判定 ax²+bx+c=0 的方程有无实数根 有的话求解程序制作
c是下面这样换了java改如何编写 请求帮助void fun0(void)
{
printf("方程无实根\n");
}
void fun1(float a,float b)
{
printf("x=%f\n",-b/(2*a));
}
void fun2(float a,float b,float delt)
{
printf("x1=%f\nx2=%f\n",(-b+sqrt(delt))/(2*a),(-b-sqrt(delt))/(2*a));
}
float dlt(float a,float b,float c)
{
return b*b-4*a*c;
}
int main(void)
{
float a,b,c,delt;
scanf("%f%f%f",&a,&b,&c);
delt=dlt(a,b,c);
if (delt<0) fun0();
else if (fabs(delt)<1e-6) fun1(a,b);
else fun2(a,b,delt); return 0;
}
c是下面这样换了java改如何编写 请求帮助void fun0(void)
{
printf("方程无实根\n");
}
void fun1(float a,float b)
{
printf("x=%f\n",-b/(2*a));
}
void fun2(float a,float b,float delt)
{
printf("x1=%f\nx2=%f\n",(-b+sqrt(delt))/(2*a),(-b-sqrt(delt))/(2*a));
}
float dlt(float a,float b,float c)
{
return b*b-4*a*c;
}
int main(void)
{
float a,b,c,delt;
scanf("%f%f%f",&a,&b,&c);
delt=dlt(a,b,c);
if (delt<0) fun0();
else if (fabs(delt)<1e-6) fun1(a,b);
else fun2(a,b,delt); return 0;
}
import java.util.Scanner;public class X { /**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub Scanner in = new Scanner(System.in); double a, b, c, d;
a = in.nextDouble();
b = in.nextDouble();
c = in.nextDouble();
d = delt(a, b, c);
if (d < 0)
function0();
else if (Math.abs(d) < 1e-6)
function1(a, b);
else
function2(a, b, d);
} public static double delt(double a, double b, double c) {
return b * b - 4 * a * c;
} public static void function0() {
System.out.println("方程无实根");
} public static void function1(double a, double b) {
System.out.println("x = " + -b / 2 / a);
} public static void function2(double a, double b, double d) {
System.out.println("x1 = " + (-b + Math.sqrt(d)) / (2 * a) + ", x2 = "
+ (-b - Math.sqrt(d)) / (2 * a));
}
}