我在获取给定整数A和整数B之间的所有素数时遇到问题。
问题在于输出结果远远超出了我为B定义的结果。我认为
if (isPrime){
count++;
将解决此问题,但输出仍然远远超过整数B的预期数量。
例如,如果int valueA = 1
和int valueB = 100
,它将在停止之前获得大约1到500的质数,而不仅仅是在100处结束支票。
谢谢您的帮助。
import java.util.*;
public class PrimeNumbersTest {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner input = new Scanner(System.in);
// Ask user to input an integer value for A and B
System.out.print("Enter the value of A (must be an integer): ");
int valueA = input.nextInt();
System.out.print("Enter the value of B (must be an integer): ");
int valueB = input.nextInt();
System.out.println("The prime numbers between " + valueA + " and " + valueB + " are:");
final int LINE = 10;
int count = valueA;
int number = 2;
while (count < valueB) {
// Assume the number is prime
boolean isPrime = true;
// Test if number is prime
for (int divisor = 2; divisor <= number / 2; divisor++) {
if (number % divisor == 0) { // If true number is not prime
isPrime = false; // Set isPrime to false
break; // Exit the for loop
}
}
if (isPrime) {
count++;
if (count % LINE == 0) {
System.out.println(number);
}
else
System.out.print(number + " ");
}
number++;
}
}
}
答案 0 :(得分:0)
当您要处理一定范围的数字时,请使用for循环而不是while循环,以减少混乱。您显然想要
for(int number = valueA; number <= valueB; number++){/*check if number is prime*/}
示例:
public static void main(String []args){
int valueA = 1;
int valueB = 100;
int count = 0;
for(int number = valueA; number <= valueB; number++)
{
if(isPrime(number))
{
count++;
System.out.println(number);
}
}
System.out.println("count = " + count);
}
public static boolean isPrime(int n)
{
for(int i = 2; i*i <= n; i++)
{
if(n % i == 0)
{
return false;
}
}
return n > 1;
}
对于您检查isPrime内联的要求:
public static void main(String []args){
int valueA = 1;
int valueB = 100;
int count = 0;
for(int number = valueA; number <= valueB; number++)
{
boolean isPrime = number > 1;
for(int i = 2; i*i <= number; i++)
{
if(number % i == 0)
{
isPrime = false;
break;
}
}
if(isPrime)
{
count++;
System.out.println(number);
}
}
System.out.println("count = " + count);
}