Question

我已完成此代码，但似乎需要的时间比允许的时间长代码是在多个两个数字之间生成素数。

以下是代码：

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.sql.Time;
import java.util.ArrayList;

public class App {

    public static boolean isPrime(int p) {
        int i;
        boolean t = false;

        if (p == 2) {
            t = true;
        } else {
            for (i = 2; i < p; i++) {
                if (p % i == 0) {
                    t = false;
                    break;
                } else {
                    t = true;
                }
            }
        }
        return t;
    }

    public static void numOfPrimes(int a, int b) {

        if (a >= 1 && a <= b && b <= 1000000000 && (b - a) <= 100000) {
            int i, prim = 0;
            boolean t = false;

            for (i = a; i <= b; i++) {
                t = isPrime(i);

                if (t) {
                    System.out.println(i + "\n");
                    prim++;
                }
            }
        } else {
            System.out.println("bad input!!");
            System.out.println("inputs must be just like (1 <= m <= n <= 1000000000, n-m<=100000) ");
        }
    }

    public static void main(String[] args) throws IOException, NumberFormatException {

        long t1 = System.currentTimeMillis();
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = 0, counter = 0;
        String str;

        try {
            str = br.readLine();
            t = Integer.parseInt(str);
            ArrayList<String> arr = new ArrayList<String>();

            while (counter < t) {
                String s = br.readLine();
                arr.add(s);
                counter++;
                System.out.println("\n");
            }

            for (int x = 0; x < arr.size(); x++) {
                StringBuffer s1 = new StringBuffer(" "), s2 = new StringBuffer("");
                int q = 0, i, j;
                int num1, num2;

                for (i = 0; i < arr.get(x).indexOf(' '); i++) {
                    q++;
                    s1 = s1.append(arr.get(x).charAt(i));
                }

                String s5 = s1.toString();

                for (j = q + 1; j < arr.get(x).length(); j++) {
                    s2 = s2.append(arr.get(x).charAt(j));
                }

                String s6 = s2.toString();

                try {
                    num1 = Integer.parseInt(s5.trim());
                    num2 = Integer.parseInt(s6.trim());
                    App.numOfPrimes(num1, num2);
                    System.out.println("\n");
                } catch (NumberFormatException e) {

               //Will Throw exception!
               //do something! anything to handle the exception.
                }
            }
        } catch (Exception ex) {
            System.out.println("IO error ");
            System.exit(1);
        }

        long t4 = System.currentTimeMillis();
        System.out.println(t4 - t1);
    }
}

Answer 1

你在这里使用了错误的算法。查看Sieve of Erasthothenes，然后根据您的需要进行修改。此外，你想找到介于1和1x10 ^ 9之间的素数，这是相当多的。也许您需要详细说明为什么需要这个。

编辑：

作弊方式：从http://compoasso.free.fr/primelistweb/page/prime/liste_online_en.php中删除数字，而不是生成它。

如何减少Java程序的执行时间？

1 个答案: