题干：
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.
解题报告：
正解的代码显然是打表预处理nlogn，单组样例o(n)。。。。
第一个的做法是nloglogn打表，然后单组样例最坏复杂度n*20000（也就是每一次都是一个素数。、。。）
AC代码1：（860ms）
#include<cstdio>
#include<cstring>
#include<iostream>
#define MAX 20000 + 18//求MAX范围内的素数
using namespace std; long long su[MAX],cnt;
bool isprime[MAX];
void prime()
{ cnt=1; memset(isprime,1,sizeof(isprime)); isprime[0]=isprime[1]=0;for(long long i=2;i<=MAX;i++) { if(isprime[i]) {su[cnt++]=i; } for(long long j=1;j<cnt&&su[j]*i<MAX;j++) { isprime[su[j]*i]=0;} }
}
int main()
{prime();int n,x,maxx;int i,ans = 0; while(~scanf("%d",&n) ) {ans = 0;maxx = -1;while(n--) {scanf("%d",&x);for(int i = x; i>=2; i--) {//说明是最大素因子。 if(isprime[i] && x%i==0) {if(i>maxx) {ans = x;maxx = i;}}}}ans>1?printf("%d\n",ans):printf("1\n");}return 0 ;} AC代码2：（0ms）
#include <cstdio>
#include <cstring>
const int MAXN = 20017;
int s[MAXN];
int main()
{int n,m;memset(s,0,sizeof(s));s[1]=1;//此题1也是素数for(int i = 2; i < MAXN; i++)//筛选所有范围内的素数{if(s[i] == 0)//没有被更新过for(int j = i; j < MAXN; j+=i){s[j]=i;}}while(~scanf("%d",&n)){int ans;int maxx = -1;for(int i = 0; i < n; i++){scanf("%d",&m);if(s[m] > maxx){maxx = s[m];ans = m;}}printf("%d\n",ans);}return 0;
}
