文字简述
1.阶乘函数
2.2阶Fiibonacci数列
3.n阶Hanoi塔问题
代码实现
1 //
2 // Created by lady on 19-4-3.
3 //
4
5 #include <stdio.h>
6 #include <stdlib.h>
7 #include <string.h>
8
9 static int Fact(int n)
10 {
11 if(n==0){
12 return 1;
13 }else{
14 return n*Fact(n-1);
15 }
16 }
17
18 static int Fibonacci(int n)
19 {
20 if(n == 0){
21 return 0;
22 }else if(n == 1){
23 return 1;
24 }else{
25 return (Fibonacci(n-1) + Fibonacci(n-2));
26 }
27 }
28
29 // 将塔座x上按直径由小到大且自上而下编号为1至n的n个圆盘按规则搬到塔座z上，y可作辅助塔座
30 // 搬动操作move(x, n, z)可定义为(c是初值为0的全局变量，对搬动计数)
31 // printf("%d. Move disk %d from %c to %c", ++c, n, x, z);
32 int C = 0;
33 static int move(char x, int n, char z)
34 {
35 printf("step %d: move disk %d from %c to %c\n", ++C, n, x, z);
36 return 0;
37 }
38 static int hanoi(int n, char x, char y, char z)
39 {
40 if(n == 1){
41 move(x, n, z);
42 }else{
43 hanoi(n-1, x, z, y);
44 move(x, n, z);
45 hanoi(n-1, y, x, z);
46 }
47 return 0;
48 }
49 int main(int argc, char *argv[])
50 {
51 printf("5! = %d\n", Fact(5));
52 printf("Fibonacci(5) = %d\n", Fibonacci(5));
53 hanoi(3, 'a', 'b', 'c');
54 return 0;
55 }
代码运行
/home/lady/CLionProjects/untitled/cmake-build-debug/untitled 5! = 120 Fibonacci(5) = 5 step 1: move disk 1 from a to c step 2: move disk 2 from a to b step 3: move disk 1 from c to b step 4: move disk 3 from a to c step 5: move disk 1 from b to a step 6: move disk 2 from b to c step 7: move disk 1 from a to c Process finished with exit code 0
来源：https://www.cnblogs.com/aimmiao/p/10718996.html