和五指山那题目基本相同、那道题我应该写了稍微详细一点的题解，是关于x1、y1、x2、y2的关系的。

import java.util.*;
//variable = A
//while(variable != B)
//variable += C
public class Main{
    static long x2;//用来保存下一层递归的结果
    static long y2;
    public static void main(String[] args){
        Scanner input = new Scanner(System.in);
        long A = input.nextLong();
        long B = input.nextLong();
        long C = input.nextLong();
        long k = input.nextLong();
        while(A != 0 || B != 0 || C != 0 || k != 0){
            // System.out.printf("%d %d %d %d \n",A,B,C,k);
            if(A == B){
                System.out.println(0);
            }else{
                //3 7 2 16
                long a = C;
                long b = (1L << k);
                long gcd = extGcd(a,b);
                // System.out.printf("%d * %d + %d * %d = %d \n",x2,a,y2,b,gcd);
                if((B - A) % gcd != 0){
                    System.out.println("FOREVER");
                }else{
                    //x2 * C + y2 * (1 << k) = (B - A)
                    x2 *= (B - A) / gcd;
                    //实际上通解的形式是+号左边加上一个最小公倍数数，加号右边减去一个最小公倍数
                    //因为是最小公倍数，所以是最小的能同时使用原数a、b表示的数，每次加上一个最小增量即是通解
                    b /= gcd;// a * b / gcd最小公倍数 a * (x + k * b / gcd )为通解
                    // System.out.println("x2:"+x2+" b:"+b);
                    long res = (x2 % b + b) % b;
                    System.out.println(res);

                }
            }
            A = input.nextLong();
            B = input.nextLong();
            C = input.nextLong();
            k = input.nextLong();
        }

    }
            public static long extGcd(long a,long b){
            if(b == 0){
                x2 = 1;
                y2 = 0;
                return a;
            }
            long gcd = extGcd(b,a%b);
            long x1 = y2;
            long y1 = x2 - (a / b) * y2;
            x2 = x1;//我老是忘了更新x2,y2。
            y2 = y1;
            return gcd;
        }
}