求方程ax2+bx+c=0的實數根
阿新 • • 發佈:2018-11-08
求方程ax2+bx+c=0的實數根。a, b, c由鍵盤輸入, a!=0。若只有一個實數根(b2-4ac=0)則只輸出x1,若無實數根(b2-4ac<0)則輸出Error。
輸入
2.5 7.5 1.0
輸出
(注意等號前面後面都有一個空格)
(注意等號前面後面都有一個空格)
x1 = -0.139853
x2 = -2.860147
import java.util.Scanner; public class ALGO160 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); double a = sc.nextDouble(); double b = sc.nextDouble(); double c = sc.nextDouble(); double d = b*b-4*a*c; double e = -b/(2*a); if(d<0){ System.out.println("Error"); }else{ if(d>0){ double x1 = e+Math.sqrt(d)/(2*a); double x2 = e- Math.sqrt(d)/(2*a); System.out.println("x1"+" = "+String.format("%.6f", x1)); System.out.println("x2"+" = "+String.format("%.6f", x2)); }else { System.out.println("x1"+" = "+String.format("%.6f", e)); } } } }