【POJ - 3048】Max Factor （数论，打表，水题）

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;
}