2Q

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==0) return 0;
    else if(n==1) return 1;
    else return fibbonacci(n-1) + fibbonacci(n-2);
}

int main()
{
    int n;
    while(scanf("%d", &n), n!=-1)
        printf("%d
", fibbonacci(n));
    return 0;
}

// 

__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=0, x2=1, x3=0;
    int i;
    for(i=2;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!=-1)
    {
        i=fibbonacci(n);
        printf("%I64d
", i);  /* __intxx io格式 */ 
    }
    return 0;
}

// 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