HDU - 2710 Max Factor（筛选质因子）

最新推荐文章于 2024-08-29 11:58:30 发布

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最新推荐文章于 2024-08-29 11:58:30 发布

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本文介绍了一个算法问题，即在一组给定的整数中找出拥有最大质因子的数。通过预处理质因子的方式，该文提供了一种高效的解决方案，利用筛法提前计算每个数的最小质因子，从而在O(N)的时间复杂度内解决此问题。

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

题意：找到数内最大的质因子，输出这个数

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define n 20017
int pri[n];
void prime()
{
    pri[1] = 1;
    for(int i = 2;i <= n;i ++)
    {
        if(!pri[i])
        {
            for(int j = i;j <= n;j = j + i)
            pri[j] = i;
        }
    }
}
int main()
{
    prime();
    int t,x,max,maxn;
    while(~scanf("%d",&t))
    {
        max = 0;
        maxn = -1;
        for(int i = 0;i < t;i ++)
        {
            scanf("%d",&x);
            if(pri[x] > max)
            {
                max =pri[x];
                maxn = x;
            }
        }
        printf("%d\n",maxn);
    }
    return 0;
}