The code works just fine for the int data type, but 600851475143 seems to be too big of a number. How can I make it work? It just keeps running and never gives me an answer.
public class Main {
public static void main(String[] args) {
long a = 600851475143L;
boolean prime = false;
long big = 0L;
for (long i = 1L; i < a; i++){
if (a % i == 0){
for (int j = 2; j < i/(float)2; j++){
if (i % j == 0){
prime = true;
break;
}
}
if(!prime){
big = i;
}
}
}
System.out.println(big);
}
}
You need to use a long
also for j
.
Your variable naming is also a bit misleading: prime
is true when the number is not a prime...
Your code has a lot of problems. My first advice would be to write clean and concise code with well-named variables. Secondly, analyze the runtime complexity of your program even if it works fast for large inputs. The fact that you run an inner loop inside if(a % i == 0)
condition, makes your program extremely inefficient.
Here I provide a refactored version of your code with runtime complexity and good variable names in mind:
public static void main(String[] args) {
System.out.println(largestPrimeFactorOf(600851475143L));
}
public static long largestPrimeFactorOf(long input)
{
List<Long> factors = new ArrayList<>();
// You start from 2, as 1 is not a prime number and also using 1 will cause an infinite loop
for (long i = 2L; i < input; i++) {
if (input % i == 0) {
factors.add(i);
while (input % i == 0) {
input /= i;
}
}
}
// As we always add a bigger number to the factor list, the last element is the greatest factor.
return factors.get(factors.size() - 1);
}
Denote that this program will still be slow when the input is a large prime number, eg 100000000019. To handle such cases efficiently, it's better to use an efficient primality test algorithm. You can search the web for that. https://en.wikipedia.org/wiki/Primality_test
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