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~), [e-mail] *   Organization:  AMS/ICT * * Fibbonacci NumberTime Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11472    Accepted Submission(s): 5820Problem DescriptionYour objective for this question is to develop a program which will generate a fibbonacci number. The fibbonacci function is defined as such:f(0) = 0f(1) = 1f(n) = f(n-1) + f(n-2)Your program should be able to handle values of n in the range 0 to 50. InputEach test case consists of one integer n in a single line where 0≤n≤50. The input is terminated by -1. OutputPrint out the answer in a single line for each test case. Sample Input345-1 Sample Output235HintNote: 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\n");            continue;        }        if(n==1)        {            printf("1\n");            continue;        }        fn  = 1;        fn1 = 0;        for(i=2;i<=n;i++)        {            fn  = fn + fn1;            fn1 = fn - fn1;        }        printf("%d\n",fn);            }    return 0;}