组合数打表

当n,m都很小的时候可以利用杨辉三角直接求(<=3000)
C(n,m)=C(n-1,m)+C(n-1,m-1)；

const int CN = 55;long long c[CN][CN] = {};void cinit(){    for (int i = 0; i < CN; i++)    {        c[i][0] = c[i][i] = 1;        for (int j = 1; j < i; j++)        {            c[i][j] = c[i - 1][j] + c[i - 1][j - 1];        }    }}

当n固定时对C(n,i)打表，n可以很大

typedef long long ll;const int M = 1e9 + 7;const ll MAXN = 1e6 + 50;ll inv[MAXN + 200];ll C[MAXN];void getinv(){    inv[1] = 1;    for (int i = 2; i <= MAXN; i++)    {        inv[i] = inv[M % i] * (M - M / i) % M;    }}void getfac(ll n){    C[0] = 1;    for (int i = 1; i <= n; i++)    {        C[i] = C[i - 1] * (n - i + 1) % M * inv[i] % M;    }}int main(){    getinv();    ll n, m;    cin >> n;    getfac(n);    while (cin >> m)    {        cout << C[m] << endl;    }    return 0;}