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本文作者：ljh2000
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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
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...
1 //It is made by ljh2000
2 #include <iostream>
3 #include <cstdlib>
4 #include <cstring>
5 #include <cstdio>
6 #include <cmath>
7 #include <algorithm>
8 #include <ctime>
9 #include <vector>
10 #include <queue>
11 #include <map>
12 #include <set>
13 #include <string>
14 #include <stack>
15 using namespace std;
16 typedef long long LL;
17 const int MAXN = 50011;
18 const int MAXM = 200011;
19 int n,N,a[MAXN],maxl[MAXN];
20 int cnt,prime[MAXM],ans;
21 bool vis[MAXM];
22
23 inline int getint(){
24 int w=0,q=0; char c=getchar(); while((c<‘0‘||c>‘9‘) && c!=‘-‘) c=getchar();
25 if(c==‘-‘) q=1,c=getchar(); while (c>=‘0‘&&c<=‘9‘) w=w*10+c-‘0‘,c=getchar(); return q?-w:w;
26 }
27
28 inline void work(){
29 n=getint(); for(int i=1;i<=n;i++) a[i]=getint(),N=max(a[i],N); ans=1; int x;
30 for(int i=2;i<=N;i++) { if(!vis[i]) prime[++cnt]=i; for(int j=1;j<=cnt && i*prime[j]<=N;j++) { vis[i*prime[j]]=1; if(i%prime[j]==0) break;} }
31 for(int i=1;i<=n;i++) {
32 x=a[i];
33 for(int j=1;j<=cnt;j++) {
34 if(x%prime[j]!=0) continue;
35 maxl[i]=prime[j]; while(x%prime[j]==0) x/=prime[j];
36 if(x==1) break;
37 }
38 }
39 for(int i=2;i<=n;i++) if(maxl[i]>maxl[ans]) ans=i;
40 printf("%d",a[ans]);
41 }
42
43 int main()
44 {
45 work();
46 return 0;
47 }