HDU 2710 Max Factor （水题）

Max Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2238    Accepted Submission(s): 684

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

 

 

Recommend

teddy

 

 

一大清早起来，刷了道水题。。。找了下感觉，试了下素数分解的模板，求下午网络赛给力啊。

 

这个素数分解的模板用了很久了，真的是屡试不爽啊。。。哈哈

 

此题注意对1的处理。把1的最大素因子当成1就可以AC了。

#include<stdio.h>

#include<iostream>

#include<algorithm>

#include<string.h>

using namespace std;



const int MAXN=20000;

int prime[MAXN+1];

int getPrime()//得到小于等于MAXN的素数，prime[0]存放的是个数

{

    memset(prime,0,sizeof(prime));

    for(int i=2;i<=MAXN;i++)

    {

        if(!prime[i]) prime[++prime[0]]=i;

        for(int j=1;j<=prime[0]&&prime[j]<=MAXN/i;j++)

        {

            prime[prime[j]*i]=1;

            if(i%prime[j]==0) break;

        }

    }

    return prime[0];

}

long long factor[100][2];

int facCnt;

int getFactors(long long x)//把x进行素数分解

{

    facCnt=0;

    long long tmp=x;

    for(int i=1;prime[i]<=tmp/prime[i];i++)

    {

        factor[facCnt][1]=0;

        if(tmp%prime[i]==0)

        {

            factor[facCnt][0]=prime[i];

            while(tmp%prime[i]==0)

            {

                   factor[facCnt][1]++;

                   tmp/=prime[i];

            }

            facCnt++;

        }

    }

    if(tmp!=1)

    {

        factor[facCnt][0]=tmp;

        factor[facCnt++][1]=1;

    }

    return facCnt;

}









int main()

{

    int n;

    getPrime();

    int num;

    while(scanf("%d",&n)!=EOF)

    {

        int ans=0;

        int temp=0;

        for(int i=0;i<n;i++)

        {

            scanf("%d",&num);

            if(num==1)//1的时候要单独处理一下

            {

                if(temp<1)

                {

                    temp=1;

                    ans=1;

                }

                continue;

            }

            getFactors(num);

            if(temp<factor[facCnt-1][0])

            {

                temp=factor[facCnt-1][0];

                ans=num;

            }

        }

        printf("%d\n",ans);

    }

    return 0;

}