2Q - Fibbonacci Number

Posted 2021-02-08 goldenretriever

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

 1 #include<stdio.h>
 2 
 3 long fibbonacci(int n)
 4 {
 5     if(n==0) return 0;
 6     else if(n==1) return 1;
 7     else return fibbonacci(n-1) + fibbonacci(n-2);
 8 }
 9 
10 int main()
11 {
12     int n;
13     while(scanf("%d", &n), n!=-1)
14         printf("%d
", fibbonacci(n));
15     return 0;
16 }

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

 1 #include<stdio.h>
 2 
 3 __int64 fibbonacci(int n)
 4 {
 5     __int64 x1=0, x2=1, x3=0;
 6     int i;
 7     for(i=2;i<=n;i++)
 8     {
 9         x3=x1+x2;
10         x1=x2; x2=x3;
11     }
12     if(x3) return x3;
13     else
14     {
15         if(n) return x2;
16         else return x1;
17     }
18 }
19 
20 int main()
21 {
22     int n; __int64 i;
23     while(scanf("%d", &n), n!=-1)
24     {
25         i=fibbonacci(n);
26         printf("%I64d
", i);  /* __intxx io格式 */ 
27     }
28     return 0;
29 }

AC

// 从第47个斐波那契数开始，其大小超过int范围