Question

我有下面的java程序，该程序打印所有素数到100，现在我想对其进行修改，以便用户输入该数字，它不会打印范围，但该范围中存在最后一个素数，请告知如何修改以下程序，我也想将返回类型从字符串更改为int

onAttach

现在输出为

 class PrimeNumbers
 {
   public static void main (String[] args)
   {        
       int i =0;
       int num =0;
       //Empty String
       String  primeNumbers = "";

       for (i = 1; i <= 100; i++)         
       {              
          int counter=0;      
          for(num =i; num>=1; num--)
      {
             if(i%num==0)
         {
        counter = counter + 1;
         }
      }
      if (counter ==2)
      {
         //Appended the Prime number to the String
         primeNumbers = primeNumbers + i + " ";
      } 
       }    
       System.out.println("Prime numbers from 1 to 100 are :");
       System.out.println(primeNumbers);
   }
}

我只希望仅打印最后一个素数

Prime numbers from 1 to 100 are :
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

Answer 1

尝试仅维护状态以获取循环发现的最新素数：

// Top-level build file where you can add configuration options common to all sub-projects/modules.

buildscript {
    ext.kotlin_version = '1.3.21'

    repositories {
        google()
        jcenter()
        mavenCentral()
        maven { url "https://dl.bintray.com/android/android-tools" }

    }
    dependencies {
        classpath 'com.google.gms:google-services:4.1.0'
        classpath 'com.android.tools.build:gradle:3.3.2'
        classpath "org.jetbrains.kotlin:kotlin-gradle-plugin:$kotlin_version"

        // NOTE: Do not place your application dependencies here; they belong
        // in the individual module build.gradle files
    }

}

allprojects {
    repositories {
        google()
        jcenter()
        maven { url "https://dl.bintray.com/android/android-tools" }


    }

}

Answer 2

您可能想引入一个cv::Vec3f getEyeball(cv::Mat &eye, std::vector<cv::Vec3f> &circles) { std::vector<int> sums(circles.size(), 0); for (int y = 0; y < eye.rows; y++) { uchar *ptr = eye.ptr<uchar>(y); for (int x = 0; x < eye.cols; x++) { int value = static_cast<int>(*ptr); for (int i = 0; i < circles.size(); i++) { cv::Point center((int)std::round(circles[i][0]), (int)std::round(circles[i][1])); int radius = (int)std::round(circles[i][2]); if ( std::pow(x - center.x, 2) + std::pow(y - center.y, 2) < std::pow(radius, 2)) { sums[i] += value; } } ++ptr; } } . . . }变量（int），该变量将保存最大质数的值。对问题代码进行了一些小的更改：

int primeNumber

Answer 3

根据我的理解，您要打印接近100的最大素数。

 class PrimeNumbers
{
   public static void main (String[] args)
   {        
   int i =0;
   int num =0;
   //Empty String
   String  primeNumbers = "";

   for (i = 1; i <= 100; i++)         
   {              
      int counter=0;      
      for(num =i; num>=1; num--)
   {
         if(i%num==0)
     {
    counter = counter + 1;
     }
  }
  if (counter ==2)
  {
     //Appended the Prime number to the String
     primeNumbers =  i+ "";
  } 
   }    
   System.out.println("Max prime number between 1 to 100 is:");
   System.out.println(primeNumbers);
  }
}

更新：

可以进一步优化该程序以改善时间复杂度

public class  PrimeNumbers{

 public static void main (String[] args)
{        
int i =0;
int num =0;

int  primeNumbers = 0;

for (i = 100; i >= 1; i--)         
{              
  int counter=0;      
  for(num =i; num>=1; num--)
{
     if(i%num==0)
 {
  counter = counter + 1;
 }
}
if (counter ==2)
 {
  primeNumbers =  i;
  break;
 } 
}    
 System.out.println("Max prime number between 1 to 100 is:");
 System.out.println(primeNumbers);
}
}

Answer 4

建议的源代码可以编写得更加结构化：

public class PrimeNumbers {
    private static boolean isPrime(int n) {
        if (n == 2) return true;
        if (n % 2 == 0) return false;
        for (int i = 3; i <= Math.sqrt(n); i = i + 2)
            if (n % i == 0) return false;
        return true;
    }

    public static void main(String[] args) {
        int biggestPrimeNumber = 0;
        String primeNumbers = "";
        for (int n = 2; n <= 100; n++)
            if (isPrime(n)) {
                biggestPrimeNumber = n;
                //Appended the Prime number to the String
                primeNumbers += n + " ";
            }
        System.out.println("Prime numbers from 1 to 100 are: " + primeNumbers);
        System.out.println("Last prime number from 1 to 100 is: " + biggestPrimeNumber);
    }
}

要检查数字的素数，有很多算法，例如Miller–Rabin primality test，但是一种简单的蛮力方法是检查数字，直到所需数字的sqrt。

更改退货类型以及最后打印的号码

4 个答案: