题意

给出 \(n,k\) ， \(n\le10^9,k\le10^6\) ，求 \(\sum_{i=1}^n i^k(mod\;10^9+7)\)

题解

自然数幂次和，是一个\(k+1\)次多项式，那么算出\(k+2\)个值然后差值就行了

//minamoto#include
    
     #define R register#define fp(i,a,b) for(R int i=a,I=b+1;i
     
      I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)using namespace std;const int N=1e6+5,P=1e9+7;inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}int ksm(R int x,R int y){    R int res=1;    for(;y;y>>=1,x=mul(x,x))if(y&1)res=mul(res,x);    return res;}int f[N],inv[N];int n,k;inline int Inv(R int x){return x<=k?inv[x]:ksm(x,P-2);}int Large(int k,int n){    if(k<=n)return f[k];    int ty=(n&1)?P-1:1,tmp=1,res=0;    fp(i,1,n)tmp=1ll*tmp*(k-i)%P*Inv(i)%P;    fp(i,0,n){        res=add(res,1ll*f[i]*tmp%P*ty%P);        tmp=1ll*tmp*(k-i)%P*Inv(k-i-1)%P*(n-i)%P*Inv(i+1)%P;        ty=P-ty;    }    return res;}int main(){//  freopen("testdata.in","r",stdin);    scanf("%d%d",&n,&k);    inv[0]=inv[1]=1;fp(i,2,k)inv[i]=1ll*inv[P%i]*(P-P/i)%P;    fp(i,1,k+1)f[i]=add(f[i-1],ksm(i,k));    printf("%d\n",Large(n,k+1));    return 0;}