求解组合数 C (n, k) % p 的三种方法:
方法1(逆元求法):
const int N = 1e5 + 10; const int MOD = 1e9 + 7; int f[N], finv[N], inv[N]; void init(void) { //要求MOD是质数,预处理时间复杂度O(n) inv[1] = 1; for (int i=2; i<N; ++i) { inv[i] = (MOD - MOD / i) * 1ll * inv[MOD%i] % MOD; } f[0] = finv[0] = 1; for (int i=1; i<N; ++i) { f[i] = f[i-1] * 1ll * i % MOD; finv[i] = finv[i-1] * 1ll * inv[i] % MOD; } } int comb(int n, int k) { //C (n, k) % MOD if (k < 0 || k > n) return 0; return f[n] * 1ll * finv[n-k] % MOD * finv[k] % MOD; }
方法2: C(n,m)= C(n,n-m)= C(n-1,m-1)+C(n-1,m)
const int N = 2000 + 10; const int MOD = 1e9 + 7; int comb[N][N]; void init(void) { //对MOD没有要求,预处理时间复杂度O(n^2) for (int i=0; i<N; ++i) { comb[i][i] = comb[i][0] = 1; for (int j=1; j<i; ++j) { comb[i][j] = comb[i-1][j] + comb[i-1][j-1]; if (comb[i][j] >= MOD) { comb[i][j] -= MOD; } } } //printf ("%d ", comb[6][3]); }
方法3(Lucas定理,大组合数取模, HDOJ 3037为例):
ll f[N]; void init(int p) { //f[n] = n! f[0] = 1; for (int i=1; i<=p; ++i) f[i] = f[i-1] * i % p; } ll pow_mod(ll a, ll x, int p) { ll ret = 1; while (x) { if (x & 1) ret = ret * a % p; a = a * a % p; x >>= 1; } return ret; } ll Lucas(ll n, ll k, int p) { //C (n, k) % p ll ret = 1; while (n && k) { ll nn = n % p, kk = k % p; if (nn < kk) return 0; //inv (f[kk]) = f[kk] ^ (p - 2) % p ret = ret * f[nn] * pow_mod (f[kk] * f[nn-kk] % p, p - 2, p) % p; n /= p, k /= p; } return ret; } int main(void) { init (p); printf ("%I64d ", Lucas (n + m, n, p)); return 0; }