POJ3048 Max Factor

时间：2022-04-24 21:17:15

本文版权归ljh2000和博客园共有，欢迎转载，但须保留此声明，并给出原文链接，谢谢合作。

本文作者：ljh2000
作者博客：http://www.cnblogs.com/ljh2000-jump/
转载请注明出处，侵权必究，保留最终解释权！

 

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

Hint

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

Source

 
 
正解：线性筛+暴力
解题报告：
　　直接筛出质数，然后暴力就可以了。

　　我连线性筛都写挂了一次，真是没救了...

 //It is made by ljh2000

 #include <iostream>

 #include <cstdlib>

 #include <cstring>

 #include <cstdio>

 #include <cmath>

 #include <algorithm>

 #include <ctime>

 #include <vector>

 #include <queue>

 #include <map>

 #include <set>

 #include <string>

 #include <stack>

 using namespace std;

 typedef long long LL;

 const int MAXN = ;

 const int MAXM = ;

 int n,N,a[MAXN],maxl[MAXN];

 int cnt,prime[MAXM],ans;

 bool vis[MAXM];

 inline int getint(){

     int w=,q=; char c=getchar(); while((c<''||c>'') && c!='-') c=getchar();

     if(c=='-') q=,c=getchar(); while (c>=''&&c<='') w=w*+c-'',c=getchar(); return q?-w:w;

 }

 inline void work(){

     n=getint(); for(int i=;i<=n;i++) a[i]=getint(),N=max(a[i],N); ans=; int x;

     for(int i=;i<=N;i++) { if(!vis[i]) prime[++cnt]=i; for(int j=;j<=cnt && i*prime[j]<=N;j++) { vis[i*prime[j]]=; if(i%prime[j]==) break;} }

     for(int i=;i<=n;i++) {

     x=a[i];

     for(int j=;j<=cnt;j++) {

         if(x%prime[j]!=) continue;

         maxl[i]=prime[j]; while(x%prime[j]==) x/=prime[j];

         if(x==) break;

     }

     }

     for(int i=;i<=n;i++) if(maxl[i]>maxl[ans]) ans=i;

     printf("%d",a[ans]);

 } 

 int main()

 {

     work();

     return ;

 }

相关文章