斐波纳契数

我不会告诉你怎么做，因为它显然是家庭作业。 Julian V. Noble 物理荣誉教授 jv*@lessspamformother.virginia.edu ^^^^^^^^^ ^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ 因为有从来没有哲学家耐心地耐受 牙痛。 - Wm。莎士比亚，很多阿多无所事事。法案诉Sc。 1. The algorithm is F(n+1) = F(n) + F(n-1) , F(0) = 0, F(1) = 1 You can do this recursively (very stupid and slow--see my article"Recurses!" in Computing in Science and Engineering) or iteratively(much better). I won''t tell you how to do it since it is evidently homework. Julian V. NobleProfessor Emeritus of Physics jv*@lessspamformother.virginia.edu^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ "For there was never yet philosopher that could endure thetoothache patiently." -- Wm. Shakespeare, Much Ado about Nothing. Act v. Sc. 1. #include< stdio.h> int main（）{ printf（" fibonacci mubers in C\ n）; 返回0; } #include <stdio.h> int main() { printf("fibonacci mubers in C\n");return 0;} 这篇关于斐波纳契数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！