luogu P4948 数列求和 推式子 简单数学推导 二项式 拉格朗日插值

LINK:数列求和

每次遇到这种题目都不太会写。但是做法很简单.

终有一天我会成功的。

考虑类似等比数列求和的东西 帽子戏法一下.

设(f(k)=sum_{i=1}^ni^ka^i)

考虑(af(k))这个式子 两式做差.

((a-1)f(k)=n^ncdot a^{n+1}-a+sum_{i=2}^n{a^i((i-1)^k-i^k)})

右边直接二项式展开 然后 交换求和顺序可得.

((a-1)f(k)=n^kcdot a^{n+1}-a+sum_{j=0}^{k-1}C(k,j)(-1)^{k-j}(f_j-a))

然后除以a-1就可以(k^2)推了.

注意a==1时 也可以类似这样推式子 不过 直接拉格朗日插值就最简单的做法。

code

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<ctime>
#include<cctype>
#include<queue>
#include<deque>
#include<stack>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<string>
#include<ctime>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<queue>
#include<deque>
#include<stack>
#include<vector>
#include<algorithm>
#include<utility>
#include<bitset>
#include<set>
#include<map>
#define ll long long
#define db double
#define INF 10000000000000000ll
#define inf 1000000000
#define ldb long double
#define pb push_back
#define put_(x) printf("%d ",x);
#define get(x) x=read()
#define gt(x) scanf("%d",&x)
#define gi(x) scanf("%lf",&x)
#define put(x) printf("%d
",x)
#define putl(x) printf("%lld
",x)
#define rep(p,n,i) for(RE ll i=p;i<=n;++i)
#define go(x) for(ll i=lin[x],tn=ver[i];i;tn=ver[i=nex[i]])
#define fep(n,p,i) for(RE ll i=n;i>=p;--i)
#define vep(p,n,i) for(RE ll i=p;i<n;++i)
#define pii pair<ll,ll>
#define mk make_pair
#define RE register
#define P 1000000007ll
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define uint unsigned long long
#define ui unsigned
#define EPS 1e-10
#define sq sqrt
#define S second
#define F first
#define mod 1000000007
using namespace std;
char *fs,*ft,buf[1<<15];
inline char gc()
{
	return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;
}
inline ll read()
{
	RE ll x=0,f=1;RE 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;

}
const ll MAXN=2010;
ll n,k,a;
ll f[MAXN],fac[MAXN],inv[MAXN];
inline ll ksm(ll b,ll p)
{
	ll cnt=1;
	while(p)
	{
		if(p&1)cnt=cnt*b%mod;
		b=b*b%mod;p=p>>1;
	}
	return cnt;
}
inline ll C(ll a,ll b){return a<b?0:fac[a]*inv[b]%mod*inv[a-b]%mod;}
inline ll lag(ll x,ll n)
{
	ll ans=0;x%=mod;
	rep(1,n,i)
	{
		ll sum=1,ww=1;
		f[i]=(f[i-1]+ksm(i,k))%mod;
		rep(1,n,j)
		{
			if(i==j)continue;
			sum=sum*(x-j)%mod;
			ww=ww*(i-j)%mod;
		}
		sum=sum*ksm(ww,mod-2)%mod;
		ans=(ans+f[i]*sum)%mod;
	}
	return (ans+mod)%mod;
}
inline void solve_1()
{
	putl(lag(n,k+2));
}
inline void solve_2()
{
	fac[0]=1;
	rep(1,k,i)fac[i]=fac[i-1]*i%mod;
	inv[k]=ksm(fac[k],mod-2);
	fep(k-1,0,i)inv[i]=inv[i+1]*(i+1)%mod;
	ll IN=ksm(a-1,mod-2),cc;
	f[0]=((cc=ksm(a,(n+1)%(mod-1)))-a)*IN%mod;
	n%=mod;
	rep(1,k,i)
	{
		cc=cc*n%mod;ll ans=cc-a;
		rep(0,i-1,j)ans=(ans+C(i,j)*((i-j)&1?-1:1)*(f[j]-a))%mod;
		f[i]=ans*IN%mod;
	}
	putl((f[k]+mod)%mod);
}
signed main()
{
	//freopen("1.in","r",stdin);
	get(n);get(a);get(k);
	if(a==1)solve_1();
	else solve_2();
	return 0;
}