解题思路: 求a的平方根的迭代公式为: X[n+1]=(X[n]+a/X[n])/2 要求前后两次求出的差的绝对值少于0.00001。输出保留3位小数
迭代公式: X[n+1]=(X[n]+a/X[n])/2
当x[n]-x[n+1]<=0.00001时,x[n+1]为输出结果
如当a=3时:
x1 = (3+(3/3))/2 = 2
x2 = (2+(3/2))/2 = 1.75
x3 = (1.75+(3/1.75))/2 = 1.7321428571428572
.....===> x[n+1]-x[n]<=0.00001输出x[n+1]
注意事项:
参考代码:
public static void main(String[] args) { DecimalFormat df = new DecimalFormat("#.000");// Scanner sc = new Scanner(System.in); double a = sc.nextDouble(); String num = df.format(t(a,a)); System.out.println(num); // System.out.println(Double.parseDouble(num)*Double.parseDouble(num)); } public static double t(double x1,double a) { double x2 = (x1+(a/x1))/2; // System.out.println("x1 = "+x1+" x2 = "+x2+" x1-x2 = "+(x1-x2)); if (x1-x2<=0.00001) { return x2; } return t((x2+(a/x2))/2,a); }
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平凡 2021-01-24 20:08:33 |
嗯....,老久之前写的。哈哈哈,你觉得合适可以写上去的,没事的