[洛谷4948]数列求和

题目传送门：https://www.luogu.org/problemnew/show/P4948

题目大意：令\(A_n=n^k\times q^n\)，求\(\sum\limits_{i=1}^nA_i\)

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其实很久以前数学课学了数列就开始想这题了……

但最近才会解法……

我们令\(S_k(n)=\sum\limits_{i=1}^ni^k\times q^i\)，对其扰动可得

\[\begin{align}S_k(n)&=\sum\limits_{i=1}^n(i+1)^k\times q^{i+1}-(n+1)^k\times q^{n+1}+q\nonumber\\&=\sum\limits_{i=1}^n\sum\limits_{j=0}^k\binom{k}{j}i^j\times q^{i+1}-(n+1)^k\times q^{n+1}+q\nonumber\\&=q\sum\limits_{j=0}^k\binom{k}{j}S_j(n)-(n+1)^k\times q^{n+1}+q\nonumber\end{align} \]

所以我们就得到了\(S_k(n)\)的递推式

\[S_k(n)=\dfrac{(n+1)^k\times q^{n+1}-\sum\limits_{j=0}^{k-1}\binom{k}{j}S_j(n)-q}{q-1} \]

但是这仅限于\(q>1\)的情况，那么\(q=1\)的情况呢，那就是幂和的形式，具体方法可以看这篇博客

/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=2e3,p=1e9+7;
int fac[N+10],inv[N+10],T[N+10];
void prepare(){
	fac[0]=inv[0]=inv[1]=1;
	for (int i=1;i<=N;i++)	fac[i]=1ll*i*fac[i-1]%p;
	for (int i=2;i<=N;i++)	inv[i]=1ll*(p-p/i)*inv[p%i]%p;
	for (int i=1;i<=N;i++)	inv[i]=1ll*inv[i-1]*inv[i]%p;
}
int C(int n,int m){return 1ll*fac[n]*inv[m]%p*inv[n-m]%p;}
int mlt(ll a,ll b){
	int res=1; a%=p; b%=(p-1);
	for (;b;b>>=1,a=1ll*a*a%p)	if (b&1)	res=1ll*res*a%p;
	return res;
}
int main(){
	prepare();
	ll n=read(0ll); int a=read(0),k=read(0);
	if (a==1){
		T[0]=n%p;
		for (int i=1;i<=k;i++){
			int res=0;
			for (int j=0;j<i;j++)	res=(1ll*C(i+1,j)*T[j]+res)%p;
			T[i]=1ll*(mlt(n+1,i+1)-res-1)*mlt(i+1,p-2)%p;
		}
		printf("%d\n",(T[k]+p)%p);
	}else{
		T[0]=1ll*(a-mlt(a,n+1))*mlt(1-a,p-2)%p;
		for (int i=1;i<=k;i++){
			int res=0;
			for (int j=0;j<i;j++)	res=(1ll*C(i,j)*T[j]+res)%p;
			res=1ll*res*a%p;
			T[i]=1ll*(1ll*mlt(n+1,i)*mlt(a,n+1)-a-res)%p*mlt(a-1,p-2)%p;
		}
		printf("%d\n",(T[k]+p)%p);
	}
	return 0;
}