#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; using lint = long long; using pint = pair; using plint = pair; 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##_begin_;i--) #define REP(i, n) FOR(i,0,n) #define IREP(i, n) IFOR(i,0,n) template void ndarray(vector& vec, const V& val, int len) { vec.assign(len, val); } template void ndarray(vector& vec, const V& val, int len, Args... args) { vec.resize(len), for_each(begin(vec), end(vec), [&](T& v) { ndarray(v, val, args...); }); } template bool chmax(T &m, const T q) { return m < q ? (m = q, true) : false; } template bool chmin(T &m, const T q) { return m > q ? (m = q, true) : false; } const std::vector> 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 T1 floor_div(T1 num, T2 den) { return (num > 0 ? num / den : -((-num + den - 1) / den)); } template std::pair operator+(const std::pair &l, const std::pair &r) { return std::make_pair(l.first + r.first, l.second + r.second); } template std::pair operator-(const std::pair &l, const std::pair &r) { return std::make_pair(l.first - r.first, l.second - r.second); } template std::vector sort_unique(std::vector vec) { sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end()); return vec; } template int arglb(const std::vector &v, const T &x) { return std::distance(v.begin(), std::lower_bound(v.begin(), v.end(), x)); } template int argub(const std::vector &v, const T &x) { return std::distance(v.begin(), std::upper_bound(v.begin(), v.end(), x)); } template IStream &operator>>(IStream &is, std::vector &vec) { for (auto &v : vec) is >> v; return is; } template OStream &operator<<(OStream &os, const std::vector &vec); template OStream &operator<<(OStream &os, const std::array &arr); template OStream &operator<<(OStream &os, const std::unordered_set &vec); template OStream &operator<<(OStream &os, const pair &pa); template OStream &operator<<(OStream &os, const std::deque &vec); template OStream &operator<<(OStream &os, const std::set &vec); template OStream &operator<<(OStream &os, const std::multiset &vec); template OStream &operator<<(OStream &os, const std::unordered_multiset &vec); template OStream &operator<<(OStream &os, const std::pair &pa); template OStream &operator<<(OStream &os, const std::map &mp); template OStream &operator<<(OStream &os, const std::unordered_map &mp); template OStream &operator<<(OStream &os, const std::tuple &tpl); template OStream &operator<<(OStream &os, const std::vector &vec) { os << '['; for (auto v : vec) os << v << ','; os << ']'; return os; } template OStream &operator<<(OStream &os, const std::array &arr) { os << '['; for (auto v : arr) os << v << ','; os << ']'; return os; } #if __cplusplus >= 201703L template std::istream &operator>>(std::istream &is, std::tuple &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); return is; } template OStream &operator<<(OStream &os, const std::tuple &tpl) { os << '('; std::apply([&os](auto &&... args) { ((os << args << ','), ...);}, tpl); return os << ')'; } #endif template OStream &operator<<(OStream &os, const std::unordered_set &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; } template OStream &operator<<(OStream &os, const std::deque &vec) { os << "deq["; for (auto v : vec) os << v << ','; os << ']'; return os; } template OStream &operator<<(OStream &os, const std::set &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; } template OStream &operator<<(OStream &os, const std::multiset &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; } template OStream &operator<<(OStream &os, const std::unordered_multiset &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; } template OStream &operator<<(OStream &os, const std::pair &pa) { return os << '(' << pa.first << ',' << pa.second << ')'; } template OStream &operator<<(OStream &os, const std::map &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; } template OStream &operator<<(OStream &os, const std::unordered_map &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 min_factor; std::vector 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 std::map factorize(T x) const { std::map 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 std::vector divisors(T x) const { std::vector 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 std::map euler_of_divisors(T x) const { assert(x >= 1); std::map ret; ret[1] = 1; std::vector 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 GenerateMoebiusFunctionTable() const { std::vector 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 std::vector enumerate_kth_pows(long long K, int nmax) const { assert(nmax < int(min_factor.size())); assert(K >= 0); if (K == 0) return std::vector(nmax + 1, 1); std::vector 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 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 Int mod_inverse(Int a, Int m) { Int x, y; extgcd(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 /* constexpr */ std::pair 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 /* constexpr */ std::pair crt(const std::vector &r, const std::vector &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(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 std::pair linear_congruence(const std::vector &A, const std::vector &B, const std::vector &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(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 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 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> and map // https://judge.yosupo.jp/problem/binomial_coefficient struct combination { std::vector cpps; std::vector ms; template 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 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; }