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 =  ;
 int prime[MAX],temp[MAX];
 void init()
  {
      memset(temp,,sizeof(temp));
      for(int i = ; i < MAX; i++)  /*先建立一个素数表*/
          if(temp[i])
          {
              temp[i] = ; //素数
              for(int j = i*; j < MAX; j+=i)
                  temp[j] = ; //非素数
         }

     prime[] = ; //素数
     int count = ;
     for(int i = ; 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 = ;
         memset(temp,,sizeof(temp));
         for(int i = ; i<= t; i++)
             scanf("%d",&temp[i]);

         for(int i = ; i <= t; i++)        /*对每个输入的数进行处理*/
             for(int j = ; prime[j] <= temp[i]; j++)  /*二营长，把老子的意大利炮拉出来*/
                 if(temp[i] % prime[j] == )         /*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 ;
 }