結果

問題 No.2645 Sum of Divisors?
ユーザー kotatsugame
提出日時 2024-02-19 22:30:37
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,903 ms / 2,000 ms
コード長 4,019 bytes
コンパイル時間 808 ms
コンパイル使用メモリ 88,588 KB
実行使用メモリ 27,668 KB
最終ジャッジ日時 2024-09-29 02:23:01
合計ジャッジ時間 13,270 ms
ジャッジサーバーID
(参考情報)
judge4 / 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 * 3
other AC * 32
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ソースコード

diff #
プレゼンテーションモードにする

#include<iostream>
#include<iomanip>
#include<cmath>
#include<vector>
#include<numeric>
#include<cassert>
using namespace std;
#line 2 "multiplicative-function/sum-of-multiplicative-function.hpp"
#line 2 "prime/prime-enumerate.hpp"
// Prime Sieve {2, 3, 5, 7, 11, 13, 17, ...}
vector<int> prime_enumerate(int N) {
vector<bool> sieve(N / 3 + 1, 1);
for (int p = 5, d = 4, i = 1, sqn = sqrt(N); p <= sqn; p += d = 6 - d, i++) {
if (!sieve[i]) continue;
for (int q = p * p / 3, r = d * p / 3 + (d * p % 3 == 2), s = 2 * p,
qe = sieve.size();
q < qe; q += r = s - r)
sieve[q] = 0;
}
vector<int> ret{2, 3};
for (int p = 5, d = 4, i = 1; p <= N; p += d = 6 - d, i++)
if (sieve[i]) ret.push_back(p);
while (!ret.empty() && ret.back() > N) ret.pop_back();
return ret;
}
#line 4 "multiplicative-function/sum-of-multiplicative-function.hpp"
const int LIM=(int)1e6+1;
double H[LIM];
double calc(long long x)
{
if(x<LIM)return H[x];
return (log(x)+0.5772156649+0.5/x-1)*2;
}
// f(p, c) : f(p^c) ソ
template <typename T, T (*f)(long long, long long)>
struct mf_prefix_sum {
using i64 = long long;
i64 M, sq, s;
vector<int> p;
int ps;
vector<T> buf;
T ans;
mf_prefix_sum(i64 m) : M(m) {
assert(m < (1LL << 42));
sq = sqrt(M);
while (sq * sq > M) sq--;
while ((sq + 1) * (sq + 1) <= M) sq++;
if (M != 0) {
i64 hls = md(M, sq);
if (hls != 1 && md(M, hls - 1) == sq) hls--;
s = hls + sq;
p = prime_enumerate(sq);
ps = p.size();
ans = T{};
}
}
// ョ prefix sum Ν
vector<T> prime_sum_table() {
if (M == 0) return {};
i64 hls = md(M, sq);
if (hls != 1 && md(M, hls - 1) == sq) hls--;
vector<T> h(s);
for (int i = 1; i < hls; i++) {
T x = md(M, i);
h[i] = calc(x);
}
for (int i = 1; i <= sq; i++) {
T x = i;
h[s - i] = calc(x);
}
for (auto& x : p) {
T xt = (T)1/(T)x;
T pi = h[s - x + 1];
i64 x2 = i64(x) * x;
i64 imax = min<i64>(hls, md(M, x2) + 1);
i64 ix = x;
for (i64 i = 1; i < imax; ++i, ix += x) {
h[i] -= ((ix < hls ? h[ix] : h[s - md(M, ix)]) - pi) * xt;
}
for (int n = sq; n >= x2; n--) {
h[s - n] -= (h[s - md(n, x)] - pi) * xt;
}
}
assert((int)h.size() == s);
return h;
}
void dfs(int i, int c, T pc, i64 prod, T cur) {
ans += cur * (c+2)*(pc/(T)p[i]);//f(p[i], c + 1);
i64 lim = md(M, prod);
if (lim >= 1LL * p[i] * p[i]) dfs(i, c + 1, pc/(T)p[i], p[i] * prod, cur);
//cur *= f(p[i], c);
cur *= (c+1)*pc;
ans += cur * (buf[idx(lim)] - buf[idx(p[i])]);
int j = i + 1;
// M < 2**42 -> p_j < 2**21 -> (p_j)^3 < 2**63
for (; j < ps && 1LL * p[j] * p[j] * p[j] <= lim; j++) {
dfs(j, 1, (T)1/(T)p[j], prod * p[j], cur);
}
for (; j < ps && 1LL * p[j] * p[j] <= lim; j++) {
T sm = (T)3/((T)p[j]*p[j]);//f(p[j], 2);
int id1 = idx(md(lim, p[j])), id2 = idx(p[j]);
sm += (T)2/(T)p[j] * (buf[id1] - buf[id2]);
ans += cur * sm;
}
}
// fprime
T run(vector<T>& fprime) {
if (M == 0) return {};
set_buf(fprime);
assert((int)buf.size() == s);
ans = buf[idx(M)] + 1;
for (int i = 0; i < ps; i++) dfs(i, 1, (T)1/(T)p[i], p[i], 1);
return ans;
}
i64 md(i64 n, i64 d) { return double(n) / d; }
i64 idx(i64 n) { return n <= sq ? s - n : md(M, n); }
void set_buf(vector<T>& _buf) { swap(buf, _buf); }
};
/**
* @brief ョprefix sum
* @docs docs/multiplicative-function/sum-of-multiplicative-function.md
*/
double f(long long p,long long c){
assert(false);
return(double)(c+1)/pow((double)p,c);
}
int main()
{
for(int i=2;i<LIM;i++)H[i]=H[i-1]+(double)2/i;
long long N;
cin>>N;
mf_prefix_sum<double,f>solve(N);
vector<double>t=solve.prime_sum_table();
cout<<fixed<<setprecision(16)<<solve.run(t)<<endl;
}
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