Fibbonacci Number

最新推荐文章于 2024-09-14 14:06:51 发布

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本文介绍了一种高效计算斐波那契数的方法，避免了传统递归方法的超时问题。通过使用数组记录每个已计算的值，使得后续计算能够复用之前的结果，大大提高了计算效率。

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

Sample Output

2
3
5

you can use 64bit integer: __int64

思路：最开始想到用递归去解，但是会超时，所以又想到利用数组。利用数组对每一个算出来的结果都记录下来，这样计算其后的一个可以调用上次计算的结果，所以时间将会被节约。代码：

#include<stdio.h>

void main() {
   long long f[55];
   int n, i;
   while (scanf_s("%d", &n) && n != -1) {
   f[0] = 0;
   f[1] = 1;
   for (i = 2; i <= n; i++) {
   f[i] = f[i - 1] + f[i - 2];
   }
   printf("%I64d\n", f[n]);
   }
}