小编Uma*_*nth的帖子

我们知道可以使用以下方法生成3以上的所有素数:

6 * k + 1
6 * k - 1

但是,我们从上面的公式生成的所有数字都不是素数.

For Example:    
6 * 6 - 1 = 35 which is clearly divisible by 5.

为了消除这些条件,我使用Sieve方法并删除了数字,这些数字是从上面公式生成的数字的因子.

使用事实:

如果没有素数因素,那么一个数字被称为素数.

1. 因为我们可以使用上面的公式生成所有素数.
2. 如果我们可以删除上述数字的所有倍数,我们只剩下素数.

生成低于1000的素数.

ArrayList<Integer> primes = new ArrayList<>();
primes.add(2);//explicitly add
primes.add(3);//2 and 3
int n = 1000;
for (int i = 1; i <= (n / 6) ; i++) {
//get all the numbers which can be generated by the formula
    int prod6k = 6 * i;
    primes.add(prod6k - 1); …

我有这种格式的JSON对象.

[
    {
        "name": "schoolname",
        "line id": "0",
        "time": "4-5",
        "minage": "15",
        "maxage": "35"
    },
    {
        "name": "studentname1",
        "line id": "1",
        "class": "A"
    },
    {
        "name": "studentname2",
        "line id": "2",
        "class": "B"
    }
]

我想做的事

从一组指定的标题中,从中获取并将其"line id" : "0"设置为其他项目.

例如: headers = ["time", "minage", "maxage"]

我从中获取这些"line id" : "0"并将其交给其他人.

[
    {
        "name": "schoolname",
        "line id": "0",
        "time": "4-5",
        "minage": "15",
        "maxage": "35"
    },
    {
        "name": "studentname1",
        "line id": "1",
        "class": "A",
        "time": "4-5",
        "minage": "15",
        "maxage": …

我正在尝试编写一个代码来计算给定整数的以下内容n:

1/1 + 1/2 + 1/3 ... + 1/n

这是我到目前为止编写的代码:

public class RecursiveSum
{
  public static double Sumto(int n)
  {
    if (n == 0) { return 0.0; }
    else if (n > 0) { return 1/n + 1/Sumto(n - 1); }
    else { throw new IllegalArgumentException("Please provide positive integers"); }
  }
  public static void main(String[] args)
  {
    System.out.println(Sumto(5));
  }
}

但是,它始终输出:

Infinity

有什么问题,我该如何解决？

谢谢

从数学原理:

数字N可表示为2个平方的总和,当且仅当在N的素数因子分解中,形式的每个素数都(4k+3)出现偶数次!

我所做的是预先计算所有4k+3数字并通过连续分割检查它的频率.

该程序是根据约束编写的:

1 < T <100
0 < N < 10 ^ 12

import java.util.Scanner;

public class TwoSquaresOrNot {
    static int max = 250000;
    static long[] nums = new long[max];

    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for (int i = 0; i < max; ++i)
            nums[i] = 4 * i + 3;
        while (T-- > 0) {
            long n = sc.nextLong();
            System.out.println((canWrite(n) ? "Yes" : …

我试图使用OpenPgpjs加密和签名消息.

但我不断收到此错误"错误加密邮件:私钥未解密"

这是我试过的:

var openpgp = require('openpgp');

var publicKey = [].join("\n"); //This has the complete key. Removed for representation
var privateKey =  [].join("\n"); //This has the complete key. Removed for representation
var publicKeys = openpgp.key.readArmored(publicKey).keys;
var privateKeys = openpgp.key.readArmored(privateKey).keys;

encryptionOptions = {
    data : 'Example Test',
    publicKeys : publicKeys,
    privateKeys : privateKeys
};

return openpgp.encrypt(encryptionOptions).then(function(ciphertext) {
    encryptedData = ciphertext.data;
    console.log(ciphertext);
    return encryptedData;
});

假设我需要继续询问用户,直到他进入double.

我做的是我使用了一个while循环并检查是否有异常.

如果有异常,我会去询问下一个输入.

double bal = 0;
Scanner sc = new Scanner(System.in);
while (true) {
    try {
        System.out.println("Enter the balance");
        bal = sc.nextDouble();
        break;
        } catch (Exception e) {
          System.out.println("That isn't a number");
    }
}
System.out.println("Bal is " + bal);
sc.close();

但是如果我输入一个非double,那么它不会要求下一个输入,继续打印以无限循环结束的那两行.

Enter the balance
XYZ
That isn't a number
Enter the balance
That isn't a number
Enter the balance
That isn't a number
Enter the balance
That isn't a number
....

我错过了什么？