Fibbonacci Number斐波那契数列

Yunpeng1119

已于 2022-09-16 10:10:09 修改

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分类专栏：算法文章标签：算法

于 2018-07-19 09:51:26 首次发布

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本文介绍了一个生成斐波那契数列的程序实现，该程序能够处理0到50之间的任意整数输入，并输出相应的斐波那契数值。通过使用64位整数类型确保了大范围内的准确性。

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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

#include <stdio.h>
int main()
{
    long long n,i, a[100]={0,1};
    for(;;)
    {
        scanf("%lld",&n);
            if(n==-1)
                break;
            if(n==0)
                printf("0\n");
            else if(n==1)
                printf("1\n");
            else
            {
                for(i=2;i<=n;i++)
                    a[i]=a[i-1]+a[i-2];
                printf("%lld\n",a[n]);
            }

    }
    return 0;
}