HDU - 2710 Max Factor-优快云博客

本文链接：https://blog.youkuaiyun.com/IT_w_TI/article/details/87091533

本文介绍了一种使用埃氏筛法解决特定数学问题的方法，即在给定范围内找到具有最大素因数的整数。通过预先计算素数，然后逐个检查每个输入数的素因数，从而高效地解决问题。

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题面

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

题目大意

给定N个数，求出其具有最大素因数的数，注意，是输出具有最大素因数的数，而不是最大素因数，若具有多个，则输出最先出现的那一个。

题目分析

这算是埃氏筛的模板题吧……

注意到范围是 $[1, 20000]$ ，这个范围我们可以先用埃氏筛（ $nlog⁡log⁡nn\log \log n$ ），先求出20000以内的素数，然后每输入一个数就枚举所有比它小的素数，找出最大素因子。然后与上一个最大素因子比较，若比原来的大，就保存下来。不比原来的大就丢弃就行了。

代码

#include <cstdio>
#include <cstring>
using namespace std;
bool fact[20001];
int prime[20001];
int a[5001];
int main(int argc, char const *argv[]) {
  memset(fact, 1,  sizeof(fact));
  for(int i = 2; i < 20001; i++){
    if(fact[i]){
      for(int j = 2 * i; j < 20001; j += i)
        fact[j] = 0;
    }
  }
  int cnt = 0;
  for(int i = 2; i < 20001; i++){
    if(fact[i])
      prime[cnt++] = i;
  }
  int n;
  while(~scanf("%d", &n)){
    int r = -1, yz = -1;
    for(int i = 0; i < n; i++){
      int t = 0, m;
      scanf("%d", &m);
      for(int j = 0; j < cnt && prime[j] <= m; j++){
        if(!(m % prime[j]))
          t = prime[j];
      }
      if(t > yz){
        r = m;
        yz = t;
      }
    }
    printf("%d\n", r);
  }

  return 0;
}