Yet Another Number Sequence CodeForces

题目链接:点我

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Everyone knows what the Fibonacci sequence is. This sequence can be defined by the recurrence relation:F1 = 1, F2 = 2, Fi = Fi - 1 + Fi - 2 (i > 2).We'll define a new number sequence Ai(k) by the formula:Ai(k) = Fi × i^k (i ≥ 1).In this problem, your task is to calculate the following sum: A1(k) + A2(k) + ... + An(k). The answer can be very large, so print it modulo 1000000007 (109 + 7).

Input

The first line contains two space-separated integers n, k (1 ≤ n ≤ 1e17; 1 ≤ k ≤ 40).

Output

Print a single integer — the sum of the first n elements of the sequence Ai(k) modulo 1000000007 (1e9 + 7).

Example

Input1 1Output1Input4 1Output34Input5 2Output316Input7 4Output73825

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题意:

如果提上公式所说:计算 A1(k) + A2(k) + … + An(k), Ai(k) = Fi × i^k.

思路:

矩阵快速幂.
首先我们应该知道,(n+1)k =(kk)nk +(kk−1) nk−1 + (kk−2)nk−2 + ….. +(k1) n +(k0) n0
我们令u(n+1,k) = (n+1)k * F(n+1), v(n+1,k) = (n+1)k * F(n)
根据斐波那契数列递推式:
u(n+1,k) = (n+1)k * F(n+1)
= (n+1)k * F(n) + (n+1)k * F(n-1)
= ∑ki=0 (ki) * ni * F(n) + ∑ki=0 (ki) * ni * F(n-1)
= ∑ki=0 (ki) * ni * F(n) + ∑ki=0 (ki) * v(n,i)
v(n+1,k) = (n+1)k * F(n)
= ∑ki=0 (ki) * ni * F(n)
= ∑ki=0 (ki) * u(n,i)
令Sn=A1(k) + A2(k) + ... + An(k)
所以 Sn+1=Sn+u(n,k),
那么我们构造一个(2K+3) * (2k+3)的矩阵即可,这个矩阵大家根据上面的递推式在纸上写写就出来了,
代码:

#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<cmath>#include<queue>using namespace std;typedef long long LL;const int mod = 1e9+7;int c[45][45];LL m;struct mat{    LL a[85][85];    mat (){memset(a,0 ,sizeof(a)); }    mat operator *(const mat q){        mat c;        for(int i = 1; i <= m; ++i)            for(int k = 1; k <= m; ++k)            if(a[i][k])        for(int j = 1; j <= m; ++j){            c.a[i][j] += a[i][k] * q.a[k][j];            if (c.a[i][j] >= mod) c.a[i][j] %= mod;        }return c;    }};mat qpow(mat x, LL n){    mat ans;    for(int i = 1; i <= 84; ++i)        ans.a[i][i] = 1;    while(n){        if (n&1) ans = ans * x;        x = x * x;        n >>= 1;    }return ans;}void init(){    for(int i = 0; i <= 41; ++i)        c[i][0] = c[i][i] = 1;    for(int i = 1; i <= 41; ++i)        for(int j = 1; j < i; ++j)        c[i][j] = (c[i-1][j] + c[i-1][j-1]) % mod;}int main(){    LL n, k;    init();    scanf("%lld %lld", &n, &k);    mat ans;    m = 2 * k + 3;    ans.a[1][1] = 1;    for(int i = 2; i <= m; ++i){        ans.a[i][1] = 0;        ans.a[1][i] = c[k][(i-2)%(k+1)];    }for(int i = 2; i <= k+2; ++i)        for(int j = 2; j <= i;++j)            ans.a[i][j] = ans.a[i][j+k+1] = ans.a[i+k+1][j] =  c[i-2][j-2];    ans = qpow(ans,n-1);    LL sum = 0;    for(int i = 1; i <= m; ++i)        sum = (sum + ans.a[1][i]) % mod;    printf("%lld\n",sum);    return 0;}