HDU-2710: Max Factor（埃式筛法）

数论

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

题意：输入n，再输入n个数，找出n个数当中拥有最大素因子的数，如果同时拥有最大的素因子，输出先输入的那个数。

分析：利用埃式筛法，预处理1-20000内的素数，然后判断…xjb写一下就可以(´・ω・`)

• 完整代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#define maxn 20005
bool isprime[maxn];  
void primeform()
{
    memset(isprime,true,sizeof(isprime));
    int m=(int)sqrt(maxn+0.5)+1;
    for(int i=2;i<=maxn/2;i++)
    {
        if(isprime[i])
            for(int j=i*2;j<=maxn;j+=i)
                isprime[j]=false;
    }
}
int main()
{
    int n,x,i,m,y;
    primeform();
    while(scanf("%d",&n)!=EOF)
    {
        m=-1;
        while(n--)
        {
            scanf("%d",&x);
            for(i=1;i<=x;i++)
            {
                if(x%i==0&&isprime[x/i])
                {
                    if(x/i>m)
                    {
                        m=x/i;
                        y=x;
                    }
                    break;
                }
            }
        }
        printf("%d\n",y);
    }
    return 0;
}