#1196075 (C++17) No.3785 Tree of Multiples

• 問題: No.3785 Tree of Multiples
• ユーザー: 👑 potato167
• 提出日時: 2026-10-30 22:21:57
• 言語: C++17 (gcc 15.3.0 + boost 1.92.0 + ACL)
• 結果: WA
• 実行時間: -
• コード長: 8,286 bytes
• コンパイル時間: 3,512 ms
• コンパイル使用メモリ: 290,816 KB
• 実行使用メモリ: 29,232 KB
• 最終ジャッジ日時: 2026-10-30 22:22:05
• 合計ジャッジ時間: 7,746 ms
• ジャッジサーバーID （参考情報）: judge3_0 / judge1_0

ファイルパターン	結果
sample	AC * 3
other	AC * 15 WA * 1

ソースコード

#line 1 "d.cpp"
// https://github.com/potato167/po167_library
#include <bits/stdc++.h>
#line 4 "d.cpp"
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 3 "/Users/Shared/po167_library/fps/FPS_add.hpp"

namespace po167{
template <class T>
// a(x) += b(x) * c * x^d
void FPS_add(std::vector<T> &a, std::vector<T> b, T c = 1, int d = 0){
    for (int i = 0; i < (int)(b.size()); i++){
        while ((int)a.size() <= i + d) a.push_back((T)0);
        a[i + d] += b[i] * c;
    }
}
}
#line 31 "d.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(){
    int N, M;
    cin >> N >> M;
    po167::Binomial<mint> table;
    vector<pair<int, int>> p(N), q(M);
    int sumP = 0, sumQ = 0;
    rep(i, 0, N) cin >> p[i].first >> p[i].second, sumP += p[i].second;
    rep(i, 0, M) cin >> q[i].first >> q[i].second, sumQ += q[i].second;
    So(p), So(q);
    vector<mint> ok = {1}, ng = {0};
    vector<tuple<int, int, int>> A;
    {
        int j = 0;
        rep(i, 0, p.size()) {
            while (j != (int)q.size() && q[j].first < p[i].first) {
                A.push_back({q[j].first, q[j].second, 0});
                j++;
            }
            if (j == (int)q.size() || q[j].first != p[i].first) {
                A.push_back({p[i].first, 0, p[i].second});
            }
            else {
                A.push_back({p[i].first, q[j].second, p[i].second});
                j++;
            }
        }
    }
    vector<vector<mint>> X(A.size()), Y(A.size());
    rep(i, 0, A.size()) {
        auto [w, a, b] = A[i];
        vector<mint> okz(min(a, b) + 1), ngz(min(a, b) + 1);
        mint pw = 1;
        rep(j, 0, min(a, b) + 1) {
            okz[j] = pw * table.invfact(j) * table.invfact(a - j) * table.invfact(b - j);
            ngz[j] = pw * table.invfact(j) * table.invfact(a - j - 1) * table.invfact(b - j - 1);
            pw *= w;
        }
        /*
        ng = atcoder::convolution(ng, okz);
        po167::FPS_add(ng, atcoder::convolution(ok, ngz));
        ok = atcoder::convolution(ok, okz);*/
        X[i] = okz;
        Y[i] = ngz;
    }
    auto f = [&](auto self, int l, int r) -> pair<vector<mint>, vector<mint>> {
        if (l + 1 == r) return {X[l], Y[l]};
        int m = (l + r) / 2;
        auto [okL, ngL] = self(self, l, m);
        auto [okR, ngR] = self(self, m, r);
        auto tmp = atcoder::convolution(okL, ngR);
        po167::FPS_add(tmp, atcoder::convolution(okR, ngL));
        return {atcoder::convolution(okL, okR), tmp};
    };
    tie(ok, ng) = f(f, 0, A.size());
    mint ans = 0;
    rep(i, 0, ok.size()) {
        ans += (ok[i] - ng[i] * table.inv(sumP - i) * table.inv(sumQ - i)) * table.fact(sumP - i) * table.fact(sumQ - i) * table.fact(i);
    }
    cout << ans.val() << "\n";
}

/*
 * これ一意なの？
 * 頂点数が最小というのがありました
 * とりあえずサンプルを見てみると、因数分解の順序ということになっていそう
 * score が z のものの、必要条件を考える
 * 頭が z の並べ方
 * 後ろが、先頭は不一致になるように並べる方法
 * 後ろが a, b 個ずつだとして、
 * a! / prod(p[i]!) * b! / prod(q[i]!)
 * i が先頭のとき
 * (p[i] / a) * (q[i] / b) だけダメである
 *
 */