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C中的斐波纳契程序

#include <stdio.h>
int factorial(int n) {
   //base case
   if(n == 0) {
      return 1;
   } else {
      return n * factorial(n-1);
   }
}
int fibbonacci(int n) {
   if(n == 0){
      return 0;
   } else if(n == 1) {
      return 1;
   } else {
      return (fibbonacci(n-1) + fibbonacci(n-2));
   }
}
int main() {
   int n = 5;
   int i;
   printf("Factorial of %d: %d\n" , n , factorial(n));
   printf("Fibbonacci of %d: " , n);
   for(i = 0;i<n;i++) {
      printf("%d ",fibbonacci(i));            
   }
}

如果我们编译并运行上面的程序，它将产生以下结果 -

输出 (Output)

Factorial of 5: 120
Fibbonacci of 5: 0 1 1 2 3