[HDU - 2710] Max Factor

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本文介绍了一种算法，用于从一组整数中找出具有最大质数因子的数。通过预处理素数并使用这些素数来确定每个输入数的最大质数因子，从而有效地解决了问题。

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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

解析

给n个数给你，让你找到一个数它的因子是某个质数（所有数的质数因子最大的一个），然后输出这个数。

38的质数因子19是4（3、19、5、7）个数里面最大的所以输出38.

注意：如果有两个质数因子相同，输出先出现的。在这个题目条件下，1也视为质数。

对于每个数，我们都需要去求质数因子，过程是重复的。

所以我们先把数据范围内的素数先求出来，然后用数余素数（素数大到小），得到的就是这个数的最大质数因子。

（避免了求素数过程的重复）

代码

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <cmath>
using namespace std;
const int ran=20005;
int prime[ran],mark[ran],cnt=2;
int main()
{
    int i,j;
    prime[0]=1;
    prime[1]=2;
    for(i=3;i<=ran;i++)//预处理素数
    {
        if(mark[i]) continue;
        int stat=1,k=2;
        for(j=2;j<sqrt(i)+1;j++)
            if(i%j==0)
            {
                stat=0;
                break;
            }
        if(stat)    prime[cnt++]=i;
        while(i*k<=ran)
            mark[i*k]=1,k++;
    }
    int T;
    while(~scanf("%d",&T))
    {
        int maxn=0,ans,n;
        while(T--)
        {
            scanf("%d",&n);
            for(i=cnt-1;i>=0;i--)//找最大质数因子
                if(n%prime[i]==0)
                {
                    if(prime[i]>maxn)//记录数和它的最大质数因子
                    {
                        maxn=prime[i];
                        ans=n;
                    }
                    break;
                }
        }
        printf("%d\n",ans);
    }

}