找出第 n 个质数

Question

I don't know why my code won't work when I input certain nthprime numbers. I have tried to alter my code a couple of times but for every one nthprime I make it work, I make it worse for others. So if I change my code to make it work for nthprime=8, I realize nthprime=7 and some others stop working. Can anyone point out a specific flaw I made or maybe I should rethink the outline of my code. Thank you.

public class NthPrime {

public static void main(String[] args) {

    int nthprime;

    System.out.println("Enter value for n:");
    nthprime=IO.readInt();

    while(nthprime <= 0){
        System.out.println("Enter a positive value for n");
        nthprime=IO.readInt();
    }

    if(nthprime == 1){
        System.out.println("The nth prime number is: "+2);
    }
    if(nthprime == 2){
        System.out.println("The nth prime number is: "+3);
    }

    if(nthprime > 2){

    int prime=2; 
    int num=3;
    int square;
    boolean nonprime=false;

    while(prime < nthprime){
        prime++;
        num+=2;
        square = (int) Math.sqrt(num);
        for (int i=3; i <= square; i++){
            if (num % i == 0){
                nonprime=true;
                num+=2;
            }
            if(nonprime==false){
                prime++;
                num+=2;
            }
        }
    }
    System.out.println("The nth prime number is: "+num);
    }
}

}

Answer 1

Check this loop:

for (int i=3; i <= square; i++){
        if (num % i == 0){
            num+=2;
        }
        else {
            prime++;
            num+=2;
        }
    }

It appears you want to loop through all odd numbers up to the square root of your number. If one of them divides, then it should STOP the loop and mark it as a non-prime. If it doesn't divide any number, you should mark it as a prime, but even that should be done after the loop has completed. (eg. when i > square)

I don't want to give you the answer as it appears you want to fix your existing loop yourself. But one strategy is to mark it as a non-prime (in a boolean) for example, and then after the loop, check the boolean and increment your prime count (prime++) if the boolean indicates it is a prime. Remember to reinitialize the boolean so that it will be set correctly next time it hits the for loop.

Answer 2

Create another method to check whether number is prime or not and use while loop to get the primes.

public static void main(String[] args) {
    ...

    if (nthprime == 1) {
        System.out.println("The nth prime number is: 2");
    } else {
        int num = 3;

        for (int i = 2; i <= nthPrime; i++) {
            while(!isPrime(num)) {
                num += 2;
            }

            num += 2;
        }

        System.out.println("The nth prime number is: "+ (num - 2));
    }
}

private static boolean isPrime(int n) {
    int sqrt = (int) Math.sqrt(n);

    for (int i = 2; i <= sqrt; i++) {
        if (n % i == 0) {
            return false;
        }
    }

    return true;
}

Answer 3

import java.util.Scanner;
public class NewNthPrime {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        for(int a0 = 0; a0 < t; a0++){
            boolean flag = false;
            int n = in.nextInt();
            int[] prime = new int[n];
            prime[0] = 2;
            int i = 3,j=1;

            while(j<n){
                for(int y=0;y<prime.length;y++){
                    if(prime[y]==0){
                        break;
                    }
                    else if (i%prime[y]==0){
                       flag = true;
                       break;
                    }
                }
                if(!flag){
                   prime[j++]=i;
                }
                flag =false;
                i+=2;
            }
            System.out.println(prime[n-1]);
        }
    }
}

here i declare an array named prime which stores all prime numbers. The while loop section checks whether any prime numbers which is stored in array-prime[] divides the number 'i' ( which are only odd no. ) or not. If it does then flag becomes true and we skip for the next number 'i'. Here inside while i also check if number inside array is zero then it break the for loop because we can't have any more prime numbers in prime array. now if the number 'i' isn't divisible by any prime number which is in array then 'flag' remains false and 'i' put inside array prime[].

Answer 4

Sorry, I haven't got time to look into your problem, but I think we can also find nth prime number this way as well:

public static void main(String[] args) {
    // TODO Auto-generated method stub
    System.out.println("Enter nth position for which you want to find prime number ");
    Scanner sc  = new Scanner(System.in);       
    int k = sc.nextInt();
    findnthprimenumber(k);  
}

private static void findnthprimenumber(int k) {
    // TODO Auto-generated method stub
    int count=0;
    int j=2;    
    int l =0;
    while(l<k){
        for(int i=2;i<=j/2;i++)
        {
            if(j%i != 0)                
                continue;               
            else 
                count++;                                                    
        } 
        if(count == 0) {
            if(l==(k-1))
            System.out.println("Prime number on "+ k +"th position is "+ j);
            l++;
        }
        count=0;
        j++;
    }
}