結果
問題 | No.2751 429-like Number |
ユーザー |
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提出日時 | 2024-05-10 21:27:15 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 594 ms / 4,000 ms |
コード長 | 3,983 bytes |
コンパイル時間 | 5,425 ms |
コンパイル使用メモリ | 317,044 KB |
実行使用メモリ | 6,820 KB |
最終ジャッジ日時 | 2024-12-20 04:18:11 |
合計ジャッジ時間 | 10,024 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 6 |
other | AC * 22 |
ソースコード
#include <bits/stdc++.h>using namespace std;#include <atcoder/all>using namespace atcoder;template<typename T> inline bool chmax(T &a, T b) { return ((a < b) ? (a = b, true) : (false)); }template<typename T> inline bool chmin(T &a, T b) { return ((a > b) ? (a = b, true) : (false)); }#define rep(i, n) for (long long i = 0; i < (long long)(n); i++)#define rep2(i, m ,n) for (int i = (m); i < (long long)(n); i++)#define REP(i, n) for (long long i = 1; i < (long long)(n); i++)typedef long long ll;#define updiv(N,X) (N + X - 1) / X#define l(n) n.begin(),n.end()#define YesNo(Q) Q==1?cout<<"Yes":cout<<"No"using P = pair<int, int>;using mint = modint;const int MOD = 998244353LL;const ll INF = 999999999999LL;vector<long long> fact, fact_inv, inv;/* init_nCk :二項係数のための前処理計算量:O(n)*/template <typename T>void input(vector<T> &v){rep(i,v.size()){cin>>v[i];}return;}void init_nCk(int SIZE) {fact.resize(SIZE + 5);fact_inv.resize(SIZE + 5);inv.resize(SIZE + 5);fact[0] = fact[1] = 1;fact_inv[0] = fact_inv[1] = 1;inv[1] = 1;for (int i = 2; i < SIZE + 5; i++) {fact[i] = fact[i - 1] * i % MOD;inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;fact_inv[i] = fact_inv[i - 1] * inv[i] % MOD;}}/* nCk :MODでの二項係数を求める(前処理 int_nCk が必要)計算量:O(1)*/long long nCk(int n, int k) {assert(!(n < k));assert(!(n < 0 || k < 0));return fact[n] * (fact_inv[k] * fact_inv[n - k] % MOD) % MOD;}long long modpow(long long a, long long n, long long mod) {long long res = 1;while (n > 0) {if (n & 1) res = res * a % mod;a = a * a % mod;n >>= 1;}return res;}ll POW(ll a,ll n){long long res = 1;while (n > 0) {if (n & 1) res = res * a;a = a * a;n >>= 1;}return res;}// Miller-Rabin 素数判定法template<class T> T pow_mod(T A, T N, T M) {T res = 1 % M;A %= M;while (N) {if (N & 1) res = (res * A) % M;A = (A * A) % M;N >>= 1;}return res;}bool is_prime(long long N) {if (N <= 1) return false;if (N == 2 || N == 3) return true;if (N % 2 == 0) return false;vector<long long> A = {2, 325, 9375, 28178, 450775,9780504, 1795265022};long long s = 0, d = N - 1;while (d % 2 == 0) {++s;d >>= 1;}for (auto a : A) {if (a % N == 0) return true;long long t, x = pow_mod<__int128_t>(a, d, N);if (x != 1) {for (t = 0; t < s; ++t) {if (x == N - 1) break;x = __int128_t(x) * x % N;}if (t == s) return false;}}return true;}// Pollard のロー法long long gcd(long long A, long long B) {A = abs(A), B = abs(B);if (B == 0) return A;else return gcd(B, A % B);}long long pollard(long long N) {if (N % 2 == 0) return 2;if (is_prime(N)) return N;auto f = [&](long long x) -> long long {return (__int128_t(x) * x + 1) % N;};long long step = 0;while (true) {++step;long long x = step, y = f(x);while (true) {long long p = gcd(y - x + N, N);if (p == 0 || p == N) break;if (p != 1) return p;x = f(x);y = f(f(y));}}}vector<long long> prime_factorize(long long N) {if (N == 1) return {};long long p = pollard(N);if (p == N) return {p};vector<long long> left = prime_factorize(p);vector<long long> right = prime_factorize(N / p);left.insert(left.end(), right.begin(), right.end());sort(left.begin(), left.end());return left;}int main() {int q;cin>>q;rep(i,q){ll n;cin>>n;auto u = prime_factorize(n);if(u.size()==3){cout<<"Yes"<<endl;}else{cout<<"No"<<endl;}}}