【 CodeForces - 392C】 Yet Another Number Sequence （二项式展开+矩阵加速）

Yet Another Number Sequence

Description

Everyone knows what the Fibonacci sequence is. This sequence can be defined by the recurrence relation:

F₁ = 1, F₂ = 2, F_i = F_i - 1 + F_i - 2 (i > 2).

We'll define a new number sequence A_i(k) by the formula:

A_i(k) = F_i × i^k (i ≥ 1).

In this problem, your task is to calculate the following sum: A₁(k) + A₂(k) + ... + A_n(k). The answer can be very large, so print it modulo1000000007 (10⁹ + 7).

Input

The first line contains two space-separated integers n, k (1 ≤ n ≤ 10¹⁷; 1 ≤ k ≤ 40).

Output

Print a single integer — the sum of the first n elements of the sequence A_i(k) modulo 1000000007 (10⁹ + 7).
Sample Input
Input
1 1
Output
1
Input
4 1
Output
34
Input
5 2
Output
316
Input
7 4
Output
73825

【分析】

　　哈哈照着上一题的方法我就弄出来了~~
　　应该是形如 x^k的形式，x很大，k较小的时候可以用二项式定理展开，求递推式然后矩阵加速。。
　　


　　就这样，qpow n次就好啦~

代码如下：

 #include<cstdio>

 #include<cstdlib>

 #include<cstring>

 #include<iostream>

 #include<algorithm>

 #include<queue>

 #include<cmath>

 using namespace std;

 #define Mod 1000000007

 #define Maxn 110

 #define LL long long

 struct node

 {

     LL a[Maxn][Maxn];

 }t[];

 LL c[Maxn][Maxn];

 LL n,k;

 void init()

 {

     memset(c,,sizeof(c));

     for(LL i=;i<=;i++) c[i][]=;

     for(LL i=;i<=;i++)

      for(LL j=;j<=;j++)

         c[i][j]=(c[i-][j-]+c[i-][j])%Mod;

 }

 void get_m()

 {

     for(LL i=k+;i<=*k+;i++)

     {

         for(LL j=;j<=i-k-;j++) t[].a[i][j]=c[i-k-][j];

         for(LL j=i+;j<=*k+;j++) t[].a[i][j]=;

     }

     for(LL i=;i<=k;i++)

     {

         for(LL j=;j<=i;j++) t[].a[i][j]=t[].a[i][j+k+]=c[i][j];

         for(LL j=i+;j<=k;j++) t[].a[i][j]=t[].a[i][j+k+]=;

         t[].a[i][*k+]=;

     }

     for(LL i=;i<=*k+;i++) t[].a[*k+][i]=;

     t[].a[*k+][*k+]=t[].a[*k+][k]=;

 }

 void get_un()

 {

     memset(t[].a,,sizeof(t[].a));

     for(LL i=;i<=*k+;i++) t[].a[i][i]=;

 }

 void mul(LL x,LL y,LL z)

 {

     for(LL i=;i<=*k+;i++)

      for(LL j=;j<=*k+;j++)

      {

          t[].a[i][j]=;

          for(LL l=;l<=*k+;l++)

             t[].a[i][j]=(t[].a[i][j]+t[y].a[i][l]*t[z].a[l][j])%Mod;

      }

     t[x]=t[];

 }

 void qpow(LL b)

 {

     get_un();

     while(b)

     {

         if(b&) mul(,,);

         mul(,,);

         b>>=;

     }

 }

 int main()

 {

     init();

     scanf("%lld%lld",&n,&k);

     get_m();

     qpow(n);

     LL ans=;

     for(LL i=;i<*k+;i++) ans=(ans+t[].a[*k+][i])%Mod;

     printf("%lld\n",ans);

     return ;

 }

2016-09-26 16:11:26