POJ-3048 求有最大的素数因子的数

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

Hint
OUTPUT DETAILS: 

19 is a prime factor of 38. No other input number has a larger prime factor.

    搞清楚题意，求有最大素数因子的数。要考虑的问题：它本身如果是一个素数，算不算呢？忘了算不算了，试一试就知道了。做之前先预处理筛素数！！

代码如下：

#include<iostream>
#include<cmath>
using namespace std;
const int MAX=20005;
int n;
int a[50010];
int prime[20010];
void Isprime()
{
prime[0]=0;
prime[1]=0;
prime[2]=1;
for(int i=3;i<MAX;i++)
{
if(i%2==0)
prime[i]=0;
else
prime[i]=1;
}
int temp=(int)sqrt(MAX*1.0);
for(int i=3;i<=temp;i++)
{
if(prime[i])
{
for(int j=i*i;j<MAX;j+=2*i)
{
prime[j]=0;
}
}
}
}
int judge(int x)
{
for(int i=x;i>0;i--)
{
if(prime[i] && x%i==0)
{
return i;
}
}
return 1;
}
int main()
{
int x;
Isprime();
while(cin>>n && n)
{
int ans=0,maxn=0,m=0;
for(int i=1;i<=n;i++)
{
cin>>x;
ans=judge(x);
if(maxn<ans)
{
maxn=ans;
m=x;
}
}
cout<<m<<endl;
}
return 0;
}