数学题

原创

锐* 2021-12-27 14:56:53 博主文章分类：数论 ©著作权

文章标签 #include 等比数列 #define 文章分类 C/C++ 后端开发

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也许更好的阅读体验

D e s c r i p t i o n \mathcal{Description} Description

给 n , b , k n,b,k n,b,k,求

∑ i = 1 n ( i k ) b i \begin{aligned}\sum_{i=1}^n\begin{pmatrix}i\\k \end{pmatrix}b^i\end{aligned} i=1∑n(ik)bi

答案对 998244353 998244353 998244353取模

1 ≤ k ≤ 500000 , 1 ≤ n ≤ 1 0 18 , 2 ≤ b ≤ 998244353 1\leq k\leq 500000,1\leq n\leq 10^{18},2\leq b\leq 998244353 1≤k≤500000,1≤n≤1018,2≤b≤998244353

S o l u t i o n \mathcal{Solution} Solution

设

f k = ∑ i = 1 n ( i k ) b i \begin{aligned}f_{k}=\sum ^{n}_{i=1}\begin{pmatrix} i \\ k \end{pmatrix}b^{i}\end{aligned} fk=i=1∑n(ik)bi

则有

f k − 1 = ∑ i = 1 n ( i k − 1 ) b i \begin{aligned}f_{k-1}=\sum ^{n}_{i=1}\begin{pmatrix} i \\ k-1 \end{pmatrix}b^{i}\end{aligned} fk−1=i=1∑n(ik−1)bi

b f k = ∑ i = 2 n + 1 ( i − 1 k ) b i \begin{aligned}bf_{k}=\sum ^{n+1}_{i=2}\begin{pmatrix} i-1 \\ k \end{pmatrix}b^{i}\end{aligned} bfk=i=2∑n+1(i−1k)bi

b f k − 1 = ∑ i = 2 n + 1 ( i − 1 k − 1 ) b i \begin{aligned}bf_{k-1}=\sum ^{n+1}_{i=2}\begin{pmatrix} i-1 \\ k-1 \end{pmatrix}b^{i}\end{aligned} bfk−1=i=2∑n+1(i−1k−1)bi

b f k + b f k − 1 = ∑ i = 2 n + 1 ( ( i − 1 k ) + ( i − 1 k − 1 ) ) b i = ∑ i = 2 n + 1 ( i k ) b i = f k + ( n + 1 k ) b n + 1 − ( 1 k ) b \begin{aligned}bf_{k}+bf_{k-1}=\sum ^{n+1}_{i=2}\left(\begin{pmatrix} i-1 \\ k \end{pmatrix}+\begin{pmatrix} i-1 \\ k-1 \end{pmatrix}\right)b^{i}=\sum ^{n+1}_{i=2}\begin{pmatrix} i \\ k \end{pmatrix}b^{i}=f_k+\begin{pmatrix}n+1\\ k\end{pmatrix}b^{n+1}-\begin{pmatrix}1\\k\end{pmatrix}b\end{aligned} bfk+bfk−1=i=2∑n+1((i−1k)+(i−1k−1))bi=i=2∑n+1(ik)bi=fk+(n+1k)bn+1−(1k)b

移项得

( b − 1 ) f k = ( n + 1 k ) b n + 1 − ( 1 k ) b − b f k − 1 \left(b-1\right)f_k=\begin{pmatrix}n+1\\k\end{pmatrix}b^{n+1}-\begin{pmatrix}1\\k\end{pmatrix}b-bf_{k-1} (b−1)fk=(n+1k)bn+1−(1k)b−bfk−1

f k = ( n + 1 k ) b n + 1 − ( 1 k ) b − b f k − 1 b − 1 f_k=\dfrac{\begin{pmatrix}n+1\\k\end{pmatrix}b^{n+1}-\begin{pmatrix}1\\k\end{pmatrix}b-bf_{k-1}}{b-1} fk=b−1(n+1k)bn+1−(1k)b−bfk−1

当 k = 0 k=0 k=0时，要求的就是一个等比数列，我们用等比数列求和公式直接算就可以了

f 0 = b n + 1 − b b − 1 f_0=\dfrac{b^{n+1}-b}{b-1} f0=b−1bn+1−b

总复杂度 O ( k ) O(k) O(k)

C o d e \mathcal{Code} Code

/*******************************
Author:Morning_Glory
LANG:C++
Created Time:2019年09月17日 星期二 10时54分50秒
*******************************/
#include <cstdio>
#include <fstream>
#define ll long long
using namespace std;
const int mod = 998244353;
const int maxn = 500005;
ll n,b,k,ans;
ll invb,mi,C;
ll inv[maxn];
//{{{ksm
ll ksm (ll a,ll b)
{
  a%=mod;
  ll s=1;
  for (;b;b>>=1,a=a*a%mod)
    if (b&1)  s=s*a%mod;
  return s;
}
//}}}
int main()
{
  scanf("%lld%lld%lld",&n,&b,&k);
  C=(n+1)%mod,inv[1]=1,invb=ksm(b-1,mod-2);
  for (int i=2;i<=k;++i)  inv[i]=(-mod/i*inv[mod%i]%mod+mod)%mod;
  ans=(ksm(b,n+1)-b+mod)%mod*invb%mod;
  mi=ksm(b,n+1);
  if (n<k){ printf("0\n");return 0;}
  if (n==k){  printf("%lld\n",ksm(b,n));return 0;}
  for (int i=1;i<=k;++i){
    ll t=0;
    if (i<=1) t=b;
    ans=((C*mi%mod-t-b*ans%mod)%mod+mod)%mod*invb%mod;
    C=(n-i+1)%mod*C%mod*inv[i+1]%mod;
  }
  printf("%lld\n",ans);
  return 0;
}