POJ - 3048 （最大素因数）

最新推荐文章于 2019-10-22 17:34:55 发布

原创最新推荐文章于 2019-10-22 17:34:55 发布 · 577 阅读

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#水题 #数论

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本文介绍了一种通过编程解决问题的方法，即找出一组整数中哪个数拥有最大的素因数，并给出了具体的实现代码。

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

题目大概：

求给出数据中，那个数有最大素因数，输出该数。

思路：

这个题本来很简单，但是它有个地方不和常理，1竟然是素数，我再怎么做也不会想到1竟然是素数，这个题，正常情况下，我确实做不出来。

代码：

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>

using namespace std;
int su[20020];
int a[5005];
int b[25005];
int sushu()
{
    for(int i=1;i<=20020;i++)
    {
        su[i]=1;
    }
    su[1]=0;
    for(int i=2;i<=20000;i++)
    {
        if(su[i]==1)
        {   b[i]=i;
            for(int j=i+i;j<=20000;j=j+i)
            {   b[j]=i;
                su[j]=0;
            }
        }
    }

}

int main()
{

     int n;
      scanf("%d",&n);
     sushu();

     su[1]=1;b[1]=1;

     memset(a,0,sizeof(a));

     for(int i=1;i<=n;i++)
     {
         scanf("%d",&a[i]);
     }

     int sum=0,l=0;
     for(int i=1;i<=n;i++)
     {
         int k=b[a[i]];
         if(k>sum){sum=k;l=a[i];}

     }

     printf("%d",l);
    return 0;
}