Java 打印从 1 到 100 的质数
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Printing prime number from 1 to 100
提问by Jai
This program is supposed to output the prime numbers between 1 and 100. Will anyone please explain me the flow the programme below? I am having difficulty in writing the programme below. I did find it on the internet but still I can't figure out how it works and how the flow of the programme will be?
这个程序应该输出 1 到 100 之间的素数。有人可以解释我下面程序的流程吗？我在编写下面的程序时遇到困难。我确实在互联网上找到了它，但我仍然无法弄清楚它是如何工作的以及程序的流程如何？
public class GeneratePrimeNumbersExample {
public static void main(String[] args) {
//define limit
int limit = 100;
System.out.println("Prime numbers between 1 and " + limit);
//loop through the numbers one by one
for(int i=1; i < 100; i++){
boolean isPrime = true;
//check to see if the number is prime
for(int j=2; j < i ; j++){
if(i % j == 0){
isPrime = false;
break;
}
}
// print the number
if(isPrime)
System.out.print(i + " ");
}
}
}
Output of Prime Numbers example would be Prime numbers between 1 and 100
素数示例的输出将是 1 到 100 之间的素数
1 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
1 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
回答by Himanshu Bhardwaj
How would you find a prime number with plain vanilla solution?If number is prime. It will not be a multiple of any number other than itself. So assume number is x. This number will not be divisible by any number when starting from 2 till x-1. Why start from 2 and not 1 because every number is divisible by 1. The above code is trying to replicate the same behavior. To find all primes between 1 till 99 (as per the loop):
你将如何用普通的解决方案找到素数？如果数是素数。它不会是除自身以外的任何数字的倍数。所以假设数字是x。当从 2 到 x-1 时，此数字不能被任何数字整除。为什么从 2 而不是 1 开始，因为每个数字都可以被 1 整除。上面的代码试图复制相同的行为。要查找 1 到 99 之间的所有素数（根据循环）：
- From 2 till number from (outer loop - 1)
- Try dividing the number and check if it's divisible. (remainder should be zero).
- If true number is not prime. Else number is prime.
- 从 2 到数字来自（外循环 - 1）
- 尝试将数字相除并检查它是否可整除。（余数应为零）。
- 如果真数不是素数。其他数是素数。
回答by sprinter
If you split the various parts out into their own methods with appropriate names it becomes a bit easier to understand:
如果您将各个部分拆分为具有适当名称的自己的方法，它会变得更容易理解：
for (int n = 1; n < 100; n++)
if (isPrime(n))
System.out.println(n);
private boolean isPrime(int n) {
for (int f = 2; f < n; f++) {
if (isFactor(f, n))
return false;
}
return true;
}
private boolean isFactor(int factor, int number) {
return number % factor == 0;
}
This is also an area where Java 8 streams can make things a bit clearer:
这也是 Java 8 流可以让事情变得更清晰的一个领域：
List<Integer> primes = IntStream.range(1, 100)
.filter(this::hasNoFactors)
.collect(Collectors.toList());
private boolean hasNoFactors(int number) {
return IntStream.range(2, number)
.noneMatch(f -> number % f == 0);
}
Also note that this is a horribly inefficient algorithm. You don't need to check every possible factor from 2 to n, just the primes. You can also take advantage of multi-processor machines:
还要注意，这是一个非常低效的算法。您不需要检查从 2 到 n 的每个可能的因子，只需检查素数即可。您还可以利用多处理器机器：
List<Integer> primes = new ArrayList<>();
IntStream.range(2, 100)
.filter(n -> primes.parallelStream().noneMatch(p -> n % p == 0))
.forEach(primes::add);
回答by Paresh Kumar
public class Prime {
public static void main(String arg[])
{
int count=0;
for(int i=2;i<100;i++)
{
for(int j=2;j<i;j++)
{
if(i%j!=0)
{
count++;
if(count==i-2)
{
System.out.print(i+" ");
}
}
}
count = 0;
}
}
}
回答by user8144104
public class primenum {
public static void main(String[] args) {
for(int i=2;i<100;i++){
int count=1;
for(int j=2;j<i;j++){
if(i%j ==0){
count=0;
break;
}
}if(count==1){
System.out.println(i);
}
}
}
回答by Dinesh Sonachalam
The number which is only divisible by itself and 1 is known as prime number. Here is the simplest version of the code for finding prime numbers between 1 to 100.
只能被它自己和1整除的数称为质数。这是用于查找 1 到 100 之间的质数的代码的最简单版本。
import java.io.*;
import java.util.*;
class solution
{
public static void main(String args[])
{
Scanner scan = new Scanner(System.in);
int n = 100;
/*
A prime number is a whole number greater than 1, whose only two whole-number factors are 1 and itself.
*/
for(int i=2;i<=n;i++) // 1.So we are starting with initialization i = 2
{
int flag = 1;
for(int j=2;j<=i/2;j++) // 2.Try dividing the number by half check whether it divisible
{
if(i%j==0) // 3. If the number is divisible by other number ->Not a prime Number
{
flag = 0;
break;
}
}
if(flag==1) // 4. If the number is not divisible by any other numbers but only by itself and 1 -> prime no
{
System.out.print(i+" ");
}
}
}
}
/*
Output:
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
*/
回答by Motolola
public static ArrayList prime(int limit){
公共静态 ArrayList 素数（整数限制）{
ArrayList<Integer> primes = new ArrayList<>();
for(int p = 2; p <= limit; p++) {
int count = 0;
for(int i=2; i < p; i++) {
if (p%i == 0) {
count++;
}
}
if (count == 0) {
primes.add(p);
}
}
return primes;
}
回答by Mahendra Sri
public static List<Integer> getPrimeNumbers(int from , int to) {
List<Integer> list = new ArrayList<>();
for (int i = from;i <= to; i++) {
int count = 0;
for (int num = i; num>=1;num--) {
if(i%num == 0){
count++;
}
}
if(count ==2) {
list.add(i);
}
}
return list;
}
