首页 > ACM题库 > HDU-杭电 > HDU 3117-Fibonacci Numbers[解题报告]HOJ
2014
03-02

HDU 3117-Fibonacci Numbers[解题报告]HOJ

Fibonacci Numbers

问题描述 :

The Fibonacci sequence is the sequence of numbers such that every element is equal to the sum of the two previous elements, except for the first two elements f0 and f1 which are respectively zero and one.
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What is the numerical value of the nth Fibonacci number?

输入:

For each test case, a line will contain an integer i between 0 and 108 inclusively, for which you must compute the ith Fibonacci number fi. Fibonacci numbers get large pretty quickly, so whenever the answer has more than 8 digits, output only the first and last 4 digits of the answer, separating the two parts with an ellipsis (“…”).

There is no special way to denote the end of the of the input, simply stop when the standard input terminates (after the EOF).

样例输入:

0
1
2
3
4
5
35
36
37
38
39
40
64
65

样例输出:

0
1
1
2
3
5
9227465
14930352
24157817
39088169
63245986
1023...4155
1061...7723
1716...7565

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=3117

解题思路:

很显然费波纳数的后四位存在周期,通过测试可以发现后四位的周期为15000,所以后四位可以打表得出,

前四位是参考了别人的思路才做出来的。

因为费波纳数f[n] = 1/sqrt(5)(((1+sqrt(5))/2)^n+((1-sqrt(5))/2)^n).

很显然当n非常大的时候(1-sqrt(5))/2)^n非常的小以至于可以忽略,假设F[n]可以表示成 t * 10^k(t是一个小数),那么对于F[n]取对数log10,答案就为log10 t + K,此时很明显log10 t<1,于是我们去除整数部分,就得到了log10 t ,再用pow(10,log10 t)我们就还原回了t。将t×1000就得到了F[n]的前四位。

题目实现:

#include<stdio.h>
#include<math.h>
#define aa (sqrt(5.0)+1.0)/2  
int fib_h[15000],Fib[40] = {0,1,1,2,3,5,8,13},n;
int main()
{
	fib_h[0] = 0 ;fib_h[1] = 1;
	//费波那数后四位
	for(int i = 2; i <= 14999; i++)
	{
		fib_h[i] = (fib_h[i-1] + fib_h[i-2])%10000;
	}
	for(i = 7; i < 40; i++)
		Fib[i] = Fib[i-1] + Fib[i-2];
	while(scanf("%d",&n) != EOF)
	{
		if(n < 40)
			printf("%d\n",Fib[n]);
		else
		{
			double	ans = -0.5*(log10(5.0))+n*log10(aa);
			ans-=(int)ans; 
			 ans=pow(10.0,ans);  
            while(ans<1000)  
                ans*=10;  
            printf("%d...",(int)ans);   
            printf("%4.4d\n",fib_h[n%15000]);  
		}
	}
	return 0;
}

 

参考:http://blog.csdn.net/luo964061873/article/details/7945134


  1. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?

  2. L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-1])这个地方也也有笔误
    应改为L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-2])

  3. L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-1])这个地方也也有笔误
    应改为L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-2])

  4. 第2题,TCP不支持多播,多播和广播仅应用于UDP。所以B选项是不对的。第2题,TCP不支持多播,多播和广播仅应用于UDP。所以B选项是不对的。