Question

把C语言程序转换成Java

C语言源程序如下：

#include <stdio.h>
#include <math.h>

void main()
{
float a, b, c, x1, x2;

printf("请输入a, b, c的值：");
scanf("%f%f%f", &a, &b, &c);

if (b*b - 4*a*c <= 0)
printf("输入的系数不对，b2-4ac不大于0！程序退出！\n");
else
{
x1 = (-b + sqrt(b*b - 4*a*c))/2*a;
x2 = (-b - sqrt(b*b - 4*a*c))/2*a;

printf("x1 = %.2f\n", x1);
printf("x2 = %.2f\n", x2);
}

}
即用Java编程：从键盘输入三个整型数，其分别为一元二次方程ax2+bx+c=0中a,b,c的值，判断此方程有无实数根，若有则输出其值。

Answer 1

package dis;
import java.io.*;
import java.math.*;
public class dis
{
public static void main(String [] s) throws Exception
{
Float a,b,c;
double x2;
double x1;
System.out.println("请输入a, b, c的值");
BufferedReader bReader=new BufferedReader(new InputStreamReader(System.in));
a=Float.parseFloat(bReader.readLine());
b=Float.parseFloat(bReader.readLine());
c=Float.parseFloat(bReader.readLine());
if (b*b - 4*a*c <= 0)
System.out.println("输入的系数不对，b2-4ac不大于0！程序退出！");
else
{
x1 = (-b + Math.sqrt(b*b - 4*a*c))/2*a;
x2 = (-b - Math.sqrt(b*b - 4*a*c))/2*a;

System.out.println("x1 = "+ x1);
System.out.println("x2 = "+x2);
}
return;
}

}

Answer 2

同上！

Answer 3

改改语法就行了