The story of the cow ACM algorithm problem

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A. Title Description

Have a cow, it is born heifer beginning of each year. Each heifers from the fourth year, but also the beginning of a heifer born each year. Please programming in year n, when the total number of cows?

Entry

A plurality of test input data examples, each test case per line, comprising an integer n (0 <n <55) , n the meanings as described in the title.
n = 0 indicates the end of input data, without processing.

Export

For each test case, the number of outputs of the n-th time in cows.
Each output per line.

Sample input

2
4
5
0

Sample Output

2
4
6

II. Solving ideas

The cows divided into four 1-year-old cow, the cow 2 years old, 3-year-old heifers, mature cows

With each passing year

      The number of adult cows for 2-year-old cow of the year, the number of adult cows +

     Number 3-year-old cow was 2 years the number of cows

     Number 2-year-old cow was 1 year old cow number

    Number 1-year-old cow is the number 1 adult cow

The total amount is the sum of the number of all cows

III. Sample Code

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		while (scanner.hasNext() ) {
		    int n = scanner.nextInt();
		    if(n == 0) return ;
			int num1 = 0;   //1岁母牛
			int num2 = 0;  //2岁母牛
			int num3 = 0;   //3岁母牛
			int num4 = 1;  //成年母牛
			for(int i=1;i<n;i++){
			    num4 = num3+num4;
			    num3 = num2;
			    num2 = num1;
			    num1 = num4;
			}
			System.out.println(num1+num2+num3+num4);
		}
	}
}