結果

問題 No.2120 場合の数の下８桁
ユーザー hitonanodehitonanode
提出日時 2022-11-04 21:41:10
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 26 ms / 2,000 ms
コード長 17,208 bytes
コンパイル時間 2,366 ms
コンパイル使用メモリ 194,896 KB
実行使用メモリ 10,984 KB
最終ジャッジ日時 2024-07-18 19:24:11
合計ジャッジ時間 3,665 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
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
other AC * 20
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ソースコード

diff # 

#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <complex>
#include <deque>
#include <forward_list>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iostream>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using namespace std;
using lint = long long;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#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, typename V>
void ndarray(vector<T>& vec, const V& val, int len) { vec.assign(len, val); }
template <typename T, typename V, typename... Args> void ndarray(vector<T>& vec, const V& val, int len, Args... args) { vec.resize(len), for_each(begin(vec), end(vec), [&](T& v) { ndarray(v, val, args...); }); }
template <typename T> bool chmax(T &m, const T q) { return m < q ? (m = q, true) : false; }
template <typename T> bool chmin(T &m, const T q) { return m > q ? (m = q, true) : false; }
const std::vector<std::pair<int, int>> grid_dxs{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int floor_lg(long long x) { return x <= 0 ? -1 : 63 - __builtin_clzll(x); }
template <class T1, class T2> T1 floor_div(T1 num, T2 den) { return (num > 0 ? num / den : -((-num + den - 1) / den)); }
template <class T1, class T2> std::pair<T1, T2> operator+(const std::pair<T1, T2> &l, const std::pair<T1, T2> &r) { return std::make_pair(l.first + r.first, l.second + r.second); }
template <class T1, class T2> std::pair<T1, T2> operator-(const std::pair<T1, T2> &l, const std::pair<T1, T2> &r) { return std::make_pair(l.first - r.first, l.second - r.second); }
template <class T> std::vector<T> sort_unique(std::vector<T> vec) { sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end()); return vec; }
template <class T> int arglb(const std::vector<T> &v, const T &x) { return std::distance(v.begin(), std::lower_bound(v.begin(), v.end(), x)); }
template <class T> int argub(const std::vector<T> &v, const T &x) { return std::distance(v.begin(), std::upper_bound(v.begin(), v.end(), x)); }
template <class IStream, class T> IStream &operator>>(IStream &is, std::vector<T> &vec) { for (auto &v : vec) is >> v; return is; }

template <class OStream, class T> OStream &operator<<(OStream &os, const std::vector<T> &vec);
template <class OStream, class T, size_t sz> OStream &operator<<(OStream &os, const std::array<T, sz> &arr);
template <class OStream, class T, class TH> OStream &operator<<(OStream &os, const std::unordered_set<T, TH> &vec);
template <class OStream, class T, class U> OStream &operator<<(OStream &os, const pair<T, U> &pa);
template <class OStream, class T> OStream &operator<<(OStream &os, const std::deque<T> &vec);
template <class OStream, class T> OStream &operator<<(OStream &os, const std::set<T> &vec);
template <class OStream, class T> OStream &operator<<(OStream &os, const std::multiset<T> &vec);
template <class OStream, class T> OStream &operator<<(OStream &os, const std::unordered_multiset<T> &vec);
template <class OStream, class T, class U> OStream &operator<<(OStream &os, const std::pair<T, U> &pa);
template <class OStream, class TK, class TV> OStream &operator<<(OStream &os, const std::map<TK, TV> &mp);
template <class OStream, class TK, class TV, class TH> OStream &operator<<(OStream &os, const std::unordered_map<TK, TV, TH> &mp);
template <class OStream, class... T> OStream &operator<<(OStream &os, const std::tuple<T...> &tpl);

template <class OStream, class T> OStream &operator<<(OStream &os, const std::vector<T> &vec) { os << '['; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <class OStream, class T, size_t sz> OStream &operator<<(OStream &os, const std::array<T, sz> &arr) { os << '['; for (auto v : arr) os << v << ','; os << ']'; return os; }
#if __cplusplus >= 201703L
template <class... T> std::istream &operator>>(std::istream &is, std::tuple<T...> &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); return is; }
template <class OStream, class... T> OStream &operator<<(OStream &os, const std::tuple<T...> &tpl) { os << '('; std::apply([&os](auto &&... args) { ((os << args << ','), ...);}, tpl); return os << ')'; }
#endif
template <class OStream, class T, class TH> OStream &operator<<(OStream &os, const std::unordered_set<T, TH> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <class OStream, class T> OStream &operator<<(OStream &os, const std::deque<T> &vec) { os << "deq["; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <class OStream, class T> OStream &operator<<(OStream &os, const std::set<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <class OStream, class T> OStream &operator<<(OStream &os, const std::multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <class OStream, class T> OStream &operator<<(OStream &os, const std::unordered_multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <class OStream, class T, class U> OStream &operator<<(OStream &os, const std::pair<T, U> &pa) { return os << '(' << pa.first << ',' << pa.second << ')'; }
template <class OStream, class TK, class TV> OStream &operator<<(OStream &os, const std::map<TK, TV> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
template <class OStream, class TK, class TV, class TH> OStream &operator<<(OStream &os, const std::unordered_map<TK, TV, TH> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
#ifdef HITONANODE_LOCAL
const string COLOR_RESET = "\033[0m", BRIGHT_GREEN = "\033[1;32m", BRIGHT_RED = "\033[1;31m", BRIGHT_CYAN = "\033[1;36m", NORMAL_CROSSED = "\033[0;9;37m", RED_BACKGROUND = "\033[1;41m", NORMAL_FAINT = "\033[0;2m";
#define dbg(x) std::cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << std::endl
#define dbgif(cond, x) ((cond) ? std::cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << std::endl : std::cerr)
#else
#define dbg(x) ((void)0)
#define dbgif(cond, x) ((void)0)
#endif

// Linear sieve algorithm for fast prime factorization
// Complexity: O(N) time, O(N) space:
// - MAXN = 10^7:  ~44 MB,  80~100 ms (Codeforces / AtCoder GCC, C++17)
// - MAXN = 10^8: ~435 MB, 810~980 ms (Codeforces / AtCoder GCC, C++17)
// Reference:
// [1] D. Gries, J. Misra, "A Linear Sieve Algorithm for Finding Prime Numbers,"
//     Communications of the ACM, 21(12), 999-1003, 1978.
// - https://cp-algorithms.com/algebra/prime-sieve-linear.html
// - https://37zigen.com/linear-sieve/
struct Sieve {
    std::vector<int> min_factor;
    std::vector<int> primes;
    Sieve(int MAXN) : min_factor(MAXN + 1) {
        for (int d = 2; d <= MAXN; d++) {
            if (!min_factor[d]) {
                min_factor[d] = d;
                primes.emplace_back(d);
            }
            for (const auto &p : primes) {
                if (p > min_factor[d] or d * p > MAXN) break;
                min_factor[d * p] = p;
            }
        }
    }
    // Prime factorization for 1 <= x <= MAXN^2
    // Complexity: O(log x)           (x <= MAXN)
    //             O(MAXN / log MAXN) (MAXN < x <= MAXN^2)
    template <class T> std::map<T, int> factorize(T x) const {
        std::map<T, int> ret;
        assert(x > 0 and
               x <= ((long long)min_factor.size() - 1) * ((long long)min_factor.size() - 1));
        for (const auto &p : primes) {
            if (x < T(min_factor.size())) break;
            while (!(x % p)) x /= p, ret[p]++;
        }
        if (x >= T(min_factor.size())) ret[x]++, x = 1;
        while (x > 1) ret[min_factor[x]]++, x /= min_factor[x];
        return ret;
    }
    // Enumerate divisors of 1 <= x <= MAXN^2
    // Be careful of highly composite numbers https://oeis.org/A002182/list
    // https://gist.github.com/dario2994/fb4713f252ca86c1254d#file-list-txt (n, (# of div. of n)):
    // 45360->100, 735134400(<1e9)->1344, 963761198400(<1e12)->6720
    template <class T> std::vector<T> divisors(T x) const {
        std::vector<T> ret{1};
        for (const 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
    }
    // Euler phi functions of divisors of given x
    // Verified: ABC212 G https://atcoder.jp/contests/abc212/tasks/abc212_g
    // Complexity: O(sqrt(x) + d(x))
    template <class T> std::map<T, T> euler_of_divisors(T x) const {
        assert(x >= 1);
        std::map<T, T> ret;
        ret[1] = 1;
        std::vector<T> divs{1};
        for (auto p : factorize(x)) {
            int n = ret.size();
            for (int i = 0; i < n; i++) {
                ret[divs[i] * p.first] = ret[divs[i]] * (p.first - 1);
                divs.push_back(divs[i] * p.first);
                for (T a = divs[i] * p.first, d = 1; d < p.second; a *= p.first, d++) {
                    ret[a * p.first] = ret[a] * p.first;
                    divs.push_back(a * p.first);
                }
            }
        }
        return ret;
    }
    // Moebius function Table, (-1)^{# of different prime factors} for square-free x
    // return: [0=>0, 1=>1, 2=>-1, 3=>-1, 4=>0, 5=>-1, 6=>1, 7=>-1, 8=>0, ...] https://oeis.org/A008683
    std::vector<int> GenerateMoebiusFunctionTable() const {
        std::vector<int> ret(min_factor.size());
        for (unsigned i = 1; i < min_factor.size(); i++) {
            if (i == 1) {
                ret[i] = 1;
            } else if ((i / min_factor[i]) % min_factor[i] == 0) {
                ret[i] = 0;
            } else {
                ret[i] = -ret[i / min_factor[i]];
            }
        }
        return ret;
    }
    // Calculate [0^K, 1^K, ..., nmax^K] in O(nmax)
    // Note: **0^0 == 1**
    template <class MODINT> std::vector<MODINT> enumerate_kth_pows(long long K, int nmax) const {
        assert(nmax < int(min_factor.size()));
        assert(K >= 0);
        if (K == 0) return std::vector<MODINT>(nmax + 1, 1);
        std::vector<MODINT> ret(nmax + 1);
        ret[0] = 0, ret[1] = 1;
        for (int n = 2; n <= nmax; n++) {
            if (min_factor[n] == n) {
                ret[n] = MODINT(n).pow(K);
            } else {
                ret[n] = ret[n / min_factor[n]] * ret[min_factor[n]];
            }
        }
        return ret;
    }
};
Sieve sieve((1 << 20));


// Solve ax+by=gcd(a, b)
template <class Int> Int extgcd(Int a, Int b, Int &x, Int &y) {
    Int d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x), y -= (a / b) * x;
    } else {
        x = 1, y = 0;
    }
    return d;
}
// Calculate a^(-1) (MOD m) s if gcd(a, m) == 1
// Calculate x s.t. ax == gcd(a, m) MOD m
template <class Int> Int mod_inverse(Int a, Int m) {
    Int x, y;
    extgcd<Int>(a, m, x, y);
    x %= m;
    return x + (x < 0) * m;
}

// Require: 1 <= b
// return: (g, x) s.t. g = gcd(a, b), xa = g MOD b, 0 <= x < b/g
template <class Int> /* constexpr */ std::pair<Int, Int> inv_gcd(Int a, Int b) {
    a %= b;
    if (a < 0) a += b;
    if (a == 0) return {b, 0};
    Int s = b, t = a, m0 = 0, m1 = 1;
    while (t) {
        Int u = s / t;
        s -= t * u, m0 -= m1 * u;
        auto tmp = s;
        s = t, t = tmp, tmp = m0, m0 = m1, m1 = tmp;
    }
    if (m0 < 0) m0 += b / s;
    return {s, m0};
}

template <class Int>
/* constexpr */ std::pair<Int, Int> crt(const std::vector<Int> &r, const std::vector<Int> &m) {
    assert(r.size() == m.size());
    int n = int(r.size());
    // Contracts: 0 <= r0 < m0
    Int r0 = 0, m0 = 1;
    for (int i = 0; i < n; i++) {
        assert(1 <= m[i]);
        Int r1 = r[i] % m[i], m1 = m[i];
        if (r1 < 0) r1 += m1;
        if (m0 < m1) {
            std::swap(r0, r1);
            std::swap(m0, m1);
        }
        if (m0 % m1 == 0) {
            if (r0 % m1 != r1) return {0, 0};
            continue;
        }
        Int g, im;
        std::tie(g, im) = inv_gcd<Int>(m0, m1);

        Int u1 = m1 / g;
        if ((r1 - r0) % g) return {0, 0};

        Int x = (r1 - r0) / g % u1 * im % u1;
        r0 += x * m0;
        m0 *= u1;
        if (r0 < 0) r0 += m0;
    }
    return {r0, m0};
}

// 蟻本 P.262
// 中国剰余定理を利用して，色々な素数で割った余りから元の値を復元
// 連立線形合同式 A * x = B mod M の解
// Requirement: M[i] > 0
// Output: x = first MOD second (if solution exists), (0, 0) (otherwise)
template <class Int>
std::pair<Int, Int>
linear_congruence(const std::vector<Int> &A, const std::vector<Int> &B, const std::vector<Int> &M) {
    Int r = 0, m = 1;
    assert(A.size() == M.size());
    assert(B.size() == M.size());
    for (int i = 0; i < (int)A.size(); i++) {
        assert(M[i] > 0);
        const Int ai = A[i] % M[i];
        Int a = ai * m, b = B[i] - ai * r, d = std::__gcd(M[i], a);
        if (b % d != 0) {
            return std::make_pair(0, 0); // 解なし
        }
        Int t = b / d * mod_inverse<Int>(a / d, M[i] / d) % (M[i] / d);
        r += m * t;
        m *= M[i] / d;
    }
    return std::make_pair((r < 0 ? r + m : r), m);
}

template <class Int = int, class Long = long long>
Int pow_mod(Int x, long long n, Int md) {
    static_assert(sizeof(Int) * 2 <= sizeof(Long), "Watch out for overflow");
    if (md == 1) return 0;
    Int ans = 1;
    while (n > 0) {
        if (n & 1) ans = (Long)ans * x % md;
        x = (Long)x * x % md;
        n >>= 1;
    }
    return ans;
}


// nCr mod m = p^q (p: prime, q >= 1)
// Can be used for n, r <= 1e18, m <= 1e7
// Complexity: O(m) (construction), O(log(n)) (per query)
// https://ferin-tech.hatenablog.com/entry/2018/01/17/010829
struct combination_prime_pow {
    int p, q, m;
    std::vector<int> fac, invfac, ppow;

    long long _ej(long long n) const {
        long long ret = 0;
        while (n) ret += n, n /= p;
        return ret;
    }

    combination_prime_pow(int p_, int q_) : p(p_), q(q_), m(1), ppow{1} {
        for (int t = 0; t < q; ++t) m *= p, ppow.push_back(m);
        fac.assign(m, 1);
        invfac.assign(m, 1);
        for (int i = 1; i < m; ++i) fac[i] = (long long)fac[i - 1] * (i % p ? i : 1) % m;
        invfac[m - 1] = fac[m - 1]; // Same as Wilson's theorem
        assert(1LL * fac.back() * invfac.back() % m == 1);
        for (int i = m - 1; i; --i) invfac[i - 1] = (long long)invfac[i] * (i % p ? i : 1) % m;
    }

    int nCr(long long n, long long r) const {
        if (r < 0 or n < r) return 0;
        if (p == 2 and q == 1) return !((~n) & r); // Lucas
        long long k = n - r;
        long long e0 = _ej(n / p) - _ej(r / p) - _ej(k / p);
        if (e0 >= q) return 0;

        long long ret = ppow[e0];
        if (q == 1) { // Lucas
            while (n) {
                ret = __int128(ret) * fac[n % p] * invfac[r % p] * invfac[k % p] % p;
                n /= p, r /= p, k /= p;
            }
            return (int)ret;
        } else {
            if ((p > 2 or q < 3) and (_ej(n / m) - _ej(r / m) - _ej(k / m)) & 1) ret = m - ret;
            while (n) {
                ret = __int128(ret) * fac[n % m] * invfac[r % m] * invfac[k % m] % m;
                n /= p, r /= p, k /= p;
            }
            return (int)ret;
        }
    }
};

// nCr mod m
// Complexity: O(m) space worst (construction), O(log(n) log(m)) (per query)
// Input: pairs of (prime, degree), such as vector<pair<int, int>> and map<int, int>
// https://judge.yosupo.jp/problem/binomial_coefficient
struct combination {
    std::vector<combination_prime_pow> cpps;
    std::vector<int> ms;

    template <class Map> combination(const Map &p2deg) {
        for (auto f : p2deg) {
            cpps.push_back(combination_prime_pow(f.first, f.second));
            ms.push_back(cpps.back().m);
        }
    }

    int operator()(long long n, long long r) const {
        if (r < 0 or n < r) return 0;
        std::vector<int> rs;
        for (const auto &cpp : cpps) rs.push_back(cpp.nCr(n, r));
        return crt(rs, ms).first;
    }
};


int main() {
    int N, M;
    cin >> N >> M;
    combination ncr(sieve.factorize(100000000));
    string ret = to_string(ncr(N, M));
    while (ret.size() < 8) ret = "0" + ret;
    cout << ret << endl;
}
0