习题6-4 使用函数输出指定范围内的Fibonacci数

比眉伴天荒 2021-12-24 16:15 197阅读 0赞

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 }

转载于:https://www.cnblogs.com/2018jason/p/10965819.html