习题6-4 使用函数输出指定范围内的Fibonacci数

 1 #include <stdio.h>
 2 
 3 int fib(int n);
 4 void PrintFN(int m, int n);
 5 
 6 int main()
 7 {
 8     int m, n, t;
 9 
10     scanf_s("%d %d %d", &m, &n, &t);
11     printf("fib(%d) = %d\n", t, fib(t));
12     PrintFN(m, n);
13 
14     return 0;
15 }
16 
17 /* 你的代码将被嵌在这里 */
18 
19 int fib(int n)        //斐波拉契数列
20 {
21     int a, b, c;
22     a = 1;
23     b = 1;
24     if (n == 1)
25     {
26         return 1;
27     }
28     else
29     {
30         if (n == 2)
31         {
32             return 1;
33         }
34         else
35         {
36             for (int i = 3; i <= n; i++)
37             {
38                 c = a + b;
39                 a = b;
40                 b = c;
41             }
42             return b;
43         }
44     }
45 }
46 
47 void PrintFN(int m, int n)
48 {
49     int i = 1;
50     int count = 0;
51     while (fib(i) <= n)
52     {
53         if (fib(i) >= m)
54         {
55             count++;
56             if (count == 1)
57             {
58                 printf("%d", fib(i));        //第一个数前面没有空格
59             }
60             else
61             {
62                 printf(" %d", fib(i));        //后面每个数的前面有空格
63             }
64         }
65         i++;
66     }
67     if (count == 0)
68     {
69         printf("No Fibonacci number\n");
70     }
71 }