結果

問題 No.881 sin(x)/xの和
ユーザー ecottea
提出日時 2022-10-03 22:21:15
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 50 ms / 2,000 ms
コード長 6,899 bytes
コンパイル時間 5,427 ms
コンパイル使用メモリ 237,584 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-12-28 02:55:27
合計ジャッジ時間 7,157 ms
ジャッジサーバーID
(参考情報)
judge4 / judge2
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ファイルパターン 結果
other AC * 23
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ソースコード

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プレゼンテーションモードにする

#ifndef HIDDEN_IN_VS //
//
#define _CRT_SECURE_NO_WARNINGS
//
#include <bits/stdc++.h>
using namespace std;
//
using ll = long long; // -2^63 2^63 = 9 * 10^18int -2^31 2^31 = 2 * 10^9
using pii = pair<int, int>; using pll = pair<ll, ll>; using pil = pair<int, ll>; using pli = pair<ll, int>;
using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;
using vl = vector<ll>; using vvl = vector<vl>; using vvvl = vector<vvl>;
using vb = vector<bool>; using vvb = vector<vb>; using vvvb = vector<vvb>;
using vc = vector<char>; using vvc = vector<vc>; using vvvc = vector<vvc>;
using vd = vector<double>; using vvd = vector<vd>; using vvvd = vector<vvd>;
template <class T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
using Graph = vvi;
//
const double PI = acos(-1);
const vi DX = { 1, 0, -1, 0 }; // 4
const vi DY = { 0, 1, 0, -1 };
int INF = 1001001001; ll INFL = 4004004004004004004LL;
double EPS = 1e-12;
//
struct fast_io { fast_io() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(18); } } fastIOtmp;
//
#define all(a) (a).begin(), (a).end()
#define sz(x) ((int)(x).size())
#define lbpos(a, x) (int)distance((a).begin(), std::lower_bound(all(a), x))
#define ubpos(a, x) (int)distance((a).begin(), std::upper_bound(all(a), x))
#define Yes(b) {cout << ((b) ? "Yes\n" : "No\n");}
#define rep(i, n) for(int i = 0, i##_len = int(n); i < i##_len; ++i) // 0 n-1
#define repi(i, s, t) for(int i = int(s), i##_end = int(t); i <= i##_end; ++i) // s t
#define repir(i, s, t) for(int i = int(s), i##_end = int(t); i >= i##_end; --i) // s t
#define repe(v, a) for(const auto& v : (a)) // a
#define repea(v, a) for(auto& v : (a)) // a
#define repb(set, d) for(int set = 0; set < (1 << int(d)); ++set) // d
#define repp(a) sort(all(a)); for(bool a##_perm = true; a##_perm; a##_perm = next_permutation(all(a))) // a
#define smod(n, m) ((((n) % (m)) + (m)) % (m)) // mod
#define uniq(a) {sort(all(a)); (a).erase(unique(all(a)), (a).end());} //
#define EXIT(a) {cout << (a) << endl; exit(0);} //
//
template <class T> inline ll pow(T n, int k) { ll v = 1; rep(i, k) v *= n; return v; }
template <class T> inline bool chmax(T& M, const T& x) { if (M < x) { M = x; return true; } return false; } // true
    
template <class T> inline bool chmin(T& m, const T& x) { if (m > x) { m = x; return true; } return false; } // true
    
//
template <class T, class U> inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template <class T> inline istream& operator>>(istream& is, vector<T>& v) { repea(x, v) is >> x; return is; }
template <class T> inline vector<T>& operator--(vector<T>& v) { repea(x, v) --x; return v; }
template <class T> inline vector<T>& operator++(vector<T>& v) { repea(x, v) ++x; return v; }
// Visual Studio
#ifdef _MSC_VER
#include "local.hpp"
// gcc
#else
inline int popcount(int n) { return __builtin_popcount(n); }
inline int popcount(ll n) { return __builtin_popcountll(n); }
inline int lsb(int n) { return n != 0 ? __builtin_ctz(n) : -1; }
inline int lsb(ll n) { return n != 0 ? __builtin_ctzll(n) : -1; }
inline int msb(int n) { return n != 0 ? (31 - __builtin_clz(n)) : -1; }
inline int msb(ll n) { return n != 0 ? (63 - __builtin_clzll(n)) : -1; }
#define gcd __gcd
#define dump(...)
#define dumpel(v)
#define dump_list(v)
#define input_from_file(f)
#define output_to_file(f)
#define Assert(b) { if (!(b)) while (1) cout << "OLE"; }
#endif
#endif //
//--------------AtCoder --------------
#include <atcoder/all>
using namespace atcoder;
//using mint = modint1000000007;
using mint = modint998244353;
//using mint = modint; // mint::set_mod(m);
istream& operator>>(istream& is, mint& x) { ll x_; is >> x_; x = x_; return is; }
ostream& operator<<(ostream& os, const mint& x) { os << x.val(); return os; }
using vm = vector<mint>; using vvm = vector<vm>; using vvvm = vector<vvm>;
//----------------------------------------
long double sinc(long double x) {
// sinc(x)
// = sin(x) / x
// = (x - (1/3!)x^3 + (1/5!)x^5 - (1/7!)x^7 + ...) / x
// = 1 - (1/3!)x^2 + (1/5!)x^4 - (1/7!)x^6 + ...
long double res = 0;
res += 1;
res -= pow(x, 2.) / 6;
res += pow(x, 4.) / 120;
res -= pow(x, 6.) / 5040;
res += pow(x, 8.) / 362880;
res -= pow(x, 10.) / 39916800;
res += pow(x, 12.) / 6227020800;
res -= pow(x, 14.) / 1.3076744e+12;
return res;
}
void zikken() {
// : https://proc-cpuinfo.fixstars.com/2017/07/introduction-of-convergence-acceleration-methods/
int n = 11;
vvd a(n + 1, vd(n + 1));
rep(i, n) a[1][i + 1] = a[1][i] + (i % 2 ? -1 : 1) * 4. / (2 * i + 1);
repi(k, 2, n) {
repi(i, 1, n - k + 1) {
a[k][i] = a[k - 2][i + 1] + 1 / (a[k - 1][i + 1] - a[k - 1][i]);
}
}
dumpel(a);
exit(0);
}
void zikken2() {
int n = 71;
vvd a(n + 1, vd(n + 1));
rep(i, n) a[1][i + 1] = a[1][i] + sin(i + 1) / (i + 1);
repi(k, 2, n) {
repi(i, 1, n - k + 1) {
a[k][i] = a[k - 2][i + 1] + 1 / (a[k - 1][i + 1] - a[k - 1][i]);
}
}
dumpel(a);
double val = 0;
rep(i, 1000000) val += sin(i + 1) / (i + 1);
dump(val);
exit(0);
}
int main() {
// input_from_file("input.txt");
// output_to_file("output.txt");
// zikken2();
int n;
cin >> n;
long double res = 0;
rep(i, n) {
long double x0, a0;
cin >> x0 >> a0;
repi(j, -10, 10) {
long double x = x0 + j;
if (abs(x) < 0.99) {
res += a0 * sinc(x);
}
else {
res += a0 * sin(x) / x;
}
}
{
int n = 47;
vector<vector<long double>> a(n + 1, vector<long double>(n + 1));
rep(i, n) a[1][i + 1] = a[1][i] + sin(x0 + i + 11) / (x0 + i + 11);
repi(k, 2, n) {
repi(i, 1, n - k + 1) {
a[k][i] = a[k - 2][i + 1] + 1 / (a[k - 1][i + 1] - a[k - 1][i]);
}
if (!isfinite(a[k][1])) {
a[n][1] = a[k - 2][1];
break;
}
}
res += a0 * a[n][1];
}
{
int n = 47;
vector<vector<long double>> a(n + 1, vector<long double>(n + 1));
rep(i, n) a[1][i + 1] = a[1][i] + sin(x0 - i - 11) / (x0 - i - 11);
repi(k, 2, n) {
repi(i, 1, n - k + 1) {
a[k][i] = a[k - 2][i + 1] + 1 / (a[k - 1][i + 1] - a[k - 1][i]);
}
if (!isfinite(a[k][1])) {
a[n][1] = a[k - 2][1];
break;
}
}
res += a0 * a[n][1];
}
}
cout << res << endl;
}
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