HDOJ/HDU 1250 Hat's Fibonacci(大数~斐波拉契)
HDOJ/HDU 1250 Hat'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(2) = 1, F(3) = 1,F(4) = 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]);
}
}
}