Max Factor（素数筛法）题解

Max Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 10245    Accepted Submission(s): 3304

Problem 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

436384042

 

Sample Output

38

思路：

忘记了memset（）只能初始化0、-1，手贱赋了个1，然后疯狂WA

这里用到了素数筛选法：单击666查看大神详解

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
#define N 20010
using namespace std;
int vis[N];

void prime(){	//素数筛选 
	memset(vis,0,sizeof(vis));
	vis[0]=1,vis[1]=0;	//这里1都算素数 
	for(int i=2;i<=20000;i++){
		if(!vis[i]){
			for(int j=i*i;j<=20000;j+=i){
				vis[j]=1;
			}
		}
	}
}

int main(){
	int n,ans,maxp,i,a;
	prime();
	while(~scanf("%d",&n)){
			maxp=0;
			ans=0;
			while(n--){
				scanf("%d",&a);
				for(i=a;i>0;i--){
					if(!vis[i]){
						if((i>maxp) && a%i==0){
							maxp=i;
							ans=a;
							break;
						}		
					}
				}
			}
			printf("%d
",ans);
	}
	return 0;
}