题解:
很多方法
斯特林数推导略麻烦但是不依赖于模数
代码:
拉格朗日插值
由于可以证明这是个K+1次多项式于是可以直接用插值
#include <bits/stdc++.h>
using namespace std;
const int mo=1e9+;
#define IL inline
#define ll long long
#define rint register int
#define rep(i,h,t) for (rint i=h;i<=t;i++)
#define dep(i,t,h) for (rint i=t;i>=h;i--)
const int N=2e6;
ll f[N],jc[N];
ll fst(ll x,ll y)
{
if (y==) return();
if (y==) return(x);
ll kk=fst(x,y/);
kk=(kk*kk)%mo;
if (y%) kk=(kk*x)%mo;
return kk;
}
int main()
{
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
ios::sync_with_stdio(false);
ll n,k;
cin>>n>>k;
rep(i,,k+)
f[i]=(f[i-]+fst(i*1ll,k))%mo;
if (n<=k+)
{
cout<<f[n]<<endl;
return ;
}
jc[]=;
rep(i,,k+) jc[i]=(jc[i-]*i)%mo;
ll now=,ans=;
rep(i,,k+) now=(now*(n-i))%mo;
rep(i,,k+)
{
ll inv1=fst(n-i,mo-);
ll inv2=fst((jc[i-]*jc[k+-i])%mo,mo-)%mo;
ll sign=(k+-i)%?-:;
ans=(ans+sign*inv1*inv2%mo*f[i]%mo*now%mo)%mo;
}
cout<<(ans+mo)%mo;
return ;
}