最近笔试期间遇到一个难题,现在终于解决了,感谢各路大佬的指点,我在这里分享一下结果。
小红拿到一个数列满足:
f(1) = a; f(2) = b; f(i) = f(i-1) * f(i-2) * c^d
题目要求计算出第n项的因子数量,因子数量对 10^9+1 取模。
输入:a, b, c, d, n 5个整数, (1 <= a, b, c, d, n <= 10^12)
例如:输入:1 2 3 4 3
输出:10
解题理论准备:快速幂矩阵、因子与质因子关系
1、因子与质因子的关系-CSDN博客
2、快速幂算法-python-CSDN博客
3、算法学习笔记(4):快速幂 - 知乎
python代码如下:
'''
解题思路:
1、分别算出 a, b, c三个数的质因数;
2、通过快速幂矩阵计算出第n项数据中a、b、c的指数(计算过程中要取模);
3、结合a、b、c的指数,以及a, b, c三个数的质因数,来求出第n项数据对应的质因数,以及对应质因数的指数;
4、最后将第n项数据的各个质因数的指数分别+1之后相乘,就得到第n项数据的因子个数。
'''import time
import numpy as npdef add2dict(dic, n):if n in dic:dic[n] += 1else:dic[n] = 1return dicdef primeFactors(num):factors = {}i = 2while i * i <= num:if num % i == 0:num //= ifactors = add2dict(factors, i)else:i += 1if num > 1:factors = add2dict(factors, num)return factors# 快速幂矩阵
def matrixFastPower(matrix, power, num_mod):matrix = np.array(matrix) % num_modres = np.eye(matrix.shape[0])while power:if power & 1:res = np.dot(res, matrix) % num_modpower >>= 1matrix = np.dot(matrix, matrix) % num_modreturn resif __name__ == '__main__':start_time = time.time()a, b, c, d, n = int(1e12), int(1e12), int(1e12), int(1e12), int(1e12)# a, b, c, d, n = 1, 2, 3, 4, 6num_mod = int(10e9 + 7)if n == 1:factors_a = primeFactors(a, num_mod)sum = 1for i in factors_a:sum *= (factors_a[i] + 1) % num_modelif n == 2:factors_b = primeFactors(b)sum = 1for i in factors_b:sum *= (factors_b[i] + 1) % num_modelse:factors_a = primeFactors(a)factors_b = primeFactors(b)factors_c = primeFactors(c)A = [[0, 1],[1, 1]]C = [[0, 1, 0],[1, 1, 1],[0, 0, 1]]A_n = matrixFastPower(A, n - 1, num_mod)C_n = matrixFastPower(C, n - 1, num_mod)# a_pow = A^(n-1) * [a(1), a(2)]a_pow = int(np.dot(A_n, np.array([1, 0]).T)[0]) % num_mod# b_pow = A^(n-1) * [b(1), b(2)]b_pow = int(np.dot(A_n, np.array([0, 1]).T)[0]) % num_mod# c_pow = C^(n-1) * [c(1), c(2), d]c_pow = int(np.dot(C_n, np.array([0, 0, d]).T)[0]) % num_mod# print(a_pow, b_pow, c_pow)for i in factors_a:factors_a[i] = ((factors_a[i] % num_mod) * a_pow) % num_modfor j in factors_b:factors_b[j] = ((factors_b[j] % num_mod) * b_pow) % num_modif j not in factors_a:factors_a[j] = factors_b[j] % num_modelse:factors_a[j] = (factors_a[j] + factors_b[j]) % num_moddel factors_bfor k in factors_c:factors_c[k] = ((factors_c[k] % num_mod) * c_pow) % num_modif k not in factors_a:factors_a[k] = factors_c[k] % num_modelse:factors_a[k] = (factors_a[k] + factors_c[k]) % num_moddel factors_csum = 1for ix in factors_a:sum *= (factors_a[ix] + 1) % num_modsum %= num_modprint(sum)print("Time:", round((time.time() - start_time) * 1000, 2), 'ms')