HDU 2709 Max Factor (素数因子)

Max Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8635    Accepted Submission(s): 2817

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

436384042

 

Sample Output

38

 

Source

USACO 2005 October Bronze

 

123

题目大意:   给你n个数,  问你这个n个数里的有  最大的质因子事那个; 

既是因子, 又得是质因数

先素数达标,然后判断,  找最大的质因子;

#include <iostream>#include <stdio.h>#include <queue>#include <cmath>#include <cstring>#include <string>#include <map>#include <algorithm>int T;typedef long long ll;const int MAXN=20010;using namespace std;int k;int pre[MAXN];int num[MAXN];int s[MAXN];int ans[MAXN];void tab(){    k=1;    pre[1]=1;    for(int i=2;i<=MAXN;i++)    {        if(!pre[i])        {            num[k++]=i;            for(int j=i+i;j<=MAXN;j+=i)                pre[j]=1;        }    }}int main(){    int n;    tab();    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)            scanf("%d",s+i);        for(int i=1;i<=n;i++)        {            if(!pre[s[i]])// 自己是素数            {                ans[i]=s[i];                continue;            }            ans[i]=1;            for(int j=1;num[j]<=s[i];j++)            {                 if(s[i]%num[j]==0)   //是因子,并且另一边是素数;                 {                     if(!pre[s[i]/num[j]])// 对应的 为素数                     {                          if(num[j]>ans[i])                                ans[i]=s[i]/num[j];                     }                 }            }        }        int mmax=ans[1];        int low=1;        for(int i=1;i<=n;i++)        {            if(ans[i]>mmax)            {                mmax=ans[i];                low=i;            }        }        printf("%d\n",s[low]);    }    return 0;}