【素数筛+合数分解】——hdu2710

                                        Max Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6386    Accepted Submission(s): 2094

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

 

题意：找到n个数里具有最大素因子的数

 

（套用邝斌模板）

 

附上代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

const int MAXN=20000;
int prime[MAXN+1];

long long factor[100][2];//[][0]位记录是因子的素数；[][1]记录可以被除几次
int facCnt;//不同素因子个数


//得到小于等于MAXN的素数，prime[0]存放的是素数的个数
void getPrime()
{
    memset(prime,0,sizeof(prime));

    for(int i=2;i<=MAXN;i++)
    {
        if(prime[i]==0)
        {
            prime[++prime[0]]=i;//往下一个空位放素数
        }

        for(int j=1 ; j<=prime[0]/*找已经找到的素数*/ && prime[j]<=MAXN/i/*这个素数还要可以倍增*/ ; j++)
        {
            prime[prime[j]*i]=1;//标记找到的非素数
            if(i%prime[j]==0)//如果i不是素数
            {
                break;
            }
        }
    }
}

//把x进行素数分解
int getFactors(long long x)
{
    facCnt=0;//初始化
    long long tmp=x;

    for(int i=1 ; prime[i]<=tmp/prime[i]/*prime[i]^2<=tmp*/ ; i++)
    {
        factor[facCnt][1]=0;

        //找到是x的因子的素数
        if(tmp%prime[i]==0)
        {
            factor[facCnt][0]=prime[i];

            while(tmp%prime[i]==0)//一直往下除到底
            {
                factor[facCnt][1]++;
                tmp/=prime[i];
            }
            facCnt++;
        }
    }

    if(tmp!=1)
    {
        factor[facCnt][0]=tmp;//最后变成一个素数
        factor[facCnt++][1]=1;
    }
    //factor[facCnt-1][0]里面存的是最大素因子
    return facCnt;//返回给全局变量
}


int main()
{
    getPrime();//打表‘

    int n;
    int num;

    while(scanf("%d",&n)!=EOF)
    {
        int ans=0;//结果
        int temp=0;//记录目前最大的因子

        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;
}