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  1. #include <stdio.h>
  2.  
  3. unsigned f_re(unsigned n,unsigned *row,unsigned deep)
  4.   {
  5.    unsigned ret=n<2?1:f_re(n-2,row,deep+1)+f_re(n-1,row,deep+1);
  6.    printf("%*u: f(%u)=%u;\n",deep+3,++*row,n,ret);
  7.    return ret;
  8.   }
  9.  
  10. unsigned f_er(unsigned n,unsigned *row,unsigned deep)
  11.   {
  12.    printf("%*u: f(%u);\n",deep+3,++*row,n);
  13.    return n<2?1:f_er(n-2,row,deep+1)+f_er(n-1,row,deep+1);
  14.   }
  15.  
  16. int main()
  17.   {
  18.    unsigned n,r=0;
  19.    scanf("%u",&n);
  20.    printf("Drzewo wywolan dla ciagu rekurencyjnego:\n");
  21.    printf("Obliczony rekurencyjnie %d wyraz ciagu Fibonacciego wynosi: %d\n",n,f_re(n,&r,0));
  22.    r=0;
  23.    printf("Obliczony rekurencyjnie %d wyraz ciagu Fibonacciego wynosi: %d\n",n,f_er(n,&r,0));
  24.    return 0;
  25.   }
Success #stdin #stdout 0s 2252KB
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stdin
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5
stdout
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Drzewo wywolan dla ciagu rekurencyjnego:
    1: f(1)=1;
     2: f(0)=1;
     3: f(1)=1;
    4: f(2)=2;
   5: f(3)=3;
     6: f(0)=1;
     7: f(1)=1;
    8: f(2)=2;
     9: f(1)=1;
     10: f(0)=1;
     11: f(1)=1;
    12: f(2)=2;
   13: f(3)=3;
  14: f(4)=5;
 15: f(5)=8;
Obliczony rekurencyjnie 5 wyraz ciagu Fibonacciego wynosi: 8
  1: f(5);
   2: f(3);
    3: f(1);
    4: f(2);
     5: f(0);
     6: f(1);
   7: f(4);
    8: f(2);
     9: f(0);
    10: f(1);
   11: f(3);
    12: f(1);
    13: f(2);
     14: f(0);
     15: f(1);
Obliczony rekurencyjnie 5 wyraz ciagu Fibonacciego wynosi: 8
https://ideone.com/aoxXRr
language:
C (gcc 8.3)
created:
9 years ago
visibility:
public

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