51nod1228 Sequence Summation (Bernoulli + Math)

beg$S(n,k)=\sum\limits_{i=1}^ni^k$
Can$O(k^2)$recursion.
It can be done with Bernoulli numbers$O(k)$
$\sum\limits_{i=1}^ni^k=\frac{1}{k+1}\sum\limits_{i=1}^{k+1}C_{k+1}^i*B_{k+1-i}*(n+1)^i$
The Bernoulli number satisfies$B_0=1$
$\sum\limits_{k=0}^nC_{n+1}^kB_k=0$
So we have the recurrence for Bernoulli numbers:
$B_n=-\frac{1}{n+1}\sum\limits_{i=0}^{n-1}C_{n+1}^i*B_i$

Therefore, the preprocessing combination number, inverse element, and Bernoulli number can be used.$O(T*K)$

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 2010
#define mod 1000000007
inline char gc(){
    static char buf[1<<16],*S,*T;
    if(S==T){T=(S=buf)+fread(buf,1,1<<16,stdin);if(T==S) return EOF;}
    return *S++;
}
inline ll read(){
    ll x=0,f=1;char ch=gc();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=gc();
    return x*f;
}
int C[N][N],inv[N],B[N],n,K;
inline int cal(){
    int res=0,tmp=n+1;
    for(int i=1;i<=K+1;++i)
        (res+=(ll)C[K+1][i]*B[K+1-i]%mod*tmp%mod)%=mod,tmp=(ll)tmp*(n+1)%mod;
    res=(ll)res*inv[K+1]%mod;return res;
}
int main(){
//  freopen("a.in","r",stdin);
    int tst=read();inv[1]=1;
    for(int i=2;i<=2001;++i) inv[i]=(ll)inv[mod%i]*(mod-mod/i)%mod;
    for(int i=0;i<=2001;++i) C[i][0]=1;
    for(int i=1;i<=2001;++i)
        for(int j=1;j<=i;++j) C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
    B[0]=1;
    for(int i=1;i<=2000;++i){
        for(int j=0;j<=i-1;++j) (B[i]+=(ll)C[i+1][j]*B[j]%mod)%=mod;
        B[i]=(ll)B[i]*inv[i+1]*-1%mod;if(B[i]<0) B[i]+=mod;
    }while(tst--){
        n=read()%mod;K=read();printf("%d\n",cal());
    }return 0;
}