小编use*_*011的帖子

我正在尝试编写一个小游戏,但是如果他们想要再次玩游戏又如何提醒用户以及如果他们不想再次玩游戏则如何退出循环...

import java.util.Random;
import java.util.Scanner;

public class Guessinggame {

public static void main(String[] args) {

    System.out.println("Welcome to guessing game! \n" + " You must guess a number between 1 and 100. ");

    while (true) {

        Random randomNumber = new Random();
        Scanner g = new Scanner(System.in);

        int number = randomNumber.nextInt(100) + 1;
        int guess = 0;
        int numberOfGuesses = 0;

        while (guess != number){

            System.out.print("Guess: ");
            guess = g.nextInt();

            if (guess > number ){
                System.out.println( "You guessed too high!");
            }else if …

我警告你,这可能令人困惑,我写的代码更多的是思维导图而不是完成代码.

我正在尝试使用Newton-Raphson方法来求解方程.我无法弄清楚的是如何写这个

Python中的等式,用于计算最后一次近似(xn)的下一个近似值(xn + 1).我必须使用循环,越来越接近真实的答案,并且当近似值之间的变化小于变量h时,循环应该终止.

1. 如何编写等式的代码？

2. 当近似值不再变化时,如何终止循环？

   计算方程f的导数,在点x,精度为h(这用于求解方程式())

   def derivative(f, x, h):
       deriv = (1.0/(2*h))*(f(x+h)-f(x-h))
       return deriv
   

   Run Code Online (Sandbox Code Playgroud)

   数值方程求解器

   假设循环直到近似值之间的差异小于h

   def solve(f, x0, h):
       xn = x0
       prev = 0
   
       while ( approx - prev > h):
            xn = xn - (f(xn))/derivative(f, xn, h)
   
       return xn
   

   Run Code Online (Sandbox Code Playgroud)

我试着创建一个简单的计数器来计算一个字符在字符串中出现的次数,但我做错了.当我尝试打印计数器时,它会重复多次.

import java.util.Scanner;

public class Uppgift5 {

public static void main(String[] args) {

    int counter = 0;

    Scanner scanner = new Scanner(System.in);

    System.out.println("What string do you want to search: ");
    String string = scanner.nextLine();

    System.out.println("What letter do you want to count? ");
    String letter = scanner.nextLine();


    for(int i=0; i<string.length(); i++ ) {
        if (string.charAt(i) == letter.charAt(0)){
            counter++; 
        }

        System.out.println("This string has " + counter + " " + letter);

    }
}
}