我的教授被指派写一个素数＆＃34; finder＆＃34;。如果输入的数字是素数或偶数，则显示输入的数字，然后显示下一个素数。他希望我们在键入错误输入时给出错误消息。我认为负整数部分很简单，但我无法弄清楚字符输入。或者如果角色不是数字。我如何阻止非数字输入？
此外，系统应该以三个连续错误输入退出。我该如何重置计数器？我编写程序的方式，如果用户犯了两个错误但下一个错误是可以接受的，那么再犯一个错误。 （因此不是连续的。）程序结束。 这是我的第一个编程课程，所以我不擅长它。任何帮助将不胜感激。
另外，我必须使用扫描仪和两种方法。
/**
*
* @param n
* @return
*/
public static boolean isPrime(int n) {
for (int i = 2; i < n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
public static int nextPrime(int n) {
n++;
isPrime(n);
while (isPrime(n) == false) {
n++;
isPrime(n);
}
int prime = n;
return prime;
}
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
int answer = 2;
int counter = 1;
boolean playAgain = false;
Scanner input = new Scanner(System.in);
Scanner reader = new Scanner(System.in);
Scanner in = new Scanner(System.in);
do {
//ask for input
System.out.print("\nEnter the integer value-> ");
//input answer
int n = input.nextInt();
{
//decide is negative
while ( n < 0){
//count counter
counter++;
//give error message
System.out.println("\nInvalid Input! Stike!");
//ask for input
System.out.print("\nEnter the integer value-> ");
//input answer
n = input.nextInt();
//decide is character
// if ( n != '.'){
//count counter
// counter++;
//give error message
// System.out.println("\nInvalid Input! Strike!");
// }
//decide if count three errors
if (counter == 3){
//display three errors message
System.out.println("Three Strikes! You're Out!");
//close program
System.exit(0);
}
}
//decide if prime
if (isPrime(n)) {
//display prime answer
System.out.println(n + " Is Prime");
//decide if even
} else {
//display even answer
System.out.println(n + " Is Even");
}
//counter input
n++;
//while input is false
while (isPrime(n) == false) {
n++;
}
//display next prime
System.out.println(n + " Next prime");
{
//ask if you want to continue
System.out.println("\nPlay Again?\n\nEnter 1)Yes or 2)No ");
//input answer
answer = in.nextInt();
//if answer is 1)yes
if (answer == 1) {
playAgain = true;
//display play again and next input
System.out.println("\nPlay Again!");
}
//if answer is no
if (answer == 2) {
playAgain = false;
System.out.println("\nGoodbye!");
//close program
System.exit(0);
}
}
}
} while (playAgain != false);
}
}
答案 0 :(得分：0)
您正尝试使用带有
的Scanner类进行输入int n = input.nextInt();
如果您在此输入一个字符代替数字，您将获得 java.util.InputMismatchException
你能做的就是
try {
int n = input.nextInt();
} catch (InputMismatchException e) {
//handle the error scenario where the input is a character
System.out.println("Enter Valid Input");
}
答案 1 :(得分：0)
import java.util.Scanner;
public class SOQ5B
{
public static boolean isPrime(int n) {
for (int i = 2; i < n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
public static int nextPrime(int n) {
n++;
isPrime(n);
while (isPrime(n) == false) {
n++;
isPrime(n);
}
int prime = n;
return prime;
}
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
int answer;
int counter = 0;
int n;
boolean playAgain = true;
boolean isNum;
boolean isNum2;
boolean continuePermitted = true;
Scanner input = new Scanner(System.in);
String s;
do {
//ask for input
System.out.print("\nEnter the integer value-> ");
s = input.nextLine();
isNum = true;
for(int i = 0; i < s.length(); i++)
{
if(!(s.charAt(i) >= '0' && s.charAt(i) <= '9'))
{
isNum = false;
}
}
if(isNum)
{
counter = 0;
n = Integer.parseInt(s);
//decide if prime
if (isPrime(n)) {
//display prime answer
System.out.println(n + " Is Prime");
//decide if even
}
else {
//display even answer
System.out.println(n + " Is Even");
}
//counter input
n++;
//while input is false
while (isPrime(n) == false) {
n++;
}
//display next prime
System.out.println(n + " Next prime");
do
{
continuePermitted = true;
//ask if you want to continue
System.out.println("\nPlay Again?\n\nEnter 1)Yes or 2)No ");
//input answer
s = input.nextLine();
isNum2 = true;
for(int i = 0; i < s.length(); i++)
{
if(!(s.charAt(i) >= '0' && s.charAt(i) <= '9'))
{
isNum2 = false;
}
}
if(isNum2)
{
answer = Integer.parseInt(s);
//if answer is 1)yes
if (answer == 1) {
playAgain = true;
//display play again and next input
System.out.println("\nPlay Again!");
}
//if answer is no
if (answer == 2) {
playAgain = false;
System.out.println("\nGoodbye!");
//close program
System.exit(0);
}
}
else
{
System.out.println("Incorrect response.");
continuePermitted = false;
//if answering the yes or no question incorrectly is part of the 3 strikes
//then uncomment the following lines of code
/*
counter++;
}
if(counter >= 3)
{
System.out.println("3 strikes you out");
System.exit(0);
*/
}
}while(!continuePermitted);
}
else
{
System.out.print("\nIncorrect input. Number must be a positive integer.\n");
counter++;
}
if(counter>=3)
{
System.out.println("3 strikes and you're out!");
System.exit(0);
}
} while (playAgain != false);
}
}
将来，我建议您在提问之前在互联网上研究您的问题。还有其他几个地方可以很容易地回答你的问题。
现在关于你的实际问题，请注意我在s = input.nextLine（）的行中如何更改代码？我在那里做的是检查字符串中的每个数字是否是0-9之间的任何数字。我不仅能够检查它们是否都是数字，而且我还能够看到它们是否都是正面的，因为你必须放一个 - 为了它是负面的。除此之外，您还有一个布尔值，只有在输入为正数时才有效。如果没有，它会检查3次，以确保您的程序不会搞砸。此外，我甚至评论了另一个允许3次罢工的部分，如果回答“是”则不包括罢工。如果还有其他问题，请询问，我会编辑我的答案。