Sum of Prime Numbers in Java

2025年5月7日 | 阅读 5 分钟

在本节中，我们将创建 Java 程序 来查找给定范围内的所有 素数的总和。在继续本节之前，让我们先了解一下素数的重要事实。

素数是大于 1 且只能被 1 和 它本身 整除而没有余数的数字。
数字 0 和 1 不是素数。
唯一的偶素数是 2。所有其他偶数都可以被 2 整除。

查找素数和的步骤

读取或初始化下限和上限。
迭代一个循环（for 或 while）以查找给定范围内的素数。
如果该数字是素数，则将其添加到变量 sum 中，并打印结果。

让我们在 Java 程序中实现上述步骤。

Java 程序：查找素数和

查找素数和的以下方法：

使用 Java for 循环
使用 Java while 循环
使用函数
使用动态规划

让我们分别为每种方法创建 Java 程序。

使用 for 循环

SumOfPrimeNumbersExample1.java

public class SumOfPrimeNumbers1 
{
public static void main(String args[]) 
{
int count, sum = 0;
//the loop executes 100 time and increments the variable number by 1 after each iteration
for(int number = 1; number <= 200; number++)
{
count = 0;
for (int i = 2; i <= number/2; i++)
{
//find the remainder and check if it is equal to 0 or not    
if(number % i == 0)
{
//increments the count variable by 1 if the above condition returns true
count++;
break;
} //end of if statement 
} //end of for loop
//returns true if both conditions are true
if(count == 0 && number != 1 )
{
//calculates the sum of prime numbers    
sum = sum + number;
}  //end of if statement 
}  //end of for loop
//prints the sum
System.out.println("The Sum of Prime Numbers from 1 to 200 is: " + sum);
} //end of main
} //end of class

输出

The Sum of Prime Numbers from 1 to 200 is: 4227

使用 while 循环

在下面的程序中，我们使用了 while 循环而不是 for 循环。逻辑与上面相同。

SumOfPrimeNumbersExample2.java

public class SumOfPrimeNumbersExample2
{
public static void main(String args[]) 
{
int number = 1, count, sum = 0;
//executes until the condition becomes false
while(number <= 100)
{
count = 0;
int i = 2;
while(i <= number/2 )
{
//find the remainder and check if it is equal to 0 or not        
if(number % i == 0)
{
//increments the count variable by 1 if the above condition returns true
count++;
break;
}
//increments the variable i by 1
i++;
} //end of while
if(count == 0 && number != 1 )		    
{
//calculates the sum of prime numbers    
sum = sum + number;
} //end of if statement
//increments the variable number by 1 after completing each iteration
number++;
}  //end of while
//prints the sum
System.out.println("The Sum of Prime Numbers from 1 to 100 is: " + sum);
}  //end of main
} //end of class

输出

The Sum of Prime Numbers from 1 to 100 is: 1060

使用函数

SumOfPrimeNumbersExample3.java

public class SumOfPrimeNumbersExample3
{
//function that checks if the given number is prime or not
static boolean isPrime(int num)
{
if(num == 1) 
{
return false;
} //end of if
for (int i = 2; i*i <= num; i++) 
{
if (num % i == 0) 
{
return false;
} //end of if statement
} //end of for
return true;
}
//method iterate over min to max
//function that calculates the sum, if the number is detected as prime
static int addPrime(int min, int max)
{
int sum = 0;
//loop starts from 50 and decrement the variable i by 1 after each iteration
for (int i = max; i >= min; i--) 
{
//calling the user-defined method to check the number is prime or not
boolean prime = isPrime(i);
if (prime) 
{
//calculates the sum 
sum = sum + i;
} //end of if statement
} //end of for loop
//returns sum
return sum;
}
//driver code
public static void main(String args[])
{
int min = 0, max = 50;
//call the user-defined method and print the sum
System.out.println("The sum of all the prime number between the given range is: " +addPrime(min, max));
} //end of main
} //end of class

输出

The sum of all the prime number between the given range is: 328

使用动态规划

使用动态规划计算素数和的逻辑有点棘手，实现起来比较困难。在这种方法中，我们将使用数组。让我们看看步骤。

声明两个数组 number[] 和 array[]。
将 array[] 初始化为 0。
循环迭代直到 sqrt(uprlimit)。如果数组元素为 0，则将其视为素数，并将其倍数视为非素数。在标记非素数时，将相应的位置设置为 1。
更新保存素数和的 number[] 数组。number[i] 的每个位置都代表指定范围 [1, i] 内所有素数的和。

让我们在 Java 程序中实现上述步骤。

SumOfPrimeNumbersExample4.java

public class SumOfPrimeNumbersExample4
{
//suppose the constraint is N<=1000
static int uprlimit = 1000;
//declaring an array
static long numbers[] = new long[uprlimit + 1];
//Method to compute the array
static void findSum()
{
//declaring another array
int array[] = new int[uprlimit + 1];
//initialize 0th and 1st element of the array to 1
array[0] = 1; 
array[1] = 1;
//2 is an even prime number so we will iterate the loop from 2 to sqrt(uprlimit)
for (int i = 2; i <= Math.sqrt(uprlimit); i++)
//if ith element of the array is 0, mark it as prime
if (array[i] == 0)
//mark all of it's multiples till uprlimit as non-prime 
for (int j = i * i; j <= uprlimit; j = j + i)
array[j] = 1;
long primeSum = 0;
//update the array number[] with the running sum of prime numbers within the range [1, uprlimt]
for (int i = 1; i <= uprlimit; i++)
{
if (array[i] == 0)
primeSum = primeSum + i;
//here, number[i] represents the sum of all the prime numbers within the range [1, i]
numbers[i] = primeSum;
} //end of for
} //end of method
//driver code
public static void main(String args[])
{
int lr = 100, ur = 1000;
//calling user-defined function
findSum();
System.out.println("The sum of all the prime number between given range is: "+(numbers[ur] - numbers[lr - 1]));
}   //end of main
}  //end of class

输出

The sum of all the prime number between the given range is: 75067

下一个主题Calculate-the-discriminant-value-in-java

Sum of Prime Numbers in Java

查找素数和的步骤

Java 程序：查找素数和

使用 for 循环

使用 while 循环

使用函数

使用动态规划

联系信息

关注我们

教程

面试题

在线编译器

Python

Java

.Net Framework

AI, ML and Data Science

Cloud Technology

B.Tech and MCA

Web Technology

PHP

Software Testing

Technical Interview

Java Interview

Python

Web Interview

Database Interview

B.Tech / MCA

Important Interview

Software Testing Interview

Company Interviews

Online Compilers

Multiple Choice Questions

Java Conversion

Java Misc

Sum of Prime Numbers in Java

查找素数和的步骤

Java 程序：查找素数和

使用 for 循环

使用 while 循环

使用函数

使用动态规划

相关帖子

哪个组件负责运行 Java 程序

LinkedHashMap vs HashMap

Java 中将 Long 转换为 Date

Java CORS Error

How to Generate Random Number in Java

Java 中的商业消费者问题

Bitonic Point in Java| Find Bitonic Point in Given Bitonic Sequence in Java

Java 中的提醒程序

Remove Special Characters from String

EJB vs. Spring

订阅 Tpoint Tech

联系信息

关注我们

教程

面试题

在线编译器