HDOJ/HDU 1250 Hat#39;s Fibonacci(大数~斐波拉契)

Problem Description
A Fibonacci sequence is calculated by adding the previous two members the sequence,with the first two members being both 1.
F(1) = 1,F(2) = 1,F(3) = 1,F(4) = 1,F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)
Your task is to take a number as input,and print that Fibonacci number.

Input
Each line will contain an integers. Process to end of file.

Output
For each case,output the result in a line.

Sample Input
100

Sample Output
4203968145672990846840663646

Note:
No generated Fibonacci number in excess of 2005 digits will be in the test data,ie. F(20) = 66526 has 5 digits.

就是根据这个公式：
F(1) = 1,F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)

输入一个n，输出f(n)的值。

注意，这是大数~答案的位数高达2005位~~~

再一次体会Java大数的强大吧~

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    static BigInteger f[] = new BigInteger[7045];
    public static void main(String[] args) {
        dabiao();
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            int n =sc.nextInt();
            System.out.println(f[n]);
            //System.out.println("---------");
            //System.out.println(f[n].toString().length());
            //开数组~看开到多少位的时候，位数大于2005
        }
    }
    private static void dabiao() {
        f[1]=new BigInteger("1");
        f[2]=new BigInteger("1");
        f[3]=new BigInteger("1");
        f[4]=new BigInteger("1");
        for(int i=5;i<f.length;i++){
            f[i]=f[i-1].add(f[i-2]).add(f[i-3]).add(f[i-4]);
        }
    }
}

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