求自然数幂和 B - The Sum of the k-th Powers CodeForces - 622F

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+7;
#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==0) return(1);
  if (y==1) return(x);
  ll kk=fst(x,y/2);
  kk=(kk*kk)%mo;
  if (y%2) 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,1,k+2)
    f[i]=(f[i-1]+fst(i*1ll,k))%mo;
  if (n<=k+2)
  {
    cout<<f[n]<<endl;
    return 0;
  }
  jc[0]=1;
  rep(i,1,k+2) jc[i]=(jc[i-1]*i)%mo;
  ll now=1,ans=0;
  rep(i,1,k+2) now=(now*(n-i))%mo;
  rep(i,1,k+2)
  {
    ll inv1=fst(n-i,mo-2);
    ll inv2=fst((jc[i-1]*jc[k+2-i])%mo,mo-2)%mo;
    ll sign=(k+2-i)%2?-1:1;
    ans=(ans+sign*inv1*inv2%mo*f[i]%mo*now%mo)%mo;
  }
  cout<<(ans+mo)%mo;
  return 0;
}