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 = 50011;
const int MAXM = 200011; 
int n,N,a[MAXN],maxl[MAXN];
int cnt,prime[MAXM],ans;
bool vis[MAXM];

inline int getint(){
    int w=0,q=0; char c=getchar(); while((c<'0'||c>'9') && c!='-') c=getchar();
    if(c=='-') q=1,c=getchar(); while (c>='0'&&c<='9') w=w*10+c-'0',c=getchar(); return q?-w:w;
}

inline void work(){
    n=getint(); for(int i=1;i<=n;i++) a[i]=getint(),N=max(a[i],N); ans=1; int x;
    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;} }
    for(int i=1;i<=n;i++) {
    x=a[i];
    for(int j=1;j<=cnt;j++) {
        if(x%prime[j]!=0) continue;
        maxl[i]=prime[j]; while(x%prime[j]==0) x/=prime[j];
        if(x==1) break;
    }
    }
    for(int i=2;i<=n;i++) if(maxl[i]>maxl[ans]) ans=i;
    printf("%d",a[ans]);
} 

int main()
{
    work();
    return 0;
}