小白求教各位大神，关于java，求余数

Question

分别输入两个数double a，b；
    r = b - count * a ;  
    求count为几的时候，r=1.0；

    例如，double a = 1.3;
         double b = 3.9;
     那么，count = 3

import java.util.Scanner;

public class Exg {

    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);
        System.out.println("Pleat enter the first number!");
        double a = s.nextDouble();

        Scanner w = new Scanner(System.in);
        System.out.println("Pleat enter the second number (larger than the first number) !");
        double b = w.nextDouble();
        double count=1.0;
        double r;
        do { 
            r = b- count * a;
            count ++;
        }while(r==1.0);

        System.out.println(count);


    }   
}

结果大概的样子为：

Pleat enter the first number!
2.3
Pleat enter the second number (larger than the first number) !
36.9
2.0

好忧伤啊, 不论添什么数，结果都是2.0

Answer 1

import java.util.Scanner;

public class Exg {

public static void main(String[] args) {

    Scanner s = new Scanner(System.in);
    System.out.println("Pleat enter the first number!");
    double a = s.nextDouble();

    Scanner w = new Scanner(System.in);
    System.out.println("Pleat enter the second number (larger than the first number) !");
    double b = w.nextDouble();

    double d = (b-1.0)/a;
    int count = (int) d;
    System.out.println(count);


}   

}

结果：
Pleat enter the first number!

2.1
Pleat enter the second number (larger than the first number) !
10.1
4

Answer 2

do {
r = b- count * a;
count ++;
}while(r==1.0);

楼主你的这一步有问题
首先count并不是整数，所以你使用count++是不考虑count++和count之间的数字的
你这个函数会先计算count==1时对应的r，然后count由1.0增加为2.0，此时r！=1.0，while循环终止，count就停止在2了
正确的应该是直接使用计算式子 count = (b-r)/a直接输出
顺便提一句，你举的例子也不对，b=3.9，a=1.3时，count应该等于（3.9-1.0）/1.3即2.9/1.3，大概在2.24左右吧