Max Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3966    Accepted Submission(s): 1289

Problem Description

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

Source

USACO 2005 October Bronze


题目大意：给你N个数，求这N个数中哪个数的最大素因子最大，

输出这个数，如果有多种结果，输出靠前边的那个数。


思路：将筛法求素数改变一下。若i为素数，则i的1、2、3…倍的

最大素因子都为i，筛的时候，赋值为Prime[j] = i，即j的最大素因

子为i。

注意：初始化时令所有数为0，Prime[0] = Prime[1] = 1。

即Prime[i]为0是素数，Prime[i]为1为素数。改变之后Prime[i]为

i的最大素因子。



#include<stdio.h>
#include<string.h>
#include<math.h>

int Prime[20005];

void IsPrime()
{
    Prime[1] = 1;
    for(int i = 2; i <= 20000; i++)
    {
        if(Prime[i]==0)
        {
            Prime[i]=i;
            for(int j = i+i; j <= 20000; j+=i)
                Prime[j] = i;
        }
    }
}

int main()
{
    int N,x;
    IsPrime();
    while(~scanf("%d",&N))
    {
        int Max = 0;
        while(N--)
        {
            scanf("%d",&x);
            if(Prime[x] > Prime[Max])
                Max = x;
        }
        printf("%d\n",Max);
    }

    return 0;
}