Max Factor（技巧题）

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最新推荐文章于 2020-05-12 22:05:24 发布

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本文介绍了一个算法问题，即从一组整数中找出拥有最大素数因子的数字，并提供了一种有效的解决方案。通过预处理的方式标记每个数的最大素因子，可以在较短时间内解决该问题。

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Max Factor

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6649		Accepted: 2609

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

Sample Output

Hint

OUTPUT DETAILS:
19 is a prime factor of 38. No other input number has a larger prime factor.

Source

USACO 2005 October Bronze

题意：

从一串数字中找到有最大素数因子的数输出。

思路：

在判断是否为素数时，改变，用点小技巧，另不是素数的等于他的一个最大的素数因子。

代码：

#include <iostream>
#include <cstring>
#include <string>
#include <stdio.h>
using namespace std;
int a[20055];
int maxx,num;
int main()
{
    int n,m;
    memset(a,0,sizeof(a));
    a[1]=1;
    for(int i=2; i<20055;i++)
    {
        if(a[i]==0)
            for(int j=i;j<20055;j+=i)
            {
                a[j]=i;  //关键点
            }
    }
    while(~scanf("%d",&n))
    {
        maxx=-1;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&m);
            if(a[m]>maxx)
            {
                maxx=a[m];
                num=m;
            }
        }
        printf("%d\n",num);
    }
    return 0;
}