結果

問題 No.1781 LCM
ユーザー shobonvipshobonvip
提出日時 2024-09-19 00:47:15
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2,167 ms / 5,000 ms
コード長 4,148 bytes
コンパイル時間 3,253 ms
コンパイル使用メモリ 261,012 KB
実行使用メモリ 23,036 KB
最終ジャッジ日時 2024-09-19 00:47:37
合計ジャッジ時間 17,049 ms
ジャッジサーバーID
(参考情報)
judge5 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 31
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 #
プレゼンテーションモードにする

#include<bits/stdc++.h>
using namespace std;
// [floor_sqrt(N)]
// 0 <= N <= 9e18
long long floor_sqrt(long long N) {
long long ub = 3'000'000'001;
long long lb = 0;
while (ub - lb > 1) {
long long t = (ub + lb) / 2;
if (t * t <= N) lb = t;
else ub = t;
}
return lb;
}
// Quotients
struct Quotients {
long long sqrtN;
long long N;
long long M;
std::vector<long long> data;
Quotients(long long _N): N(_N) {
sqrtN = floor_sqrt(N);
data = std::vector<long long>(sqrtN);
for (int i=1; i<=sqrtN; i++) {
data[i-1] = i;
}
for (int i=sqrtN-1; i>=0; i--){
if (N / data[i] == data[i]) continue;
data.emplace_back(N / data[i]);
}
M = data.size();
}
int size() {
return M;
}
long long operator[](int i) {
if (i < 0 || i >= M){
throw std::out_of_range("Index out of range");
}
return data[i];
}
int ind(long long x) {
if (x <= sqrtN) return x - 1;
return M - N / x;
}
std::pair<long long, long long> interval(long long v) {
return std::pair(N / (v+1) + 1, N / v + 1);
}
};
// Lucy DP
// returns (dp, prime_count)
template <typename T, T (*g)(long long), T (*G)(long long)>
std::pair<std::vector<T>, std::vector<long long>> lucy_dp(long long N) {
Quotients Q(N);
int m = Q.size();
std::vector<T> dp(m);
std::vector<long long> dp2(m);
for (int i=0; i<m; i++){
dp[i] = G(Q[i]) - g(1);
dp2[i] = Q[i] - 1;
}
for (int x=2; x<=Q.sqrtN; x++){
if (dp2[x-1] > dp2[x-2]) {
long long border = (long long)x * x;
for (int i=m-1; i>=0; i--){
if (Q[i] < border) break;
int tmpind = Q.ind(Q[i]/x);
dp[i] -= g(x) * (dp[tmpind] - dp[x-2]);
dp2[i] -= dp2[tmpind] - dp2[x-2];
}
}
}
return std::pair(dp, dp2);
}
// enumerate primes <= N
std::vector<long long> enumerate_prime(long long N) {
std::vector<bool> p(N+1);
std::vector<long long> ret(0);
for (long long i=2; i<=N; i++){
if (p[i]) continue;
ret.emplace_back(i);
for (long long j=2*i; j<=N; j+=i){
p[j] = true;
}
}
return ret;
}
// black algorithm
// recommended for N <= 1e12
template<typename T, T(*f) (long long, long long, long long)>
T black_algorithm(
std::vector<T> GF,
long long N
) {
if (N == 1) return 1;
Quotients Q(N);
std::vector<long long> prime_list = enumerate_prime(Q.sqrtN);
int pm = prime_list.size();
auto dfs = [&](
auto self,
long long num,
T f_old,
long long gpf,
long long c,
long long num_tamari
) -> T {
T ret = 0;
if (gpf >= 0
&& num * prime_list[gpf] <= N) {
ret += f_old * f(prime_list[gpf], c+1, num_tamari * prime_list[gpf]);
if (num * prime_list[gpf] * prime_list[gpf] <= N){
ret += self(self,
num * prime_list[gpf],
f_old,
gpf,
c + 1,
num_tamari * prime_list[gpf]
);
}
}
if (gpf >= 0){
int lft = Q.ind(prime_list[gpf]);
int rgt = Q.ind(N / num);
if (lft < rgt){
ret += f_old * f(prime_list[gpf], c, num_tamari) * (GF[rgt] - GF[lft]);
}
}else{
int lft = 0;
int rgt = Q.ind(N / num);
if (lft < rgt){
ret += GF[rgt] - GF[lft];
}
}
for (long long i=gpf+1; i<pm; i++){
if (num * prime_list[i] <= N
&& num * prime_list[i] * prime_list[i] <= N) {
if (gpf >= 0){
ret += self(self,
num * prime_list[i],
f_old * f(prime_list[gpf], c, num_tamari),
i,
1,
prime_list[i]
);
}else{
ret += self(self,
num * prime_list[i],
f_old,
i,
1,
prime_list[i]
);
}
}else{
break;
}
}
return ret;
};
return dfs(dfs, 1, 1, -1, 0, 1) + T(1);
}
#include<atcoder/modint>
typedef atcoder::modint998244353 mint;
typedef long long ll;
mint t[60];
mint f(ll p, ll e, ll n) {
return t[e+1];
}
mint g(ll p) {
return 1;
}
mint G(ll p){
return p;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n, m; cin >> n >> m;
for (int i=1; i<60; i++){
t[i] = mint(i).pow(n);
}
auto [a,b] = lucy_dp<mint,g,G>(m);
vector<mint> x((int)a.size());
mint tmp = mint(2).pow(n);
for (int i=0; i<(int)a.size(); i++){
x[i] = a[i] * tmp;
}
mint ans = black_algorithm<mint,f>(x,m);
cout << ans.val() << '\n';
}
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