问：import java.util.InputMismatchException;
import java.util.Scanner;
import java.io.IOException;
import java.util.Random;
public class StringVariables {
public static void main(String[] args) throws NumberFormatException,
IOException {
// user inputs their name in this section
Scanner user_input = new Scanner(System.in);
//enter their first name
String first_name;
System.out.print("Enter Your First Name: ");
while
(!user_input.hasNext("[A-Za-z]+")) {
System.out.println("Please only enter alphabet characters. Try again.");
user_input.next();
}
first_name = user_input.next();
//enter their last name
String last_name;
System.out.print("Enter Your Last Name: ");
while
(!user_input.hasNext("[A-Za-z]+")) {
System.out.println("Please only enter alphabet characters. Try again.");
user_input.next();
}
last_name = user_input.next();
//full name printed together
String full_name;
full_name = first_name + " " + last_name;
System.out.println(full_name + " Is Now Playing");
// this is the shuffle portion as well as something to see if a number
int numShuffles = -1;
while (numShuffles < 0) {
System.out.println("How many times do you want the numbers shuffled? ");
try {
numShuffles = user_input.nextInt();
} catch (InputMismatchException inputException) {
System.out.print("Please enter a valid number. \n");
//this is the buffer that resets if the user types a letter instead of a number, or any other character
user_input.next();
}
}
// here is going to be the loop for shuffles
// we are now going to generate their random number and add a delay
// after completing their name fields
delay(3000);
System.out
.println(" You will be given " + numShuffles + " hand(s) of 3 random numbers between 7-13" );
delay(2000);
System.out
.println(" Then, the computer will add the random numbers and if it is equal to 31, you win!");
/*
* end of explanation of the game, next i will create a new screen with
* the user's name and numbers
*/
delay(4000);
// printing 25 blank lines
for (int i = 0; i < 25; i++)
System.out.println(" ");
System.out.println("User playing: " + full_name);
System.out.println("Number of times shuffled: " + numShuffles);
System.out.println("Your lucky numbers are...");
// random number generator
Arraylist numberStore = new Arraylist();
Random random = new Random();
while (true) {
// the shuffle loop
boolean isWinner = false;
for (int i = 0; i < numShuffles; i++) {
int num1 = 7 + random.nextInt(7);
int num2 = 7 + random.nextInt(7);
int num3 = 7 + random.nextInt(7);
System.out.println(num1 + " + " + num2 + " + " + num3 + " = " + (num1 + num2 + num3));
numberStore.add(num1 + num2 + num3);
int lastNumber = (numberStore.size() - 31);
if (lastNumber == 31) {
isWinner = true;
System.out.println("Congratulations !! You are the Lucky Winner !!!!");
break;
//if you loose every shuffle
}
}
if (!isWinner) {
System.out.println("Better Luck Next Time");
}
// play again prompt
System.out
.println(" Do you want to play again? (If you do enter y or yes) \n To exit press any other key ");
String input = user_input.next();
if (!"y".equalsIgnoreCase(input) && !"yes".equalsIgnoreCase(input)) {
break;
}
}
// if pressed y or yes the program will run again with the same number of shuffles entered from before
user_input.close();
}
// delay field
public static void delay(int millis) {
try {
Thread.sleep(millis);
} catch (InterruptedException exp) {
// delay field
}
}
}
你好！这是我对该网站的第一篇文章。我是Java新手，有一个问题。我有一个询问用户名称的程序，然后询问您要循环多少次（所以我的程序是它会生成3个介于7和13之间的随机数，如果加起来等于31，它们就是获胜者），我的问题是，我只希望最后打印出的数字在玩家赢或输的情况下计入，其他数字仅用于展示或挑逗。问题在于，无论玩家赢还是输，无论输赢，总是会输出失败的陈述。以上是我的整个代码。如果有人能帮上忙，那将是一种荣幸。
答：我已经解决了您的问题，请测试以下代码：
导入java.util.ArrayList;
导入java.util.InputMismatchException;
导入java.util.Scanner;
导入java.io.IOException;
导入java.util.Random;
公共类StringVariables {
公共静态void main（String [] args）引发NumberFormatException，
IOException {
//用户在此部分输入他们的姓名
扫描仪用户输入=新的扫描仪（System.in）;
//输入名字
字符串名字；
System.out.print（“输入您的名字：”）;
而
（！userinput.hasNext（“ [A-Za-z] +”））{
System.out.println（“请仅输入字母字符。再试一次。”）;
userinput.next（）;
}
名字= userinput.next（）;
//输入姓氏
字符串姓氏；
System.out.print（“输入您的姓氏”）；
而
（！userinput.hasNext（“ [A-Za-z] +”））{
System.out.println（“请仅输入字母字符。再试一次。”）;
userinput.next（）;
}
姓= userinput.next（）;
//全名打印在一起
字符串全名；
全名=名+“” +姓；
System.out.println（全名+“正在播放”）;
//这是随机播放部分，也是看数字
int numShuffles = -1;
while（numShuffles <0）{
System.out.println（“您希望将数字重新排列多少次？”）;
尝试{
numShuffles = userinput.nextInt（）;
} catch（InputMismatchException inputException）{
System.out.print（“请输入一个有效的数字。\ n”）;
//如果用户键入字母而不是数字或任何其他字符，这是重置的缓冲区
userinput.next（）;
}
}
//这将是洗牌的循环
//现在我们将生成其随机数并添加一个延迟
//完成其姓名字段后
延迟（3000）;
系统输出
.println（“您将获得” + numShuffles +“ 7至13之间的3个随机数的手”）；
delay（2000）;
系统输出
.println（“然后，计算机将添加随机数，如果等于31，则您赢了！”）;
/ *
*游戏说明结束，接下来我将创建一个新屏幕
*用户名和电话号码
* /
延迟（4000）;
//打印25个空白行
对于（int i = 0; i <25; i ++）
System.out.println（“”）;
System.out.println（“用户正在播放：” +全名）;
System.out.println（“洗牌次数：” + numShuffles）;
System.out.println（“您的幸运数字在...”）;
//随机数生成器
ArrayList <Integer> numberStore =新的ArrayList <Integer>（）;
随机random = new Random（）;
而（true）{
//随机循环
boolean isWinner = false;
for（int i = 0; i <numShuffles; i ++）{
int num1 = 7 + random.nextInt（7）;
int num2 = 7 + random.nextInt（7）;
int num3 = 7 + random.nextInt（7）;
System.out.println（num1 +“ +” + num2 +“ +” + num3 +“ =” +（num1 + num2 + num3））;
numberStore.add（num1 + num2 + num3）;
int lastNumber =（int）numberStore.get（numberStore.size（）-1）;
如果（lastNumber == 31）{
isWinner = true;
System.out.println（“恭喜！您是幸运儿！!!!”）;
打破;
//如果您松开所有混洗
}
}
如果（！isWinner）{
System.out.println（“下次好运”）;
}
//再次播放提示
系统输出
.println（“是否要再次播放？（如果输入y或yes）\ n要退出，请按任何其他键”）;
字符串输入= userinput.next（）;
if（！“ y” .equalsIgnoreCase（input）&&！“ yes” .equalsIgnoreCase（input））{
打破;
}
}
//如果按y或yes，则程序将再次运行，其运行次数与之前相同
userinput.close（）;
}
//延迟字段
公共静态无效延迟（int millis）{
尝试{
Thread.sleep（millis）;
} catch（InterruptedException exp）{
//延迟字段
}
}
}
答：我在第105行看到了问题
int lastNumber =（numberStore.size（）-31）;
使用int lastNumber =（int）numberStore.get（numberStore.size（）-1）;
