SZU:A25 Favorite Number

Judge Info

• Memory Limit: 32768KB
• Case Time Limit: 10000MS
• Time Limit: 10000MS
• Judger: Number Only Judger

Description

Frog Frank likes 30 more than likes 40, yet he likes 12 and 39 equally. This is because he likes numbers that have a lot of different prime factors. For example, 30 have 3 prime factors (2,3 and 5) and 40 have 2(2 and 5) only. A prime number is a number that can be divided evenly only by itself and 1.

Task

You are given a list of numbers, find out which of the numbers Frank likes most. If there are more than one solutions, output the smallest.

Input

The first line of input contains , the number of test cases. First line of each test case contains a integers . The following line contains N positive integers, all of them are not greater than 100, 000.

Output

For each test case, print a line contains the solution.

Sample Input

2
10
3 5 7 9 11 13 15 17 19 21
11
2 4 6 8 10 13 39 105 200 201 143

Sample Output

15
105


解题思路：素数筛选

算法 from dd：

#include <stdio.h>
#include <string.h>
#include <math.h>

#define N   100000
#define yes '1'
#define no  '0'
char flag[N+1];

void is_prime(int n)
{
    int i, j;
    memset(flag, yes, sizeof(flag));
    flag[0]=flag[1]=no;
    int len=sqrt(n)+1;
    for(i=2; i<len; i++)
    {
        if(flag[i]==no) continue;
        for(j=i+i; j<=n; j+=i)
            flag[j]=no;
    }
}

int main(void)
{
    int n;
    ///find the primes from 1 to n(n<=N)
    scanf("%d", &n);
    is_prime(n);
    for(int i=1; i<=n; i++)
        if(flag[i]==yes)
            printf("%8d", i);
    printf("
");
    return 0;
}

解题：

#include <stdio.h>
#include <string.h>
#include <math.h>
 
#define N   100050
#define yes '1'
#define no  '0'
char flag[N];
 
void is_prime(int n)
{
    int i, j;
    memset(flag, yes, sizeof(flag));
    flag[0]=flag[1]=no;
    int len=sqrt(n)+1;
    for(i=2; i<len; i++)
    {
        if(flag[i]==no) continue;
        for(j=i+i; j<=n; j+=i)
            flag[j]=no;
    }
}



int main()
{
    int n;
    int count;
    int t,i,k,num;
    int max;
    scanf("%d",&t);
    while (t--) {
        scanf("%d", &n);
        max=0;
        for (i=0;i<n;++i) {
            count=0;
            scanf("%d",&num);
 
            is_prime(num);
            for(i=1; i<=num; i++)
                if(flag[i]==yes){
                    if(num%i==0)
                    count++;
            }
            if(count>max) 
            {
                max=count;
                k=num;
            }
            if (count==max)
            {
                   if (k>num) k=num;
            }
        }
        printf("%d
",k);
    }
    return 0;
}

Winifred:

#include <stdio.h>
#include <string.h>
bool f[100025];
int T,p[100025];

void getprime()
{
    int i,j;
    memset(f,true,sizeof(f));
    f[1]=false;
    for (i=2;i<=100000;i++)
    if (f[i])
    {
        for (j=i+i;j<=100000;j+=i) f[j]=false;
    }
    T=0;
    for (i=2;i<=100000;i++)
    if (f[i])
    {
        T++;
        p[T]=i;
    }
}

int getdivnum(int x)
{
    int ans=0,tmp=x;
    for (int i=1;i<=T;i++)
    {
        if (tmp%p[i]==0)
        {
            ans++;
            while (tmp%p[i]==0) tmp/=p[i];
        }
        if (p[i]>tmp) break;
    }
    return ans;
}
int main()
{
    getprime();
    int cas;
    scanf("%d",&cas);
    while (cas--)
    {
        int n;
        scanf("%d",&n);
        int Max=0,u;
        for (int i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            int t=getdivnum(x);
            if (t>Max)
            {
                Max=t;
                u=x;
            }
            else if (t==Max)
            {
                if (u>x) u=x;
            }
        }
        printf("%d
",u);
    }
    getprime();
}