POJ3048 Max Factor

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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

USACO 2005 October Bronze

 

 

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 ;
 }