我想找到从1到1000的素数,以及从1到10000的完美数字。最后我得到了代码输出的工作原理。但我想在一条线上打印10个素数。
例如:
2 3 5 7 11 13 17 19 23 29然后
31 37 41 43 47 53 59 61 67 71
然而,我的输出就像这样
235791113171923293137414347
有两组代码,一组是方法,是调用方法的实际程序
public class MyMath
{//class start
//method to define prime number
public static boolean isPrime( long num )
{//start isPrime method
//declare and initialize prime variable as true
boolean isPrime = true;
if ( num >= 2)
{//start if statement
for ( long counter = 2; counter < num; counter++)
{//start for loop
//define whether num is prime or not with
// a reminder operator
if( num%counter == 0)
isPrime = false;//if not initialize variable to false
}//end for loop
}//end if
//check for negative number and two
if( num <= 1 )
isPrime = false;
if ( num == 2)
isPrime = true;
return isPrime;//send back prime variable
}//end method
//method to define perfect numbers
public static boolean isPerfect( long num )
{//start method
//declare and initialize perferct number variable as true
boolean isPerfect = true;
long sum = 0;
//check for input
if ( num <= 1)
isPerfect = false;
//if num is larger than one
if ( num > 1)
{//start if statement
//use a for loop to find the factor
for(long factor = 1; factor <= num/2; factor++ )
{//start for loop
if( num%factor == 0 )
sum += factor;
}//end for loop
}//end if statement
//if the sum of the factor equal to num
//initialize perfect number variable as true
//else perfect number variable is false
if ( num == sum )
isPerfect = true;
else
isPerfect = false;
return isPerfect;//send back perfect number variable
}//end method
}//end class
这是实际的代码
import javax.swing.JOptionPane;//import JOptionPane for dialog boxes
public class MyMathTest
{//start MyMathTest class
public static void main (String args [])
{//start main
//a string to prompt the user in a dialog box
String choice = "Enter 1 to display the prime numbers from 1 - 1,000\n" +
"Enter 2 to display perfect numbers from 1 - 10,000\n" +
"Enter 3 to quit\n" + "written by blabla";
int userInput;//declare a variable for user's input
do
{//start do while loop to check for input range
String prompt = JOptionPane.showInputDialog( choice );
userInput = Integer.parseInt ( prompt );
}
while( userInput < 1 ^ userInput > 3);//repeat the prompt if input is larger than 3 or smaller than 1
//declare and initialize a counter for line format of the string
int lineCounter = 1;
switch ( userInput )
{//start switch
case 1:
//declare and initialize a string for header and hold the numbers of prime
String primeNumberString = String.format ("The prime number from 1 - 1000 are\n");
for (int primeCounter = 1; primeCounter < 1000; primeCounter++ )
{//start loop
//print a new line after 2 prime number
//and restart to count the counter from 1
if ( lineCounter == 11 )
{//start if
primeNumberString += "\n";
lineCounter = 1;
}//end if
//if prime number is true
if (MyMath.isPrime( primeCounter ))
{//start if
//add number to the string
primeNumberString += String.format ("%d", primeCounter );
//add tone to lineCounter after every prime number
++lineCounter;
}//end if
}//end loop
//display prime number message within a dialog box
JOptionPane.showMessageDialog ( null,primeNumberString );
//exit the switch
break;
case 2:
//declare and initialize a string for header and hold the perfect numbers
String perfectNumberString = "Perfect numbers from 1 - 10,000 are:\n";
for (int perfectCounter = 1; perfectCounter < 10000; perfectCounter++)
{//start the loop
//if the perfect number is true
if(MyMath.isPerfect( perfectCounter ))
//add perfect number into the string
perfectNumberString += String.format ( "%d ", perfectCounter);
}//end for loop
//display perfect number message within a dialog box
JOptionPane.showMessageDialog ( null, perfectNumberString );
break;//exit switch
case 3:
break;//exit switch
}//end switch
}//end main method
}//end class
答案 0 :(得分:0)
如果条件
,再添加一个if(lineCounter == 10){
primeNumberString += "\n";
lineCounter = 1;
}
这将创建一个新行并休息lineConter。
如果你可以将primeNumberString更改为StringBuffer,那么你可以将+
保存为错误。