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259 lines
6.5 KiB
Python
Executable file
259 lines
6.5 KiB
Python
Executable file
"""#
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### 谜题描述
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Everyone knows what the Fibonacci sequence is. This sequence can be defined by the recurrence relation:
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F1 = 1, F2 = 2, Fi = Fi - 1 + Fi - 2 (i > 2).
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We'll define a new number sequence Ai(k) by the formula:
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Ai(k) = Fi × ik (i ≥ 1).
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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).
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Input
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The first line contains two space-separated integers n, k (1 ≤ n ≤ 1017; 1 ≤ k ≤ 40).
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Output
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Print a single integer — the sum of the first n elements of the sequence Ai(k) modulo 1000000007 (109 + 7).
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Examples
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Input
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1 1
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Output
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1
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Input
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4 1
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Output
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34
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Input
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5 2
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Output
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316
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Input
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7 4
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Output
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73825
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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#include <bits/stdc++.h>
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using namespace std;
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struct matrix {
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long long m[90][90];
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matrix() { memset(m, 0, sizeof m); }
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long long *operator[](int i) { return m[i]; }
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friend matrix operator*(const matrix &a, const matrix &b) {
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matrix r;
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for (int i = (0); i < (90); ++i)
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for (int j = (0); j < (90); ++j)
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for (int k = (0); k < (90); ++k)
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r[i][j] =
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(r[i][j] + ((long long)a.m[i][k]) * b.m[k][j]) % (int)(1e9 + 7);
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return r;
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}
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};
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matrix pow(matrix a, long long e) {
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if (e == 1) return a;
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if (e % 2) return a * pow(a, e - 1);
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a = pow(a, e / 2);
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return a * a;
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}
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long long n;
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int k;
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long long C[90][90];
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int main() {
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scanf(\"%I64d %d\", &n, &k);
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if (n == 1) {
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printf(\"1\n\");
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return 0;
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}
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for (int i = 0; i <= k; ++i) {
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C[i][0] = 1;
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for (int j = 1; j <= i; ++j)
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C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % (int)(1e9 + 7);
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}
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matrix a, b;
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k++;
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for (int i = 0; i < k; ++i) {
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a[i][0] = 2;
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a[i + k][0] = 1;
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for (int j = 0; j < i; ++j) a[i][0] = (2L * a[i][0]) % (int)(1e9 + 7);
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}
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a[2 * k][0] = 1;
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for (int i = 0; i < k; ++i) {
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for (int j = 0; j <= i; ++j) {
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b[i][j] = b[i][j + k] = C[i][j];
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for (int d = 0; d < i - j; ++d)
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b[i][j + k] = (2L * b[i][j + k]) % (int)(1e9 + 7);
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}
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b[i + k][i] = 1;
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}
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b[2 * k][k - 1] = 1;
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b[2 * k][2 * k] = 1;
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matrix ret = pow(b, n - 1) * a;
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printf(\"%d\n\", (int)ret[2 * k][0]);
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return 0;
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}
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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from bootcamp import Basebootcamp
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import random
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import sys
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MOD = 10**9 + 7
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class Eyetanothernumbersequencebootcamp(Basebootcamp):
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def __init__(self, max_n=10**17, max_k=40, random_seed=None):
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super().__init__()
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self.max_n = max_n
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self.max_k = max_k
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self.rng = random.Random(random_seed)
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def case_generator(self):
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# 生成满足 1 ≤ n ≤ 1e17 的真实大数
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exp = self.rng.randint(0, 17)
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n = self.rng.randint(10**exp, min(10**(exp + 1), self.max_n))
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k = self.rng.randint(1, self.max_k)
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return {
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'n': n,
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'k': k,
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'correct_answer': self._solve(n, k)
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}
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@staticmethod
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def _solve(n, k):
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if n == 1:
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return 1 % MOD
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k_size = k + 1
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matrix_dim = 2 * k_size + 1 # 修正矩阵维度计算
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# 修正组合数表维度 (+2保证边界)
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C = [[0]*(k_size+2) for _ in range(k_size+2)]
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for i in range(k_size+1):
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C[i][0] = 1
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for j in range(1, i+1):
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C[i][j] = (C[i-1][j] + C[i-1][j-1]) % MOD
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class Matrix:
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__slots__ = ['data']
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def __init__(self):
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self.data = [[0]*matrix_dim for _ in range(matrix_dim)]
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def __matmul__(self, other):
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res = Matrix()
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for i in range(matrix_dim):
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for k in range(matrix_dim):
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if self.data[i][k]:
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for j in range(matrix_dim):
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res.data[i][j] = (res.data[i][j] + self.data[i][k] * other.data[k][j]) % MOD
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return res
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def fast_pow(mat, power):
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result = Matrix()
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for i in range(matrix_dim):
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result.data[i][i] = 1
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while power > 0:
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if power & 1:
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result = result @ mat
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mat = mat @ mat
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power >>= 1
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return result
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# 初始化矩阵A
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a = Matrix()
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for i in range(k_size):
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a.data[i][0] = 2
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a.data[i + k_size][0] = 1
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for _ in range(i):
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a.data[i][0] = (a.data[i][0] << 1) % MOD
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a.data[2*k_size][0] = 1
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# 初始化转移矩阵B
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b = Matrix()
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for i in range(k_size):
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for j in range(i+1):
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b.data[i][j] = C[i][j]
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b.data[i][j + k_size] = C[i][j]
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pw = pow(2, i-j, MOD)
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b.data[i][j + k_size] = (b.data[i][j + k_size] * pw) % MOD
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b.data[i + k_size][i] = 1
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b.data[2*k_size][k_size-1] = 1
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b.data[2*k_size][2*k_size] = 1
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# 矩阵快速幂计算
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res_matrix = fast_pow(b, n-1)
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final = res_matrix @ a
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return final.data[2*k_size][0] % MOD
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@staticmethod
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def prompt_func(case):
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return f"""Calculate S = Σ(F_i × i^{case['k']}) for i=1..{case['n']} mod 1e9+7
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Where Fibonacci sequence is defined as:
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- F₁ = 1, F₂ = 2
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- Fₙ = Fₙ₋₁ + Fₙ₋₂ for n > 2
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Input constraints:
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- 1 ≤ n ≤ 1e17
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- 1 ≤ k ≤ 40
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Output the answer within [answer] tags, e.g. [answer]123456789[/answer]"""
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@staticmethod
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def extract_output(output):
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import re
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answers = re.findall(r'\[answer\]\s*(\d+)\s*\[/answer\]', output)
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return int(answers[-1]) if answers else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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try:
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return int(solution) % MOD == identity['correct_answer']
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except:
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return False
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# 验证测试用例
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if __name__ == "__main__":
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test_cases = [
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(1, 1, 1),
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(4, 1, 34),
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(5, 2, 316),
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(7, 4, 73825),
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(10**17, 40, 370483476) # 预计算验证值
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]
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bootcamp = Eyetanothernumbersequencebootcamp()
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for n, k, expect in test_cases:
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result = bootcamp._solve(n, k)
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assert result == expect, f"Failed: n={n},k={k} expect {expect} got {result}"
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print("All test cases passed!")
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