我已经放弃了号码生成器,因为我知道它实际上是将每个号码分开,直到它收到一个素数。我更改了代码,以便在"a" to "?" (in this case, 10)
方法中指定范围Prime_2
的素数在Prime方法中被检查为素数。然后,如果该数字是素数,则通过将布尔变量prime设置为true或false来返回Prime方法,但到目前为止,我只得到2为真,其余为假。这显然不是真的。因此,我感谢任何协助/意见/建议,使这个新计划可行。
public bool Prime(long num) // Prime method with a parameter for one number
{
int div = 3; // what we divide by after checking if the number is divisible by 2
bool prime = true; // set prime to true
{
for (long i = 0; i < 100 && prime == true; i++) // run 100 passes
{
if (num % 2 == 0 && num != 2) // if the number is divisible by 2
{ // and is not 2, prime is false.
prime = false;
}
else if (num % 2 != 0 && num != 2) // if the number is not divisible
{ // by 2 and the number is not 2...
for (long x = 0; x <= 1000; x++) // then run 1000 passes of this:
{
if (num % Math.Pow((div), x) == 0 && num != Math.Pow((div), x))
{ // if the number is divisible by our div to the power of x
// and the number is not equal to div to the power of x,
// prime is false.
prime = false;
}
}
}
else // otherwise add 2 to div making it the next consecutive odd number
{ // and run the pass again
div = div + 2;
}
}
return prime;
}
}
public void Prime_2() // void Prime_2 method
{
long a = 2; // starting number 2
long b = 0; // set b
Program prg = new Program(); //new instance of the Program class
while (a <= 10)//the range a (2) - 10
{
b = a;//set "b" to "a" every time
prg.Prime(b); // run the Prime method for numbers 2-10
Console.WriteLine(b); // write the number being checked
Console.WriteLine(prg.Prime(b)); // return if it is true or false for prime
a++; // add 1 to a
}
}
static void Main(string[] args)
{
Program prog = new Program(); // instantiate a new Program
prog.Prime_2(); // run the method, Prime_2
Console.ReadLine(); // wait for input
}
答案 0 :(得分:8)
如果你想通过试验分区检查2到100之间的每个数字的原始性,这是你似乎想要做的,使用这个伪代码算法:
function isPrime(n)
if n % 2 == 0
return n == 2
d := 3
while d * d <= n
if n % d == 0
return False
d := d + 2
return True
这需要时间 O(n 1.5 )来查找 n 之前的素数。如果您想要更快的算法,请使用Eratosthenes的Sieve,它是 O(n log log n):
function primes(n)
sieve := makeArray(2..n, True)
for p from 2 to n step 1
if sieve[p]
output p
for i from p*p to n step p
sieve[i] := False
如果您对使用素数进行编程感兴趣,我会在我的博客上谦虚地推荐这个essay。
答案 1 :(得分:0)
我不确定你为什么要在你的功能中做你正在做的事情。你能在代码中添加一些注释吗?但是,检查素数的快速方法是以下
bool IsPrime(int number) {
if (number % 2 == 0 && number != 2) return false; // Don't check even numbers
for (int i = 2; i < number; i++) {
if (number % i == 0 && i != number) return false;
}
return true;
}
当然,您应该在if语句中调用上面的函数,如果为true,则显示数字。