Max Factor

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6322 Accepted Submission(s): 2077

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

Sample Output

38

題目大意：給出多個數，問哪個數的最大素因子最大；

解題思路：先打表，篩出個素數，再用篩素數的思路求各個數的最大素因子；

代碼如下：

#include"iostream"
#include"cstdio"
#include"cstring"
using namespace std;
const int maxn = 20005;

int factor[maxn];
int isprime[maxn];

void Eratosthenes(){
	memset(isprime, true, sizeof isprime);
	isprime[1] = true;
	for(int i = 2;i*i <= maxn;i ++){
		if(!isprime[i]) continue;
		for(int j = i<<1;j < maxn;j += i)
			isprime[j] = false;
	}
}

void MaxFactor(){ //產生各個數的最大素因子 
	Eratosthenes();
	for(int i = 1;i < maxn;i ++)
		if(isprime[i]){
			for(int j = i;j < maxn;j += i)
				factor[j] = i;
		}
	/*for(int i = 0;i < 100;i ++)
		printf("%d %d\n", i, factor[i]);*/
}

int main(){
	int n, t, max_factor, max_num;
	MaxFactor(); //打表 
	while(scanf("%d", &n) != EOF){
		max_factor = 0; 
		while(n --){
			scanf("%d", &t);
			if(max_factor  < factor[t]){
				max_factor = factor[t];
				max_num = t;
			} 
		}
		printf("%d\n", max_num);
	}
	return 0;
}