結果

問題 No.990 N×Mマス計算(Kの倍数)
ユーザー hitonanode
提出日時 2020-02-14 21:34:39
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 97 ms / 2,000 ms
コード長 6,712 bytes
コンパイル時間 2,383 ms
コンパイル使用メモリ 192,892 KB
実行使用メモリ 14,692 KB
最終ジャッジ日時 2024-11-16 00:27:38
合計ジャッジ時間 3,726 ms
ジャッジサーバーID
(参考情報)
judge3 / judge5
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 * 2
other AC * 19
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
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 /////
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
using namespace __gnu_pbds; // find_by_order(), order_of_key()
template<typename TK> using pbds_set = tree<TK, null_type, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename TK, typename TV> using pbds_map = tree<TK, TV, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
*/
// Sieve of Eratosthenes
// (*this)[i] = (divisor of i, greater than 1)
// Example: [0, 1, 2, 3, 2, 5, 3, 7, 2, 3, 2, 11, ...]
// Complexity: Space O(MAXN), Time (construction) O(MAXNloglogMAXN)
struct SieveOfEratosthenes : std::vector<int>
{
std::vector<int> primes;
SieveOfEratosthenes(int MAXN) : std::vector<int>(MAXN + 1) {
std::iota(begin(), end(), 0);
for (int i = 2; i <= MAXN; i++) {
if ((*this)[i] == i) {
primes.push_back(i);
for (int j = i; j <= MAXN; j += i) (*this)[j] = i;
}
}
}
using T = long long int;
// Prime factorization for x <= MAXN^2
// Complexity: O(log x) (x <= MAXN)
// O(MAXN / logMAXN) (MAXN < x <= MAXN^2)
std::map<T, int> Factorize(T x) {
assert(x <= 1LL * (int(size()) - 1) * (int(size()) - 1));
std::map<T, int> ret;
if (x < int(size())) {
while (x > 1) {
ret[(*this)[x]]++;
x /= (*this)[x];
}
}
else {
for (auto p : primes) {
while (!(x % p)) x /= p, ret[p]++;
if (x == 1) break;
}
if (x > 1) ret[x]++;
}
return ret;
}
std::vector<T> Divisors(int x) {
std::vector<T> ret{1};
for (auto p : Factorize(x)) {
int n = ret.size();
for (int i = 0; i < n; i++) {
for (T a = 1, d = 1; d <= p.second; d++) {
a *= p.first;
ret.push_back(ret[i] * a);
}
}
}
return ret; // Not sorted
}
// Moebius function Table
// return: [0=>0, 1=>1, 2=>-1, 3=>-1, 4=>0, 5=>-1, 6=>1, 7=>-1, 8=>0, ...]
std::vector<int> GenerateMoebiusFunctionTable() {
std::vector<int> ret(size());
for (int i = 1; i < int(size()); i++) {
if (i == 1) ret[i] = 1;
else if ((i / (*this)[i]) % (*this)[i] == 0) ret[i] = 0;
else ret[i] = -ret[i / (*this)[i]];
}
return ret;
}
};
SieveOfEratosthenes sieve(100000);
int main()
{
lint N, M, K;
cin >> N >> M >> K;
char c;
cin >> c;
vector<lint> B(M), A(N);
cin >> B >> A;
vector<lint> divs = sieve.Divisors(K);
sort(divs.begin(), divs.end());
lint ret = 0;
if (c == '*')
{
unordered_map<lint, int> dinv;
REP(i, divs.size())
dinv[divs[i]] = i;
vector<lint> coua(divs.size()), coub(divs.size());
for (auto a : A)
coua[dinv[__gcd(a, K)]]++;
for (auto b : B)
coub[dinv[__gcd(b, K)]]++;
REP(i, divs.size())
{
FOR(j, i + 1, divs.size())
{
if (divs[j] % divs[i] == 0)
coua[i] += coua[j];
}
}
REP(i, divs.size()) {
ret += coua[i] * coub[divs.size() - 1 - i];
}
}
else {
unordered_map<lint, int> coua;
for (auto a : A) coua[a % K]++;
for (auto b : B) {
lint tgt = (K - b % K) % K;
ret += coua[tgt];
}
}
cout << ret << endl;
}
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