请查看以下代码
public class Prime
{
public static void main(String[]args)
{
int i = 2;
int counter = 0;
while(true)
{
if(counter==6)//Count still 6
{
break;
}
else
{
if(getPrimes(i)==true)
{
i++;
counter++;
System.out.println("Counter: "+counter);
}
else
{
System.out.println("No");
}
}
}
}
static boolean getPrimes(int num)
{
boolean result = false;
int i = 2;
while(true)
{
if((num%i) != 0) //if the number cannot be divided by any other number (except 1 and it self) it is prime
{
result = true;
System.out.println(num);
System.out.println("I is: "+i);
i=2;
break;
}
else //Not a prime. Repeat the process
{
result = false;
i++;
}
}
return result;
}
}
在这里,我试图获得0-6之间的所有素数。这是从一个非常大的数字中获取数千个素数的测试案例。但是,它没有显示唯一的素数,它显示每个数字!
我在这做错了什么?请帮忙!
答案 0 :(得分:4)
在你的循环中试试这个答案
static boolean getPrimes(int num)
{
boolean result=true; // incase u gave 1 or 2 as input.....
int i = 2;
int mid=num/2;
while(i<mid)
{
if(num%i==0)
{
result=false; // not a prime and breaks...
break;
}
else
{
result=true; // the loop make result as true.....
}
i++;
}
return result;
}
答案 1 :(得分:2)
我认为你需要这样的东西:
public void getPrimes(int a){
for(int i = 2; i < a; i++){
int inCounter = 0;
if(counter%i==0){
System.out.println("false");
inCounter++;
}
if(inCounter == 0){
System.out.println("Prime: "+counter);
}
}
}
答案 2 :(得分:0)
使用筛子,Eratosthenes或Atkins
以下是Robert Sedgewick在Java中实施的Eratosthenes:
http://introcs.cs.princeton.edu/java/14array/PrimeSieve.java.html
答案 3 :(得分:0)
我想我找到了解决方案。至少,我找到了我需要的答案。这是我的答案
import java.math.BigInteger;
public class Problem7
{
public static void main(String[]args)
{
int i = 0;
int counter = 0;
while(true)
{
if(i>6)
{
break;
}
if(i>1)
{
String str = String.valueOf(i);
if (new BigInteger(str).isProbablePrime(i/2))
{
System.out.println(str);
counter++;
}
}
i++;
}
}
}
我想这是最简单的方法......
答案 4 :(得分:0)
import java.util.Scanner;
class PrimeNumbers2
{
public static void main (String[] args)
{
Scanner scanner = new Scanner(System.in);
int i =0;
int num =0;
//Empty String
String primeNumbers = "";
System.out.println("Enter the value of n:");
int n = scanner.nextInt();
for (i = 1; i <= n; 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 n are :");
System.out.println(primeNumbers);
}
}
它将显示所有素数。
答案 5 :(得分:0)
以最小迭代次数查找素数
boolean IsPrimeNumber(int num){
boolean isPrime = true;
if (num == 1 || num ==0)
isPrime = false;
else{
int limit = (int) Math.sqrt(num);
for (long i = 2L; i <=limit ; i++)
if (num % i == 0) {
isPrime = false;
break;
}
}
return isPrime;
}