2Q - Fibbonacci Number

Your objective for this question is to develop a program which will generate a fibbonacci number. The fibbonacci function is defined as such: 
f(0) = 0 
f(1) = 1 
f(n) = f(n-1) + f(n-2) 
Your program should be able to handle values of n in the range 0 to 50. 

Input

Each test case consists of one integer n in a single line where 0≤n≤50. The input is terminated by -1.

Output

Print out the answer in a single line for each test case.

Sample Input

3
4
5
-1

Sample Output

2
3
5

Hint

you can use 64bit integer: __int64

// 递归

 #include<stdio.h>

 long fibbonacci(int n)
 {
     if(n==) return ;
     else if(n==) return ;
     else return fibbonacci(n-) + fibbonacci(n-);
 }

 int main()
 {
     int n;
     while(scanf("%d", &n), n!=-)
         printf("%d\n", fibbonacci(n));
     return ;
 }

Time Limit Exceeded

// 

__intx

int/long	__int64
signed: -2^31 ~ 2^31-1 ~ 2.1*10^9 unsigned:       0 ~ 2^32-1 ~ 4.29*10^9	signed:-2^63 ~ 2^63-1 ~ 9.2*10^18 unsigned:    0 ~ 2^64-1 ~ 1.8*10^19

// signed: scanf("%I64d",&a);    printf("%I64d",a);

　　unsigned: scanf("%I64u",&a);    printf("%I64u",a);

// 说明：
　　1、int64不能用作为循环变量
　　2、int64的操作速度较慢

 #include<stdio.h>

 __int64 fibbonacci(int n)
 {
     __int64 x1=, x2=, x3=;
     int i;
     for(i=;i<=n;i++)
     {
         x3=x1+x2;
         x1=x2; x2=x3;
     }
     if(x3) return x3;
     else
     {
         if(n) return x2;
         else return x1;
     }
 }

 int main()
 {
     int n; __int64 i;
     while(scanf("%d", &n), n!=-)
     {
         i=fibbonacci(n);
         printf("%I64d\n", i);  /* __intxx io格式 */
     }
     return ;
 }

AC

// 从第47个斐波那契数开始，其大小超过int范围；
// 从第48个斐波那契数开始，其大小超过unsigned int范围;
// 从第93个斐波那契数开始，其大小超过__int64范围；
// 从第94个斐波那契数开始，其大小超过unsigned __int64范围