hdu2070

说明：

1. 最基本的一个函数

2. 如果写成函数形式，不会通过

3. 写成for循环时，用__int64

4. 还有一个公式方法

/*
 * =====================================================================================
 *
 *       Filename:  hdu2070.c
 *
 *        Version:  1.0
 *        Created:  2013年11月19日 16时46分39秒
 *       Revision:  none
 *       Compiler:  gcc
 *         Author:  Wenxian Ni (Hello World~), niwenxianq@qq.com
 *   Organization:  AMS/ICT
 *
 * Fibbonacci Number

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11472    Accepted Submission(s): 5820


Problem Description
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
Note:
 
you can use 64bit integer: __int64    Description:  
 *
 *
 * =====================================================================================
 */

#include <stdio.h>

int main()
{
    int i, n;
    int fn, fn1; 
    while(~scanf("%d",&n)&&n!=-1)
    {
        if(n==0)
        {
            printf("0
");
            continue;
        }
        if(n==1)
        {
            printf("1
");
            continue;
        }
        fn  = 1;
        fn1 = 0;
        for(i=2;i<=n;i++)
        {
            fn  = fn + fn1;
            fn1 = fn - fn1;
        }
        printf("%d
",fn);

        
    }
    return 0;
}