結果
問題 | No.1048 Zero (Advanced) |
ユーザー |
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提出日時 | 2020-05-08 23:14:36 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 3 ms / 2,000 ms |
コード長 | 2,271 bytes |
コンパイル時間 | 2,890 ms |
コンパイル使用メモリ | 215,232 KB |
最終ジャッジ日時 | 2025-01-10 09:18:08 |
ジャッジサーバーID (参考情報) |
judge5 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 15 |
ソースコード
#include <bits/stdc++.h>#define M_PI 3.14159265358979323846 // piusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef vector<ll> VI;typedef pair<ll, ll> P;typedef tuple<ll, ll, ll> t3;typedef tuple<ll, ll, ll, ll> t4;#define rep(a,n) for(ll a = 0;a < n;a++)#define repi(a,b,n) for(ll a = b;a < n;a++)#include <bits/stdc++.h>using namespace std;const ll mod = 1e9 + 7;const ll INF = 1e15;class Primes {private:vector<int> Prime_Number;vector<bool> is_prime_;public:Primes(int N) {is_prime_.resize(N + 1, true);is_prime_[0] = is_prime_[1] = false;for (int i = 0; i < N + 1; i++) {if (is_prime_[i]) {Prime_Number.push_back(i);for (int j = 2 * i; j <= N; j += i) is_prime_[j] = false;}}}int operator[](int i) { return Prime_Number[i]; }int size() { return Prime_Number.size(); }int back() { return Prime_Number.back(); }bool isPrime(int q) { return is_prime_[q]; }};class Divisor {private:vector<ll> F;vector<pair<ll, ll>> pfactorize;public:Divisor(ll N) {for (ll i = 1; i * i <= N; i++) {if (N % i == 0) {F.push_back(i);if (i * i != N) F.push_back(N / i);}}sort(begin(F), end(F));Primes p((ll)sqrt(N) + 1);for (int i = 0; i < p.size(); i++) {pfactorize.emplace_back(p[i], 0);while (N % p[i] == 0) {N /= p[i];pfactorize.back().second++;}if (pfactorize.back().second == 0) pfactorize.pop_back();}if (N > 1) pfactorize.emplace_back(N, 1);}int size() { return F.size(); }const vector<pair<ll, ll>>& pfac() { return pfactorize; } constll operator[](int k) { return F[k]; }};int main() {ll l, r, m, k;cin >> l >> r >> m >> k;if (k == 0 || l == 0) {cout << "Yes" << endl;return 0;}if (l == 1) {if (m <= k) {cout << "Yes" << endl;return 0;}}Divisor md(m);Divisor kd(k);map<ll, ll> mmap;for (auto p : md.pfac()) {mmap[p.first] = p.second;}ll unit = 1;for (auto p : kd.pfac()) {auto a = p.second - mmap[p.first];for (int j = 0; j < a; j++) {unit *= p.first;}}auto i = m * unit / k;auto left = (l-1) / i;auto right = r / i;if (left != right) {cout << "Yes" << endl;}else {cout << "No" << endl;}return 0;}