组合数

\(O(n^2)\)：

ll c[1005][1005]; 
const int mod=1e9+7;
    
for (int i=0; i<=1000; ++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])%mod;
}

\(O(n+log(mod))\)：

const int maxn=2e5+5;
const int mod=998244353;
ll fac[maxn];//阶乘
ll qpow(ll x, ll n)
{
    ll ans=1;
    while(n>0){
        if(n&1) ans=(ans*x)%mod;
        x=(x*x)%mod,n>>=1;
    }
    return ans;
}
void init()
{
    fac[0]=1;
    for(int i=1;i<maxn;i++)
        fac[i]=(fac[i-1])*i%mod;
}
ll C(ll n, ll k)//从n个数里选k个数
{
    return fac[n]*qpow(fac[k]*fac[n-k]%mod, mod-2)%mod;
}