(d + n) * C(2 * n + d - 1, n);
    }
    // retrun [x^a] 1/(1-x)^b
    T ruiseki(int a,int b){
        if (a < 0 || b < 0) return 0;
        if (a == 0){
            return 1;
        }
        return C(a + b - 1, b - 1);
    }
    // (a, b) -> (c, d)
    // always x + e >= y
    T mirror(int a, int b, int c, int d, int e = 0){
        if (a + e < b || c + e < d) return 0;
        if (a > c || b > d) return 0;
        a += e;
        c += e;
        return C(c + d - a - b, c - a) - C(c + d - a - b, c - b + 1); 
    }
    // return sum_{i = 0, ... , a} sum_{j = 0, ... , b} C(i + j, i)
    // return C(a + b + 2, a + 1) - 1;
    T gird_sum(int a, int b){
        if (a < 0 || b < 0) return 0;
        return C(a + b + 2, a + 1) - 1;
    }
    // return sum_{i = a, ..., b - 1} sum_{j = c, ... , d - 1} C(i + j, i)
    // AGC 018 E
    T gird_sum_2(int a, int b, int c, int d){
        if (a >= b || c >= d) return 0;
        a--, b--, c--, d--;
        return gird_sum(a, c) - gird_sum(a, d) - gird_sum(b, c) + gird_sum(b, d);
    }

    // the number of diagonal dissections of a convex n-gon into k+1 regions.
    // OEIS A033282
    // AGC065D
    T diagonal(int n, int k){
        if (n <= 2 || n - 3 < k || k < 0) return 0;
        return C(n - 3, k) * C(n + k - 1, k) * inv(k + 1);
    }
};
}
#line 29 "b.cpp"

void solve();
// DEAR MYSTERIES / TOMOO
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t = 1;
    // cin >> t;
    rep(i, 0, t) solve();
}

void solve(){
    po167::Binomial<mint> table;
    int N, M;
    cin >> N >> M;
    vector<int> C(N);
    int allQ = 0;
    rep(i, 0, N) cin >> C[i], C[i]--;
    for (auto x : C) if (x == -1) allQ++;
    if (M == 1) {
        cout << ((mint(N) * (mint(N + 1))) / 2).val() << "\n";
        return;
    }
    mint ans = 0;
    vector pw(M + 1, vector<mint>(N + 1));
    rep(i, 0, M + 1) {
        pw[i][0] = 1;
        rep(j, 0, N) pw[i][j + 1] = pw[i][j] * i;
    }
    auto f = [&](auto self, int l, int r) -> void {
        if (l + 1 == r) return;
        int m = (l + r) / 2;
        self(self, l, m);
        self(self, m, r);
        // cout << l << " " << r << " " << ans.val() << "\n";
        vector<pair<int, int>> p, q;
        {
            int a = 0, b = 0;
            rep(i, m, r) {
                if (C[i] == -1) b++;
                else a = (a | (1 << C[i]));
                p.push_back({a, b});
            }
        }
        {
            int a = 0, b = 0;
            for (int i = m - 1; i >= l; i--) {
                if (C[i] == -1) b++;
                else a = (a | (1 << C[i]));
                q.push_back({a, b});
            }
        }
        for (int a = 0, b = 0; a < (int)p.size(); a = b) {
            ll baseC = p[a].first;
            ll baseQ = p[a].second;
            vector<mint> tmp1;
            while (b != (int)p.size() && p[b].first == baseC) {
                if ((int)tmp1.size() == p[b].second - p[a].second) {
                    tmp1.push_back(0);
                }
                tmp1.back() += 1;
                b++;
            }
            for (int c = 0, d = 0; c < (int)q.size(); c = d) {
                ll baseC2 = (q[c].first | baseC);
                ll baseQ2 = q[c].second;
                vector<mint> tmp2;
                while (d != (int)q.size() && (baseC | q[d].first) == baseC2) {
                    if ((int)tmp2.size() == q[d].second - q[c].second) {
                        tmp2.push_back(0);
                    }
                    tmp2.back() += 1;
                    d++;
                }
                auto tmp = atcoder::convolution(tmp1, tmp2);
                int remC = M - pop_count(baseC2 | baseC);
                rep(i, 0, tmp.size()) {
                    // cout << i + baseQ + baseQ2 << " " << tmp[i].val() << " " << remC << " " << ans.val() << endl;
                    rep(k, 0, remC + 1) {
                        ans += table.C(remC, k) * (k & 1 ? -1 : 1) * tmp[i] * pw[M - k][baseQ + baseQ2 + i] * pw[M][allQ - baseQ - baseQ2 - i];
                    }
                }
            }
        }
    };
    f(f, 0, N);
    cout << ans.val() << "\n";
}

/*
 * ある区間について、? が K 個、使われていない色の数が c としたとき、
 * 答えは包除原理でいける
 * これを列挙しよう？
 * 無理そうなので、別の問題を考える
 * ある色の集合を ban した時の　valid な区間の数を考える
 * 先に包除原理をするパターン
 * これは、行列累乗で求められる
 *
 *
 * 分割統治を考えるか
 * いけそう(?)
 *
 */