我对通过命令行运行参数有点陌生。我在命令行中实现最后 4 个参数时遇到问题，因此命令行上的示例输入将是 java newton 2 4 .005 50 1 2 0 5 ，其中 1 2 0 5 是 return catch 底部的 static double 中的多项式系数。
它应该是1x^3 + 2x^2 + 0^2 + 5。一切似乎都有效，但我无法让参数粘在底部，也不知道为什么。如果有人可以帮助我，我已经花了近 10 个小时尝试研究，但似乎无法在任何地方找到任何有关此问题的帮助。
import java.util.Scanner;
import java.text.DecimalFormat;
public class newton {
public static void main(String[] args) {
double x0, xnew, xxnew;// Initiating double
double x1, p1;
double fx0, fx1;
double delta, delta1; // amount added to get next iterate
double error; // error estimate
double tol = Double.parseDouble(args[2]);// tolerance (max error)
int i, maxIts, j; // iteration count and maximum number of
// iteraterations made
x0 = Integer.parseInt(args[0]);
x1 = Integer.parseInt(args[1]);
p1 = Integer.parseInt(args[4]);
maxIts = Integer.parseInt(args[3]);
DecimalFormat fmt = new DecimalFormat("0.############");
System.out.println("\n");
System.out.println("Polynomail Root Finder By [Gilbert Jimenez]" + "\n");
System.out.println("Initial Perameters :" + "\n");
System.out.println("P0 : = " + args[0]);
System.out.println("p1 : = " + args[1]);
System.out.println("Tol = " + tol);
System.out.println("Maximum = " + maxIts + "\n");
System.out.println("Polynomial is of order: 4 ");
System.out.println("Terms of polynomial: " + args[4] + "x^3" + "+" + args[5] + "x^2" + "+" + args[6] + "x" + "+"
+ args[7]);
{
// Performing Newton's method
i = 1;
error = 100;
System.out.println("Newtons Method:\t " + "\n");
while (i <= maxIts && error > tol) {
delta = -(f(x0) / fprime(x0));
error = Math.abs(delta);
xnew = x0 + delta;
System.out.println("p" + i + "\t" + fmt.format(xnew));
i++;
x0 = xnew;
}
System.out.println("\n");
System.out.println("Solution found after " + i + " " + "itterations :" + fmt.format(x0) + "\n");
}
{
// Performing
j = 1;
error = 100;
System.out.println("Secant Method:\t " + "\n");
fx0 = f(x0);
while (j <= maxIts && error > tol) {
fx1 = f(x1);
delta1 = (-fx1 * (x1 - x0) / (fx1 - fx0));
error = Math.abs(delta1);
xxnew = x1 + delta1;
System.out.println("p" + j + "\t" + fmt.format(xxnew));
j++;
x0 = x1;
fx0 = fx1;
x1 = xxnew;
}
System.out.println("\n");
System.out.println("Solution found after " + j + " " + "itterations :" + fmt.format(x1) + "\n");
}
}
// function of f
public static double f(double x) {
return (x * x * x - 2.0 * x * x + 0 * x - 5);
}
// derivative of f
public static double fprime(double x) {
return (3.0 * x * x - 4.0 * x);
}
}
最佳答案
假设多项式采用以下形式:ax^3 + bx^2 + c*x + d
所做的更改列表:
- 将类名称更改为 Newton(从 newton)
- 将多项式系数保存到 a、b、c、d 中
- 修改函数 f() 和 fprime() 以使用 a、b、c、d
请尝试看看这是否有帮助。
public class Newton {
static int a = 0;
static int b = 0;
static int c = 0;
static int d = 0;
public static void main(String[] args) {
double x0, xnew, xxnew;// Initiating double
double x1, p1;
double fx0, fx1;
double delta, delta1; // amount added to get next iterate
double error; // error estimate
double tol = Double.parseDouble(args[2]);// tolerance (max error)
int i, maxIts, j; // iteration count and maximum number of
// iteraterations made
x0 = Integer.parseInt(args[0]);
x1 = Integer.parseInt(args[1]);
p1 = Integer.parseInt(args[4]);
maxIts = Integer.parseInt(args[3]);
DecimalFormat fmt = new DecimalFormat("0.############");
System.out.println("\n");
System.out.println("Polynomail Root Finder By [Gilbert Jimenez]" + "\n");
System.out.println("Initial Perameters :" + "\n");
System.out.println("P0 : = " + args[0]);
System.out.println("p1 : = " + args[1]);
System.out.println("Tol = " + tol);
System.out.println("Maximum = " + maxIts + "\n");
System.out.println("Polynomial is of order: 4 ");
a = Integer.valueOf(args[4]);
b = Integer.valueOf(args[5]);
c = Integer.valueOf(args[6]);
d = Integer.valueOf(args[7]);
System.out.println("Terms of polynomial: " + a + "x^3" + "+" + b + "x^2" + "+" + c + "x" + "+" + d);
{
// Performing Newton's method
i = 1;
error = 100;
System.out.println("Newtons Method:\t " + "\n");
while (i <= maxIts && error > tol)
{
delta = -(f(x0) / fprime(x0));
error = Math.abs(delta);
xnew = x0 + delta;
System.out.println("p" + i + "\t" + fmt.format(xnew));
i++;
x0 = xnew;
}
System.out.println("\n");
System.out.println("Solution found after " + i + " " + "itterations :" + fmt.format(x0) + "\n");
}
{
// Performing
j = 1;
error = 100;
System.out.println("Secant Method:\t " + "\n");
fx0 = f(x0);
while (j <= maxIts && error > tol)
{
fx1 = f(x1);
delta1 = (-fx1 * (x1 - x0) / (fx1 - fx0));
error = Math.abs(delta1);
xxnew = x1 + delta1;
System.out.println("p" + j + "\t" + fmt.format(xxnew));
j++;
x0 = x1;
fx0 = fx1;
x1 = xxnew;
}
System.out.println("\n");
System.out.println("Solution found after " + j + " " + "itterations :" + fmt.format(x1) + "\n");
}
}
// function of f
public static double f(double x)
{
return (a * x * x * x + b * x * x + c * x + d);
}
// derivative of f
public static double fprime(double x)
{
return (3 * a * x * x + 2 * b * x + c);
}
}
关于java - 尝试将参数(参数)(args[])实现为静态 double ，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/35518495/