判断另一种斐波那契系列是否是3 的倍数问题

该题如果用递归求出F(n),然后判

断将会超时。

注意到F(n) = F(n-1) + F(n-2)根据和的模等于模的和。

即有F(n)％3=(F(n-1)％3+F(n-2)％3)％3 F(0)％3=1,F(1)％3=2

计算F(n)％3 : 1,2,0,2,2,1,0,1,1,2,0,2,2,1……………

可以看出F(n)％3 以1,2,0,2,2,1,0,1, 重复出现

F(n)％3=0 即F(n)是3 的倍数此时的n 等于8k+2 或8k+6-Fibonacci Series 3 is whether the multiplier issue that if the calculated using recursive F (n), The judgment will then overtime. Notes F (n) = F (n-1) F (n-2) and the modulus and the same model. That is, F (n) = 3% (F (n-1)% F 3 (n-2)% 3)% 3 F (0%) 3 = 1, F (1)% 2 3 = calculated F (n)% 3 : 1,2,0,2,2,1,0,1,1,2,0. 2,2,1 ... ... ... can be seen F (n) 3% to 1,2,0,2,2,1,0,1, repeated F (n) = 0% 3 F (n) 3 is a multiple of n time to 8k or 8k 6 2