結果
| 問題 |
No.854 公平なりんご分配
|
| コンテスト | |
| ユーザー |
hitonanode
|
| 提出日時 | 2019-07-26 22:23:59 |
| 言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 4,780 bytes |
| コンパイル時間 | 1,636 ms |
| コンパイル使用メモリ | 181,836 KB |
| 実行使用メモリ | 40,320 KB |
| 最終ジャッジ日時 | 2024-07-02 07:45:29 |
| 合計ジャッジ時間 | 9,331 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 |
| other | AC * 50 TLE * 1 -- * 41 |
ソースコード
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx")
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
///// This part below is only for debug, not used /////
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
///// END /////
int main()
{
vector<int> primes{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43};
int N;
cin >> N;
vector<int> A(N);
cin >> A;
vector<vector<int>> rc(2000);
set<int> ps;
REP(i, N)
{
int a = A[i];
for (auto p : primes)
{
while (a % p == 0)
{
if (rc[p].size() == 0)
{
rc[p].resize(N + 1);
ps.insert(p);
}
rc[p][i + 1]++;
a /= p;
}
}
if (a == 1) continue;
if (rc[a].size() == 0)
{
rc[a].resize(N + 1);
ps.insert(a);
}
rc[a][i + 1]++;
}
REP(d, 2000) if (rc[d].size())
{
REP(i, N) rc[d][i + 1] += rc[d][i];
}
int Q;
cin >> Q;
REP(_, Q)
{
int P, L, R;
cin >> P >> L >> R;
bool flg = true;
for (auto d : ps)
{
if (P < d) break;
int n = rc[d][R] - rc[d][L - 1];
while (P % d == 0)
{
if (n) n--;
else
{
flg = false;
break;
}
P /= d;
}
}
if (!flg)
{
puts("NO");
continue;
}
if (P == 1)
{
puts("Yes");
continue;
}
if (P > 2000 or rc[P].size() == 0 or P < 45 or rc[P][R] - rc[P][L - 1] == 0)
{
puts("NO");
continue;
}
puts("Yes");
}
}
hitonanode