POJ 3048 Max Factor (筛素数）

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


题意：给你n个数，让你求出其中质因数最大的数
思路：这是筛素数的思想，在筛素数的时候，记录每个数对应的最大质因数，这样在访问该数的时候可以以o（1）的复杂度找到它所对应的最大质因数。再加上贪心的想法，在每次访问的时候找到当前最大质因数，并且记录该数。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const int N=20010;
bool a[N];
int ans[N];

void isprime() //作用：找到每个数所对应的最大素数
{
　　int k=0;
　　memset(a,true,sizeof(a));
　　a[0]=a[1]=false;
　　for(int i=2;i<N;i++)
　　{
　　　　if(a[i])
　　　　{
　　　　　　ans[i]=i;
　　　　　　for(int j=i*2;j<N;j+=i)
　　　　　　ans[j]=i；

　　　　　　a[j]=false;
　　　　}
　　}
}

int main()
{
　　isprime();
　　int m;
　　while(~scanf("%d",&m))
　　{
　　　　int maxn=-N;
　　　　int x=-N;
　　　　while(m--)
　　　　{
　　　　　　int a;
　　　　　　scanf("%d",&a);
　　　　　　if(ans[a]>maxn)
　　　　　　{
　　　　　　　　maxn=ans[a];
　　　　　　　　x=a;
　　　　　　}
　　　　}
　　　　printf("%d
",x);
　　}
　　return 0;
}