結果

問題 No.3505 Sum of Prod of Root
コンテスト
ユーザー 👑 potato167
提出日時 2026-04-17 22:02:12
言語 C++17
(gcc 15.2.0 + boost 1.89.0)
コンパイル:
g++-15 -O2 -lm -std=c++17 -Wuninitialized -DONLINE_JUDGE -o a.out _filename_
実行:
./a.out
結果
AC  
実行時間 364 ms / 2,000 ms
コード長 14,236 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 3,956 ms
コンパイル使用メモリ 292,608 KB
実行使用メモリ 33,264 KB
最終ジャッジ日時 2026-04-17 22:02:28
合計ジャッジ時間 6,063 ms
ジャッジサーバーID
(参考情報)
judge2_0 / judge1_0
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 1
other AC * 13
権限があれば一括ダウンロードができます

ソースコード

diff #
raw source code

#line 1 "d.cpp"
#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;}

#line 3 "/Users/Shared/po167_library/fps/FPS_Product_Sequence.hpp"
#include <atcoder/convolution>

namespace po167{
template<class T>
std::vector<T> FPS_Product_Sequence(std::vector<std::vector<T>> f){
    if (f.empty()) return {1};
    auto op = [&](auto self,int l, int r) -> std::vector<T> {
        if (l + 1 == r) return f[l];
        int m = (l + r) / 2;
        return atcoder::convolution(self(self, l, m), self(self, m, r));
    };
    return op(op, 0, f.size());
}
}
#line 4 "/Users/Shared/po167_library/fps/FPS_inv.hpp"

namespace po167{
// return 1 / f
template <class T>
std::vector<T> FPS_inv(std::vector<T> f, int len = -1){
    if (len == -1) len = f.size();
    assert(f[0] != 0);
    std::vector<T> g = {1 / f[0]};
    int s = 1;
    while(s < len){
        // g = 2g_s - f(g_s)^2 (mod x ^ (2 * s))
        // g = g - (fg - 1)g
        // (fg - 1) = 0 (mod x ^ (s))
        std::vector<T> n_g(s * 2, 0);
        std::vector<T> f_s(s * 2, 0);
        g.resize(s * 2);
        for (int i = 0; i < s * 2; i++){
            if (int(f.size()) > i) f_s[i] = f[i];
            n_g[i] = g[i];
        }
        atcoder::internal::butterfly(g);
        atcoder::internal::butterfly(f_s);
        for (int i = 0; i < s * 2; i++){
            f_s[i] *= g[i];
        }
        atcoder::internal::butterfly_inv(f_s);
        T iz = 1 / (T)(s * 2);
        for (int i = s; i < s * 2; i++){
            f_s[i] *= iz;
        }
        for (int i = 0; i < s; i++){
            f_s[i] = 0;
        }
        atcoder::internal::butterfly(f_s);
        for (int i = 0; i < s * 2; i++){
            f_s[i] *= g[i];
        }
        atcoder::internal::butterfly_inv(f_s);
        for (int i = s; i < s * 2; i++){
            n_g[i] -= f_s[i] * iz;
        }
        std::swap(n_g, g);
        s *= 2;
    }
    g.resize(len);
    return g;
}
}
#line 5 "/Users/Shared/po167_library/fps/Multipoint_Evaluation.hpp"

namespace po167{
// return {f(p[0]), f(p[1]), f(p[2]), ... }
// https://codeforces.com/blog/entry/100279
template<class T>
std::vector<T> Multipoint_Evaluation(std::vector<T> f, std::vector<T> p){
    std::reverse(f.begin(), f.end());
    int M = f.size();
    int N = p.size();
    std::vector<std::vector<T>> g(N);
    for (int i = 0; i < N; i++){
        g[i] = {1, p[i] * -1};
    }
    auto tmp = po167::FPS_Product_Sequence(g);
    tmp = po167::FPS_inv(tmp, M);
    tmp = atcoder::convolution(tmp, f);
    tmp.resize(M);
    int size = 1;
    while (size < N) size *= 2;
    std::vector<std::vector<T>> seg(size * 2, {1});
    for (int i = 0; i < N; i++){
        seg[i + size] = g[i];
    }
    for (int i = size - 1; i > 0; i--){
        seg[i] = atcoder::convolution(seg[i * 2], seg[i * 2 + 1]);
    }

    std::vector<T> inv = {1, (T)(1) / (T)(2)};
    // a, b の convolution のうち、l, r の間だけ欲しい
    auto calc = [&](std::vector<T> a, std::vector<T> b, int l, int r) -> std::vector<T> {
        if (l == r) return {};
        if (l < 0) l = 0;
        while ((int)a.size() > r) a.pop_back();
        while ((int)b.size() > r) b.pop_back();
        int lim = (int)a.size() + (int)b.size() - 1;
        lim -= std::min(lim - r, l);
        int z = 0;
        while ((1 << z) < lim) z++;
        a.resize((1 << z), 0);
        b.resize((1 << z), 0);
        atcoder::internal::butterfly(a);
        atcoder::internal::butterfly(b);
        for (int i = 0; i < (1 << z); i++) a[i] *= b[i];
        atcoder::internal::butterfly_inv(a);
        while ((int)inv.size() <= z){
            inv.push_back(inv.back() * inv[1]);
        }
        std::vector<T> d(r - l);
        for (int i = l; i < r; i++) d[i - l] = a[i] * inv[z];
        return d;
    };

    std::vector<T> res(N);
    auto rec = [&](auto self, int ind, std::vector<T> v) -> void {
        if ((int)seg[ind].size() == 1){
            return;
        }
        if (size <= ind){
            res[ind - size] = v.back();
            return;
        }
        self(self, ind * 2, calc(v, seg[ind * 2 + 1], (int)(v.size() - seg[ind * 2].size() + 1) ,(int)v.size()));
        self(self, ind * 2 + 1, calc(v, seg[ind * 2], (int)(v.size() - seg[ind * 2 + 1].size() + 1) ,(int)v.size()));
    };
    rec(rec, 1, tmp);
    return res;
}
}
#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 5 "/Users/Shared/po167_library/fps/Polynomial_Interpolation.hpp"

namespace po167{
template<class T>
// ラグランジュの多項式補完
// f(X[i]) = Y[i] である f を返す
std::vector<T> Polynomial_Interpolation(std::vector<int> X, std::vector<T> Y){
    int N = X.size();
    assert(Y.size() == X.size());
    if (N == 0) return {};
    {
        auto Z = X;
        std::sort(Z.begin(), Z.end());
        for (int i = 0; i < N - 1; i++){
            assert(Z[i] != Z[i + 1]);
        }
    }
    std::vector<std::vector<T>> p(N);
    for (int i = 0; i < N; i++) p[i] = {-X[i], 1};
    auto g = FPS_Product_Sequence(p);
    for (int i = 0; i < N; i++){
        g[i] = g[i + 1] * (i + 1);
    }
    g.pop_back();
    std::vector<T> xt(N);
    for (int i = 0; i < N; i++) xt[i] = X[i];
    auto Z = Multipoint_Evaluation(g, xt);
    std::vector<T> inv = {1, (T)(1) / (T)(2)};
    auto rec = [&](auto self, int l, int r) -> std::pair<std::vector<T>, std::vector<T>> {
        if (l + 1 == r){
            return {{Y[l] / Z[l]}, {-X[l], 1}};
        }
        int m = (l + r) / 2;
        auto [Lf, Ls] = self(self, l, m);
        auto [Rf, Rs] = self(self, m, r);
        int mx_size = r - l + 1;
        if (mx_size < 128){
            auto D = atcoder::convolution(Ls, Rs);
            auto U = atcoder::convolution(Ls, Rf);
            FPS_add(U, atcoder::convolution(Lf, Rs));
            return {U, D};
        }
        int z = 0;
        while ((1 << z) < mx_size) z++;
        while (int(inv.size()) <= z) inv.push_back(inv.back() * inv[1]);
        Lf.resize(1 << z, 0);
        Ls.resize(1 << z, 0);
        Rf.resize(1 << z, 0);
        Rs.resize(1 << z, 0);
        atcoder::internal::butterfly(Lf);
        atcoder::internal::butterfly(Ls);
        atcoder::internal::butterfly(Rf);
        atcoder::internal::butterfly(Rs);
        std::vector<T> D(1 << z), U(1 << z);
        for (int i = 0; i < (1 << z); i++){
            D[i] = Ls[i] * Rs[i] * inv[z];
            U[i] = (Ls[i] * Rf[i] + Lf[i] * Rs[i]) * inv[z];
        }
        atcoder::internal::butterfly_inv(D);
        atcoder::internal::butterfly_inv(U);
        D.resize(r - l + 1);
        U.resize(r - l);
        return {U, D};
    };
    return rec(rec, 0, N).first;
}
}
#line 26 "d.cpp"
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 28 "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(){
    ll N;
    cin >> N;
    vector<pair<ll, ll>> p;
    rep(k, 3, 61) {
        rep(v, 2, INF) {
            __int128_t tmp = 1;
            rep(rp, 0, k) tmp *= v;
            if (tmp > N) break;
            p.push_back({tmp, k});
        }
    }
    N++;
    po167::Binomial<mint> table;
    vector<int> now_val(66, 1);
    mint tmp = 1, ans = 0;
    auto f = [&](ll l, ll r) -> mint {
        mint res = 1;
        res *= r - l;
        res *= (r + l - 1);
        res *= table.inv(2);
        return res;
    };
    const int L = 6;
    vector<mint> base(L);
    {
        vector<int> X(L);
        vector<mint> Y(L);
        rep(i, 1, L) {
            X[i] = i;
            Y[i] = Y[i - 1] + f(i * i, (i + 1) * (i + 1)) * i;
        }
        base = po167::Polynomial_Interpolation(X, Y);
        // cout << base.size() << endl;
    }
    auto g = [&](ll r) -> mint {
        ll l_val = 1, r_val = INF;
        while (r_val - l_val > 1) {
            ll mid = (l_val + r_val) / 2;
            if (mid * mid <= r) l_val = mid;
            else r_val = mid;
        }
        mint res = 0;
        {
            mint tmpG = 1;
            rep(i, 0, L) {
                res += base[i] * tmpG;
                tmpG *= l_val - 1;
            }
        }
        res += f(l_val * l_val, r) * l_val;
        // cout << r << " " << l_val << " " << res.val() << endl;
        return res;
    };
    So(p);
    p.push_back({N, 1});
    ll l = 1;
    for (auto [a, id] : p) {
        ans += tmp * (g(a) - g(l));
        // cout << l << " " << a << " " << tmp.val() << "\n";
        l = a;
        tmp *= table.inv(now_val[id]);
        tmp *= now_val[id] + 1;
        now_val[id]++;
    }
    cout << ans.val() << "\n";
}

/*
 *
 * k >= 3 を全て列挙する
 * そうすると、境目ができて、
 * その中で、1 <= k <= 2 バージョンが解ければいいということになる
 * 結局 1 <= k <= 2 の場合を高速に解ければいい
 * sqrt(x) = y の範囲だと?
 * y * (y ^2 + (y + 1)^2 - 1) * (2 * y - 1) / 2
 * y * (y^2 + y) * (2*y - 1)
 * これのアレなので、
 * yosupo judge にだした問題になる
 * なんで整備していないのかな?
 * 多項式補完します()
 */
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