結果

問題 No.2798 Multiple Chain
ユーザー pitPpitP
提出日時 2024-06-29 01:01:20
言語 C++23(gcc13)
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 6 ms / 2,000 ms
コード長 4,493 bytes
コンパイル時間 8,798 ms
コンパイル使用メモリ 336,912 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-06-29 01:01:37
合計ジャッジ時間 10,428 ms
ジャッジサーバーID
(参考情報)
judge1 / judge3
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
6,816 KB
testcase_01 AC 2 ms
6,944 KB
testcase_02 AC 2 ms
6,944 KB
testcase_03 AC 2 ms
6,940 KB
testcase_04 AC 2 ms
6,944 KB
testcase_05 AC 2 ms
6,940 KB
testcase_06 AC 2 ms
6,940 KB
testcase_07 AC 2 ms
6,944 KB
testcase_08 AC 2 ms
6,944 KB
testcase_09 AC 2 ms
6,940 KB
testcase_10 AC 2 ms
6,940 KB
testcase_11 AC 2 ms
6,940 KB
testcase_12 AC 2 ms
6,944 KB
testcase_13 AC 2 ms
6,944 KB
testcase_14 AC 2 ms
6,944 KB
testcase_15 AC 2 ms
6,944 KB
testcase_16 AC 2 ms
6,944 KB
testcase_17 AC 2 ms
6,940 KB
testcase_18 AC 2 ms
6,940 KB
testcase_19 AC 2 ms
6,944 KB
testcase_20 AC 2 ms
6,944 KB
testcase_21 AC 2 ms
6,940 KB
testcase_22 AC 2 ms
6,944 KB
testcase_23 AC 2 ms
6,944 KB
testcase_24 AC 2 ms
6,940 KB
testcase_25 AC 2 ms
6,940 KB
testcase_26 AC 2 ms
6,940 KB
testcase_27 AC 2 ms
6,944 KB
testcase_28 AC 2 ms
6,944 KB
testcase_29 AC 2 ms
6,940 KB
testcase_30 AC 2 ms
6,944 KB
testcase_31 AC 2 ms
6,944 KB
testcase_32 AC 6 ms
6,944 KB
testcase_33 AC 2 ms
6,944 KB
testcase_34 AC 5 ms
6,940 KB
testcase_35 AC 2 ms
6,944 KB
testcase_36 AC 3 ms
6,944 KB
testcase_37 AC 1 ms
6,940 KB
testcase_38 AC 2 ms
6,940 KB
testcase_39 AC 2 ms
6,940 KB
testcase_40 AC 2 ms
6,940 KB
testcase_41 AC 2 ms
6,944 KB
testcase_42 AC 2 ms
6,940 KB
testcase_43 AC 2 ms
6,944 KB
testcase_44 AC 2 ms
6,940 KB
testcase_45 AC 2 ms
6,940 KB
testcase_46 AC 2 ms
6,944 KB
testcase_47 AC 2 ms
6,944 KB
testcase_48 AC 2 ms
6,944 KB
testcase_49 AC 1 ms
6,940 KB
testcase_50 AC 2 ms
6,940 KB
testcase_51 AC 2 ms
6,940 KB
testcase_52 AC 3 ms
6,944 KB
testcase_53 AC 2 ms
6,944 KB
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 #

#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
istream &operator>>(istream &is, modint &a) { long long v; is >> v; a = v; return is; }
ostream &operator<<(ostream &os, const modint &a) { return os << a.val(); }
istream &operator>>(istream &is, modint998244353 &a) { long long v; is >> v; a = v; return is; }
ostream &operator<<(ostream &os, const modint998244353 &a) { return os << a.val(); }
istream &operator>>(istream &is, modint1000000007 &a) { long long v; is >> v; a = v; return is; }
ostream &operator<<(ostream &os, const modint1000000007 &a) { return os << a.val(); } 

typedef long long ll;
typedef vector<vector<int>> Graph;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define FOR(i,l,r) for (int i = l;i < (int)(r); i++)
#define rep(i,n) for (int i = 0;i < (int)(n); i++)
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define my_sort(x) sort(x.begin(), x.end())
#define my_max(x) *max_element(all(x))
#define my_min(x) *min_element(all(x))
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF = (1<<30) - 1;
const ll LINF = (1LL<<62) - 1;
const int MOD = 998244353;
const int MOD2 = 1e9+7;
const double PI = acos(-1);
vector<int> di = {1,0,-1,0};
vector<int> dj = {0,1,0,-1};

#ifdef LOCAL
#  include <debug_print.hpp>
#  define debug(...) debug_print::multi_print(#__VA_ARGS__, __VA_ARGS__)
#else
#  define debug(...) (static_cast<void>(0))
#endif

// https://algo-method.com/tasks/553/editorial
// 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(){
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    ll N; cin >> N;
    vector<long long> ps = prime_factorize(N);
    debug(ps);
    vector<long long> p, e;

    long long now = ps[0];
    long long cnt = 0;
    for(auto &prime : ps){
        if(now == prime) cnt++;
        else {
            p.push_back(now);
            e.push_back(cnt);
            now = prime;
            cnt = 1;
        }
    }
    p.push_back(now);
    e.push_back(cnt);
    int c = p.size();
    long long mx = *max_element(all(e));

    vector P(mx + 1, vector<ll>(mx + 1));
    P[0][0] = 1;
    for(int n = 1; n <= mx; n++){
        for(int k = 0; k <= mx; k++){
            if(k > 0) P[n][k] += P[n - 1][k - 1];
            if(n > k) P[n][k] += P[n - k][k];
        }
    }
    debug(P);

    ll ans = 1;
    rep(i, c){
        ll tmp = 0;
        FOR(len, 1, e[i] + 1) tmp += P[e[i]][len];
        ans *= tmp;
    }
    cout << ans << endl;
}
0