제출 #942051

#제출 시각아이디문제언어결과실행 시간메모리
942051juliany2Tents (JOI18_tents)C++17
100 / 100
344 ms35852 KiB
#include<bits/stdc++.h> using namespace std; using ll = long long; #define all(x) (x).begin(), (x).end() template<int MD> struct mi { int v; explicit operator int() const { return v; } mi(long long _v = 0) { v = (-MD <= _v && _v < MD) ? _v : _v % MD; if (v < 0) v += MD; } mi &operator+=(const mi &o) { if ((v += o.v) >= MD) v -= MD; return *this; } mi &operator-=(const mi &o) { if ((v -= o.v) < 0) v += MD; return *this; } mi &operator*=(const mi &o) { v = 1LL * v * o.v % MD; return *this; } mi &operator/=(const mi &o) { return (*this) *= inv(o); } friend mi operator+(mi a, const mi &b) { return a += b; } friend mi operator-(mi a, const mi &b) { return a -= b; } friend mi operator*(mi a, const mi &b) { return a *= b; } friend mi operator/(mi a, const mi &b) { return a /= b; } friend mi inv(const mi &m) { return binpow(m, MD - 2); } friend mi binpow(mi b, long long e) { mi ret = 1; for (; e; e >>= 1, b *= b) if (e & 1) ret *= b; return ret; } friend bool operator==(const mi &a, const mi &b) { return a.v == b.v; } friend bool operator!=(const mi &a, const mi &b) { return !(a == b); } friend bool operator<(const mi &a, const mi &b) { return a.v < b.v; } friend istream &operator>>(istream &i, mi &m) { long long x; i >> x; m.v = x; return i; } friend ostream &operator<<(ostream &o, const mi &m) { o << m.v; return o; } }; const int N = 3007, mod = 1e9 + 7; using Mint = mi<mod>; int n, m; Mint fac[N], ifac[N]; Mint dp[N][N]; Mint nck(int n, int k) { return fac[n] * ifac[n - k] * ifac[k]; } int main() { cin.tie(0)->sync_with_stdio(false); fac[0] = 1; for (int i = 1; i < N; i++) fac[i] = fac[i - 1] * i; ifac[N - 1] = binpow(fac[N - 1], mod - 2); for (int i = N - 2; ~i; i--) ifac[i] = ifac[i + 1] * (i + 1); cin >> n >> m; for (int i = 0; i <= n; i++) { for (int j = 0; j <= m; j++) { if (i == 0 || j == 0) dp[i][j] = 1; else dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] * j * 4; } } Mint ans = 0; for (int i = 0; i <= min(n / 2, m); i++) { Mint w = nck(m, i) * nck(n, i * 2) * fac[i * 2] / binpow((Mint) 2, i); int nn = n - i * 2, nm = m - i; for (int j = 0; j <= min(nm / 2, nn); j++) { Mint cur = w * nck(nn, j) * nck(nm, j * 2) * fac[j * 2] / binpow((Mint) 2, j); ans += cur * dp[nn - j][nm - j * 2]; } } cout << ans - 1 << '\n'; return 0; }
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