令\[S_i=\sum_{k=1}^n k^i m^k\]我们有\[\begin{eqnarray*}(m-1)S_i & = & mS_i - S_i \\& = & \sum_{k=1}^n k^i m^{k+1} - \sum_{k=1}^n k^i m^k \\& = & \sum_{k=2}^{n+1} (k-1)^i m^k - \sum_{k=1}^n k^i m^k \\& = & n^i m^{n+1} + \sum_{k=1}^n m^k\big( (k-1)^i - k^i\big) \\& = & n^i m^{n+1} + \sum_{k=1}^n \bigg( \sum_{j=1}^{i-1} (-1)^{i-j}{i \choose j}k^j m^k \bigg) \\& = & n^i m^{n+1} + \sum_{j=0}^{i-1} (-1)^{i-j}{i \choose j} S_j\end{eqnarray*}\]直接按照这条式子\(O(m^2)\)递推即可。

P.S. 来自

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 using namespace std; 6 typedef long long LL; 7 const LL mod=1000000007; 8 inline LL sqr(LL num){
   return num*num;} 9 LL mexp(LL a,LL b,LL p){
   return b?sqr(mexp(a,b>>1,p))%p*((b&1)?a:1)%p:1;}10 LL fac[1100],facinv[1100],powm1[1100];11 LL C(LL n,LL r){
   return fac[n]*facinv[r]%mod*facinv[n-r]%mod;}12 LL n,m,s[1100];13 int main(int argc, char *argv[])14 {15     cin>>n>>m;16     if(m==1){cout<
         
          =1;i--)facinv[i]=facinv[i+1]*(i+1)%mod;20     powm1[0]=1;for(int i=1;i<=m;i++)powm1[i]=powm1[i-1]*-1;21     s[0]=((mexp(m,n+1,mod)-m)%mod+mod)%mod*mexp(m-1,mod-2,mod);22     for(int i=1;i<=m;i++)23     {24         s[i]=mexp(n,i,mod)*mexp(m,n+1,mod)%mod;25         for(int j=0;j