Solution : Recursion নিয়া এর চাইতে মজার প্রবলেম আমি এখনও পাইনি :D ।
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| /// Author : Md .shohanur Rahaman | |
| /// UriOj Problem No : 1029 | |
| /// Problem Name : Fibonacci, how many calls? | |
| /// Problem Type : Recursion | |
| #include<stdio.h> | |
| int fib(int n); | |
| int a=0,num_call=0; | |
| int main() | |
| { | |
| int n,tmp,cl,fn,ncl,t,i; | |
| while(scanf("%d",&t)==1){ | |
| for(i=1;i<=t;i++){ | |
| scanf("%d",&n); | |
| if(n>=1 && n<=39){ | |
| tmp=n; | |
| fib(tmp); | |
| ncl=num_call; | |
| cl=a; | |
| printf("fib(%d) = %d calls = %d\n",n,ncl-1,cl); | |
| a=0; | |
| num_call=0; | |
| } | |
| } | |
| } | |
| return 0; | |
| } | |
| int fib(int n) | |
| { | |
| int f; | |
| num_call++; | |
| if(n==0 || n==1 ){ | |
| a=a+n; | |
| return n; | |
| } | |
| f= fib(n-1)+fib(n-2); | |
| } | |
