ACM Max Factor

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 lineOutput* 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

/*
    Name: WTZPT    
    Copyright: 
    Author: 
    Date: 14/08/17 11:13
    Description:   先输入数字的个数n，然后n次输入数字，输出其中拥有最大素数因子的数字 
*/
#include<bits/stdc++.h>
using namespace  std;
const int MAX =  50005;
int prime[MAX],temp[MAX];
void init()
 {
     memset(temp,1,sizeof(temp));
     for(int i = 2; i < MAX; i++)  /*先建立一个素数表*/
         if(temp[i])
         {
             temp[i] = 1; //素数 
             for(int j = i*2; j < MAX; j+=i)
                 temp[j] = 0; //非素数 
        }
    
    prime[1] = 1; //素数      
    int count = 2; 
    for(int i = 2; i < MAX; i++)   /*把素数存储到一个容器内*/ 
    {
        if(temp[i])
            prime[count++] = i;
    }
             
 }
int main()
{
    init();
    int t,ans,max;
    while(cin>>t)
    {
        max = 0;
        memset(temp,0,sizeof(temp));    
        for(int i = 1; i<= t; i++)
            scanf("%d",&temp[i]);
            
        for(int i = 1; i <= t; i++)        /*对每个输入的数进行处理*/ 
            for(int j = 1; prime[j] <= temp[i]; j++)  /*二营长，把老子的意大利炮拉出来*/ 
                if(temp[i] % prime[j] == 0)         /*应题目要求 因子为素数的情况*/ 
                    if(max < prime[j])                /*当目前的素数因子最大时*/ 
                    {
                        max = prime[j];
                        ans = i;                    /*存储原数据的位置*/ 
                    }
        cout<<temp[ans]<<endl;    
    }
    return 0;
}