Question

大家下午好，

我目前正在尝试创建一个执行以下操作的程序：

开发一个代码，打印所有素数，直到输入的用户为止数。输出的一个例子：

 Enter an integer (2 or above): 19

 The prime numbers up to you integer are:

 2

 3

 5

 7

 11

 13

 17

 19

不幸的是，还有一个＆＃34;要求列表＆＃34;这也需要得到满足：

如果用户输入的数字低于2，您的程序应打印一条消息，表明该号码无效，然后停止。
如果数字不能被除1和它本身之外的任何数字整除，则数字是素数。
对于这个程序，为了测试一个数字是否为素数，你应该尝试将数字除以2到数字1的每个值，看看它是否均匀分开。例如：

- 要查看5是素数还是： 5不均匀地除以2 5不均匀分为3 5不均匀分为4 因此5是素数

- 要查看9是否为素数： 9不均匀地除以2 9平均分为3 因此9不是素数
此程序要求您编写嵌套循环（即循环内的循环）。将使用一个循环从2到用户的数字进行计数，这样您就可以测试这些数字中的每个数字，看它是否为素数。对于这些数字中的每一个，x：
嵌套循环将检查从2到x-1的所有值，以查看x是否均匀划分。
您需要使用布尔变量（也称为标志变量）来帮助您确定是否在屏幕上打印数字

上面的问题是关于我目前的代码：

import java.util。*;

public class Something3 {

public static void main(String[] args) {

    Scanner kbd = new Scanner(System.in);

    //Variable declaration.
    int limit, number;

    //get input till which prime number to be printed
    System.out.print("Enter an integer (2 or above): ");
    limit = kbd.nextInt();
    kbd.close();

    //Will print prime numbers till the limit (user entered integer).
    number = 2;

    if (limit >=2) {
        System.out.println("The prim numbers up to your interger are: " + limit+"\n");
        for(int i = 0; i <= limit;){         
            //print prime numbers only
            if(isPrime(number)){
                System.out.println(number +"\n");
                i++;
            } 
            number = number + 1;
        }
    }
    else
        System.out.println("Number is not vaild");

}

//Prime number is not divisible by any number other than 1 and itself
//return true if number is prime. 
public static boolean isPrime(int number){
    for(int i=2; i==number; i++){
        if(number%i == 0){
            return false; //Number is divisible, thus not prime.
        }
    }
    return true;//The number is prime.
}

}

极限变量I＆＃39; m是否使用此问题？任何帮助将非常感激。

Answer 1

Your actual problem was number variable. Here's your solution. There's no need of variable number. Below is your solution

import java.util.*;

public class Something3 {

public static void main(String[] args) {

    Scanner kbd = new Scanner(System.in);

    // Variable declaration.
    int limit;

    // get input till which prime number to be printed
    System.out.print("Enter an integer (2 or above): ");
    limit = kbd.nextInt();
    kbd.close();

    if (limit >= 2) {
        System.out.println("The prim numbers up to your interger are: "
                + limit + "\n");
        for (int i = 1; i <= limit; i++) {
            // print prime numbers only
            if (isPrime(i)) {
                System.out.println(i);
            }
        }
    } else
        System.out.println("Number is not vaild");

}

// Prime number is not divisible by any number other than 1 and itself
// return true if number is prime.
public static boolean isPrime(int n) {
    if (n % 2 == 0)
        // The only even prime is 2.
        return (n == 2);
    for (int i = 3; i * i <= n; i += 2) {
        if (n % i == 0)
            return false;
    }
    return true;
}
}

Answer 2

your isPrime() method is returning wrong result. for loop condition is wrong i==number it should be i < number

public static boolean isPrime(int number){
    for(int i=2; i*i <= number; i++){
        if( number % i == 0){
            return false; // Number is divisible, thus not prime.
        }
    }
    return true; //The number is prime.
}

To check a number prime or not you don't have to check divisbility from 2 to N, all you need to check upto sqrt(N). To find prime numbers in a range(N) use Sieve of Eratosthenes

Answer 3

import java.util.Scanner;
class prime
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
System.out.println("enter the value of n");
int n=sc.nextInt();
for(int i=2;i<=n;i++)
{
    int count=0;
    for(int j=1;j<=i;j++)
    {
        if(i%j==0)
        {
            count++;
        }
    }
    if(count==2)
    {
        System.out.println(i+" ");
    }
}
}
}

如何打印素数到用户输入的整数？

3 个答案: