Ada*_*les 9 java primes number-theory
下面的代码片段检查给定的数字是否为素数.有人可以向我解释为什么这有效吗?这段代码是我们为Java考试提供的学习指南.
public static void main(String[] args)
{
int j = 2;
int result = 0;
int number = 0;
Scanner reader = new Scanner(System.in);
System.out.println("Please enter a number: ");
number = reader.nextInt();
while (j <= number / 2)
{
if (number % j == 0)
{
result = 1;
}
j++;
}
if (result == 1)
{
System.out.println("Number: " + number + " is Not Prime.");
}
else
{
System.out.println("Number: " + number + " is Prime. ");
}
}
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Ric*_*gle 25
条件if (number % j == 0)询问是否可以number被整除j
素数的定义是
一个只能被自身整除的数字和1
因此,如果你测试2和数字之间的所有数字,并且它们都不是完全可分的那么它就是素数,否则就不是.
当然,你实际上并没有全力以赴number,因为number不能完全被上半部分整除number.
这部分运行增加j的值,如果我们假装number= 12那么它将运行j= 2,3,4,5,6
int j = 2;
.....
while (j <= number / 2)
{
........
j++;
}
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result如果在任何点number上可以被整除,则此部分设置为1 j.一旦设置为1,它result永远不会重置为0.
......
if (number % j == 0)
{
result = 1;
}
.....
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当然,你可以进一步改进,因为你实际上需要不高于sqrt(number)此但是这个片段决定不这样做; 你不需要更高的原因是因为如果(例如)40可以被4整除,则它是4*10,你不需要测试4和10.并且在那些对中,总是在下面sqrt(number).
值得注意的是,它们似乎打算result用作布尔值,但实际上使用整数0和1代表真和假.这不是好习惯.
我试图评论每一行来解释正在进行的过程,希望它有所帮助!
int j = 2; //variable
int result = 0; //variable
int number = 0; //variable
Scanner reader = new Scanner(System.in); //Scanner object
System.out.println("Please enter a number: "); //Instruction
number = reader.nextInt(); //Get the number entered
while (j <= number / 2) //start loop, during loop j will become each number between 2 and
{ //the entered number divided by 2
if (number % j == 0) //If their is no remainder from your number divided by j...
{
result = 1; //Then result is set to 1 as the number divides equally by another number, hergo
} //it is not a prime number
j++; //Increment j to the next number to test against the number you entered
}
if (result == 1) //check the result from the loop
{
System.out.println("Number: " + number + " is Not Prime."); //If result 1 then a prime
}
else
{
System.out.println("Number: " + number + " is Prime. "); //If result is not 1 it's not a prime
}
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