HDOJ/HDU 2710 Max Factor(素数快速筛选~)

Problem Description
To improve the organization of his farm, Farmer John labels each of his N (1 <= N <= 5,000) cows with a distinct serial number in the range 1..20,000. Unfortunately, he is unaware that the cows interpret some serial numbers as better than others. In particular, a cow whose serial number has the highest prime factor enjoys the highest social standing among all the other cows.

(Recall that a prime number is just a number that has no divisors except for 1 and itself. The number 7 is prime while the number 6, being divisible by 2 and 3, is not).

Given a set of N (1 <= N <= 5,000) serial numbers in the range 1..20,000, determine the one that has the largest prime factor.

Input
* Line 1: A single integer, N

• Lines 2..N+1: The serial numbers to be tested, one per line

Output
* Line 1: The integer with the largest prime factor. If there are more than one, output the one that appears earliest in the input file.

Sample Input
4
36
38
40
42

Sample Output
38

哎~又一个英文题~

题意：
输入一个正整数n，然后输入n个正整数（1<=a[i]<=20000），要你求这n个数里哪个数的最大素因数（即能被该数整除(包括这个数本身！)的最大素数）最大，然后输出这个数。若有两个数的最大素因数相同，则输出前面那个。

用到了素数快速筛选~不然会超时的~

import java.util.Arrays;
import java.util.Scanner;

/**
 * @author 陈浩翔
 */
public class Main{
    static boolean db[] = new boolean[20005];
    public static void main(String[] args) {
        dabiao();
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            int n = sc.nextInt();
            int a[] = new int[n];
            int prime[] = new int[n];
            for(int i=0;ifor(int k=1;k<=a[i];k++){
                    if(db[k]&&a[i]%k==0){
                        prime[i]=k;
                    }
                }
            }
            int max=prime[0];
            int con=0;
            for(int i=1;iif(prime[i]>max){
                    max=prime[i];
                    con=i;
                }
            }
            System.out.println(a[con]);
        }
    }
    private static void dabiao() {
        Arrays.fill(db, true);
        for(int i=2;i<=Math.sqrt( db.length);i++){
            for(int j=i+i;jif(db[j]==true){
                    db[j]=false;
                }
            }
        }
    }
}