小球下落可看做取极限的过程

S=A+2∗A/2+2∗A/22+...+2∗A/2nS=A+2*A/2 + 2*A/2^2+...+2*A/2^n
 =A+A(1+1/2+1/22+...+1/2n−1)=A+A(1+1/2+1/2^2+...+1/2^{n-1})
 =A+A∗1−2−n1−2−1=A+A * \frac{1-2^{-n}}{1-2^{-1}}
 =A(3−1/2n−1)=A(3-1/2^{n-1})

x→∞S=3∗A{\lim_{x \to \infty}}S = 3*A

import java.util.*;

public class Balls {
    public int calcDistance(int A, int B, int C, int D) {
        return (A+B+C+D) * 3;
    }
}