Question

我的程序不适用于日食..而且我不知道它有什么问题.. Plz帮助.. 并且PLZ也说明原因。

 public class ProblemThree {

        public static void main(String args[])
        {
            long a=0L, z=0L;
            long n=600851475143L;

            for(long i=2;i<=n ;++i)
            {
                if(600851475143L % i==0)
                {
                    a=i;

                    if(a%2==0)
                        {;  }
                    else if(a%3==0)
                        { ;}
                    else if(a%5==0)
                        { ;}
                    else if(a%7==0)
                        { ;}
                    else if (a>z)
                    { 
                        z=a;
                    }

                }

            }

            System.out.println(z);

        }
    }

Answer 1

谢谢你们的反馈，但我已经用以下代码解决了这个问题.. :)

公共类ProblemThree {

public static void main(String args[])
{

    long n=600851475143L;

    for(long i=2;i<n ;++i)
    {
        while(n % i==0)
        {//for yes
            n=n/i;

        }   

    }
    System.out.println(n);
}

}

Answer 2

你的节目不是＆＃34;返回＆＃34;什么，因为它仍然在运行。你的循环仍然是迭代的。您的代码需要修改才能更高效。这是我对同一问题的解决方案。

long testNum = 600851475143l; 
int largestFactor = 0; 
long loopMax = 17425170l; //largest known prime  
for (int i = 3; i * i <= loopMax; i++) { 
    boolean isPrime = true;             
    for (int j = 2; j < i; j++) { 
       if (i % j == 0) { 
          isPrime = false; 
          break; 
       } 
    }                         
    if (isPrime && testNum % i == 0) { 
        System.out.println("prime factor: " + i); 
        largestFactor = i; 
        loopMax = (testNum / i) + 1; 
    } 
} 
System.out.println("result is: " + largestFactor);

Answer 3

package Euler;

import java.util.Scanner;

public class euler3 {
    public static void main(String args[])
    {
        long num;
        Scanner sc = new Scanner(System.in);
        num= sc.nextLong();

        for(int i=2;i<num; i++)
        {
            while(num%i == 0)
            {
                //System.out.println(i);
                num=num/i;
            }
        }
        if(num>2)
            System.out.println(num);

    }

}

13195的主要因子是5、7、13和29。
600851475143的最大素数是6857。

Answer 4

#include <bits/stdc++.h>
using namespace std;
int SieveOfEratosthenes(int x,long long signed int n)
{
    bool prime[x];                                 //use sieve upto square root of 
                                                       // required number
    memset(prime, true, sizeof(prime));
 
    for (long long signed int p = 2; p * p <= x; p++)    
    {
        if (prime[p] == true) 
        {
            for (long long signed int i = p * p; i <= x; i += p)
                prime[i] = false;
        }
    }
    int res;
    for (long long signed int p = 2; p <= x; p++)
        if (prime[p])
            {
                if(n%p==0)   //if number is divisble by prime number than update with 
                res=p;        //latest value of prime number
            }
            return res;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout<<SieveOfEratosthenes(775146,600851475143);//Return the answer

    // long long signed int n;cin>>n;   for value of n
   //cout<<SieveOfEratosthenes(sqrt(n),n);
    return 0;
}

13195的主要因素是5,7,13和29. 600851475143中最大的素数因子是什么？

4 个答案: