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爲我的學校項目我必須創建一個程序,根據用戶(老師)想要的完美數字輸出完美的數字。用戶可以從1-4中選擇任何數字,但應顯示用戶選擇的許多號碼。這是我目前的代碼。請忽略sumupTo,factorial,isprime和testGoldbach方法,請僅查看完美數字方法/代碼。爲什麼我的程序不能正確運行?
import java.util.Scanner;
public class MyMathB
{
public static int sumUpTo(int n)
{
int sum = 0;
for (int k = 1; k <= n; k++)
sum += k;
return sum;
}
public static long factorial(int n)
{
long f = 1;
for (int k = 2; k <= n; k++)
f *= k;
return f;
}
public static boolean isPrime(int n)
{
if (n <= 1)
return false;
int m = 2;
while (m * m <= n)
{
if (n % m == 0)
return false;
m++;
}
return true;
}
public static void PerfectNumbers(int number)
{
System.out.println("How many perfect numbers would you like to see? Please enter an integer from 1 to 4");
Scanner s = new Scanner(System.in);
int numbersToSee = s.nextInt();
int counts = 0;
for(counts = 0; counts <= numbersToSee; counts++)
{
for (int n = 5; n <= 10000; n++)
{
int temp = 0;
for(int i = 1; i <= number/2; i++)
{
if (number % i == 0)
{
temp += i;
}
if (temp == number)
{
System.out.println(number);
}
}
}
}
}
public static boolean testGoldbach(int bigNum)
{
for (int n = 6; n <= bigNum; n += 2)
{
boolean found2primes = false;
for (int p = 3; p <= n/2; p += 2)
{
if (isPrime(p) && isPrime(n - p))
found2primes = true;
}
if (!found2primes)
{
System.out.println(n + " is not a sum of two primes!");
return false;
}
}
return true;
}
public static void main(String[] args)
{
Scanner kb = new Scanner(System.in);
int n;
do
{
System.out.print("Enter an integer from 4 to 20: ");
n = kb.nextInt();
} while (n < 4 || n > 20);
kb.close();
System.out.println();
System.out.println("1 + ... + " + n + " = " + sumUpTo(n));
System.out.println(n + "! = " + factorial(n));
System.out.println("Primes: ");
for (int k = 1; k <= n; k++)
if (isPrime(k))
System.out.print(k + " ");
System.out.println();
System.out.println("Goldbach conjecture up to " + n + ": " + testGoldbach(n));
}
}
對不起,它的「爲什麼我的程序不能正常運行?」 –
編輯標題,請勿將其添加爲評論。並告訴我們會發生什麼和預期的事情。 – AntonH
*「請忽略sumupTo,factorial,isprime和testGoldbach方法」*如果那些確實與再現問題行爲無關,那麼您應該從示例中刪除它們。請參閱http://stackoverflow.com/help/mcve。 – Radiodef