java中1到100之間的素數

Question

伙計們，我一直在試圖弄清楚為什么這段代碼在將近 5 個小時內沒有顯示適當的結果！！！ 這應該是一項非常簡單的任務，但它讓我發瘋。 我必須讓項目使用包含的代碼而不是另一個類似的代碼。 我遇到的問題是我的軟件底部沒有顯示素數。 根據我的理解，這段代碼應該顯示 1-100 的質數，並且除了最后一個數字外，所有的數字都是 5 行。 非常感謝一些指導。

public class Prime numbers 1-100{
public static void main (String[] args) {

    int number = 100;

    // Assume the number is prime
    boolean isPrime = true; // Is the current number prime?

    // Test if number is prime
    for (int divisor = 2; divisor <= number / 2; divisor++) {
        if (number % divisor == 0) { // If true, number is not prime
            isPrime = false; // Set isPrime to false
            break; // Exit for loop
        }
        // print prime numbers
        if(isPrime)
            System.out.println("Prime numbers between 1 and " + number);
    }
}

}

Answer 1

您的代碼應該具有嵌套循環和一個數組，該數組存儲要在最后顯示的素數。

要在最后打印它們，您應該首先打印您的語句，然后使用循環打印包含找到的素數的數組。

Answer 2

如果你想打印從 1 到 100 的所有質數，那么你必須遍歷所有這些數字（第一個循環），並為它們中的每一個迭代所有可能的除數（嵌套循環）：

int number = 100;

boolean isPrime = false;

System.out.println("2");
System.out.println("3");

for (int i = 5; i <= number; i++) {
    for (int divisor = 2; divisor <= Math.sqrt(i); divisor++) {
        isPrime = !(i % divisor == 0);
        if (!isPrime)
            break;
    }

    if (isPrime)
        System.out.println("" + i);
}

請注意，對於 2 和 3，無法應用條件，因此首先打印它們。
我使用了@selbie 的優化提示<= sqrt(number)

Answer 3

public class Prime{
public static void main (String[] args) {
    boolean isPrime = true; 
    // iterate over all numbers from 1 to 100
    for (int number = 1; number <= 100; number++) {
        // We want to believe it's a prime
        isPrime = true;
        // 1 is not a prime
        if (number == 1) isPrime = false;
        // iterate over all numbers which are lower then 'number'
        for (int divisor = 2; divisor < number && isPrime; divisor++) {
            // if we can divide by it, it's not a prime
            if (number % divisor == 0) {
                isPrime = false;
            }
        }
        // print if it's a prime
        if (isPrime) System.out.println(number+", ");
    }
}}

Answer 4

我要感謝大家的建議和提示。 以及推動我的動力。 我能夠讓程序運行。 見下文

public class primes {
public static void main (String[] args) {
    final int NUMBER_OF_PRIMES = 26; // Number of primes to display
    final int NUMBER_OF_PRIMES_PER_LINE = 5; // Display 5 numbers per line
    int count = 0; // Count the number of prime numbers
    int number = 1; // A number to be tested for primeness

    System.out.println("The prime numbers between 1 and 100 are \n");

    // Repeatedly find prime numbers
    while (count < NUMBER_OF_PRIMES) {
        // Assume the number is prime
        boolean isPrime = true; // Is the current number prime?

        // Test whether number is prime
        for (int divisor = 2; divisor <= number / 2; divisor++) {
            if (number % divisor == 0) { // If true, number is not prime
                isPrime = false; // Set isPrime to false
                break; // Exit the for loop
            }
        }

        // Display the prime number and increase the count
        if (isPrime) {
            count++; // Increase the count

            if (count % NUMBER_OF_PRIMES_PER_LINE == 0) {
                // Display the number and advance to the new line
                System.out.println(number);
            }
            else
                System.out.print(number + " ");     
            }

        //Check if the next number is prime
        number++;
        }
    }
}