[zerojudge]a042. 平面圓形切割

a042. 平面圓形切割

    a0 = 1
    a1 = 2
    a2 = 4
    a3 = 8
    a4 = 14
    a5 = 22
    

a_n - a_n-1 = 2 * (n - 1), if n > 1
=> a_n = 2 * (n - 1) + a_n-1

寫成遞迴就OK了，也可以寫成公式

a_n = n² - n + 2, if n > 0

程式碼如下：

/* a042. 平面圓形切割  
*
* 2020/3/3
*/

import java.util.Scanner;

public class Pa042{

    public static void main(String[] args){
        Scanner scanner = new Scanner(System.in);

        while(scanner.hasNext()){
            System.out.println(circleCut(scanner.nextInt())); 
        }
    }

    public static int circleCut(int n){
        return n == 1? 2: 2 * (n - 1) + circleCut(n - 1);
    }

}