Fibonacci Numbers-hdu3117

最新推荐文章于 2021-10-10 20:54:43 发布

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最新推荐文章于 2021-10-10 20:54:43 发布

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分类专栏：数学文章标签： fibonacci 数学

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Fibonacci Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1285 Accepted Submission(s): 543

Problem Description

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.

What is the numerical value of the nth Fibonacci number?

Input

For each test case, a line will contain an integer i between 0 and 10 ⁸ 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).

Sample Input

Sample Output

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

思路：

//1.后4位15000次循环一次
//2.前四位用数学公式推导可求
//3.前四位具体推导方法如下：

fib(n)= ([(1+√5)/2]ⁿ -[(1-√5)/2]ⁿ)√5

fib(n)= [(1+√5)/2]ⁿ /√5//当n很大的时候[(1-√5)/2]ⁿ)很小，可忽略

log₁₀^fib(n)=log₁₀([(1+√5)/2]ⁿ/√5)

log₁₀^fib(n)=nlog₁₀(1+√5)-nlog₁₀2-log₁₀√5

例如log₁₀^fib(n)=3.123123

则 ₁₀log₁₀^fib(n)=10^3.123123

显然fib（n）=0.123123*1000

只需要求0.123123小数部分前4位即可

import java.util.Scanner;

public class Main{

	public static int arr[] = new int[1000000];
	public static int fib[] = new int[41];
	public static int T = 0;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		arr[1] = arr[2] = 1;
		//测试后4位循环(15000循环一次)
		for(int i=3; i<1000000; i++) {
			arr[i] = (arr[i-1]+arr[i-2])%10000;
			if(arr[i]==arr[2] && arr[i-1]==arr[1]) {
				T = i-2;
				break;
			}
		}
		//fib的前40位
		fib[1] = fib[2] = 1;
		for(int i=3; i<41; i++) {
			fib[i] = fib[i-1]+fib[i-2];
		}
		while(sc.hasNextInt()) {
			int n = sc.nextInt();
			if(n<40) {
				System.out.println(fib[n]);
			} else {
				double ans;
				ans = -0.5*Math.log10(5)+n*Math.log10(1+Math.sqrt(5))-n*Math.log10(2);
				ans -= (int)(ans);
				ans = Math.pow(10, ans);
				while(ans < 1000) {
					ans *= 10;
				}
				System.out.print((int)ans+"...");
				String str = arr[n%T]+"";
				while(str.length()<4) {
					str = "0"+str;
				}
				System.out.println(str);
			}
		}
	}
}