So Easy! HDU - 4565

Posted 2021-03-22 yangxuejian

易知，有(S_n = lceil{a + sqrt{b}} ceil ^ n)
(ecause a ^ 2 - 1 < b < a ^ 2)
( herefore a - sqrt{b} in [0, 1])
( herefore (a - sqrt{b}) ^ n in [0, 1])
( herefore lceil{a + sqrt{b}} ceil ^ n = (a + sqrt{b}) ^ n + (a - sqrt{b}) ^ n)
( herefore lceil{a + sqrt{b}} ceil ^ n = x_1 ^ n + x_2 ^ n)
( herefore 有特征方程: x ^ 2 - (x_1 + x_2) * x + x_1 * x_2 = 0)
( herefore 此特征方程为 t ^ 2 - 2at + a ^ 2 - b = 0 的特征方程)
( herefore 构造F_n = 2a * F_{n - 1} + (b - a ^ 2)F_{n - 2}, 矩阵快速幂即可。)