一、递推$O(nm)$

$C_a^b = C_{a-1}^{b-1}+C_{a-1}^{b}$

即可以将从$a$个苹果中选$b$个分为选/不选苹果$A$，前者是选(只需从剩余$a-1$个中选$b-1$个，后者是不选(还需选$b$个)

for(int i=0;i<=2000;i++)
    {
        for(int j=0;j<=i;j++)
        {
            if(!j) f[i][j]=1;
            else f[i][j]=(f[i-1][j-1]+f[i-1][j])%mod;
        }
    }

二、快速幂$O(logk)$

$C_a^b=\frac{a!}{b!×(a-b)!}=a!×infact(b!)×infact((a-b)!)$
$fact[i]$数组记录$i!$、$infact[i]$记录$(i!)^{-1}$
逆元

封装

typedef long long LL;
const int N = 1e5+10, mod = 1e9+7;
int fact[N], infact[N];
LL qmi(int a, int b, int p)
{
    LL res = 1 % p;
    while(b){
        if(b & 1) res = (LL)res * a % p;
        a = (LL)a * a % p;
        b >>= 1;
    }
    return res;
}
int C(int a, int b)
{
    return ((LL)fact[a] * infact[b] % mod * infact[a - b]) % mod;
}
void solve()
{
    memset(fact, 0, sizeof fact);
    memset(infact, 0, sizeof infact);
    fact[0] = 1, infact[0] = 1;
    for (int i = 1; i < N; i ++)
    {
        fact[i] = (LL)i * fact[i - 1] % mod;
        infact[i] = (LL)infact[i - 1] * qmi(i, mod - 2, mod) % mod;
    }

    // 调用C(a,b)
    int n;
    scanf("%d", &n);
    while (n --)
    {
        int a, b;
        scanf("%d %d", &a, &b);
        printf("%lld\n", C(a, b));
    }
}

套板子题：$lc2541$