【 CodeForces

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[5];

LL c[Maxn][Maxn];
LL n,k;

void init()
{
    memset(c,0,sizeof(c));
    for(LL i=0;i<=80;i++) c[i][0]=1;
    for(LL i=1;i<=80;i++)
     for(LL j=1;j<=80;j++)
        c[i][j]=(c[i-1][j-1]+c[i-1][j])%Mod;
}

void get_m()
{
    for(LL i=k+1;i<=2*k+1;i++)
    {
        for(LL j=0;j<=i-k-1;j++) t[0].a[i][j]=c[i-k-1][j];
        for(LL j=i+1;j<=2*k+2;j++) t[0].a[i][j]=0;
    }
    for(LL i=0;i<=k;i++)
    {
        for(LL j=0;j<=i;j++) t[0].a[i][j]=t[0].a[i][j+k+1]=c[i][j];
        for(LL j=i+1;j<=k;j++) t[0].a[i][j]=t[0].a[i][j+k+1]=0;
        t[0].a[i][2*k+2]=0;
    }
    for(LL i=0;i<=2*k+1;i++) t[0].a[2*k+2][i]=0;
    t[0].a[2*k+2][2*k+2]=t[0].a[2*k+2][k]=1;
}

void get_un()
{
    memset(t[1].a,0,sizeof(t[1].a));
    for(LL i=0;i<=2*k+2;i++) t[1].a[i][i]=1;
}

void mul(LL x,LL y,LL z)
{
    for(LL i=0;i<=2*k+2;i++)
     for(LL j=0;j<=2*k+2;j++)
     {
         t[2].a[i][j]=0;
         for(LL l=0;l<=2*k+2;l++)
            t[2].a[i][j]=(t[2].a[i][j]+t[y].a[i][l]*t[z].a[l][j])%Mod;
     }
    t[x]=t[2];
}

void qpow(LL b)
{
    get_un();
    while(b)
    {
        if(b&1) mul(1,0,1);
        mul(0,0,0);
        b>>=1;
    }
}

int main()
{
    init();
    scanf("%lld%lld",&n,&k);
    get_m();
    qpow(n);
    LL ans=0;
    for(LL i=0;i<2*k+2;i++) ans=(ans+t[1].a[2*k+2][i])%Mod;
    printf("%lld
",ans);
    return 0;
}

2016-09-26 16:11:26