Codecraft-17 and Codeforces Round #391 (Div. 1 + Div. 2, combined) B. Bash's Big Day（分拆素因子+求最多的公因子）

题目链接：http://codeforces.com/contest/757/problem/B

B. Bash’s Big Day
time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
Bash has set out on a journey to become the greatest Pokemon master. To get his first Pokemon, he went to Professor Zulu’s Lab. Since Bash is Professor Zulu’s favourite student, Zulu allows him to take as many Pokemon from his lab as he pleases.

But Zulu warns him that a group of k > 1 Pokemon with strengths {s1, s2, s3, …, sk} tend to fight among each other if gcd(s1, s2, s3, …, sk) = 1 (see notes for gcd definition).

Bash, being smart, does not want his Pokemon to fight among each other. However, he also wants to maximize the number of Pokemon he takes from the lab. Can you help Bash find out the maximum number of Pokemon he can take?

Note: A Pokemon cannot fight with itself.

Input
The input consists of two lines.

The first line contains an integer n (1 ≤ n ≤ 105), the number of Pokemon in the lab.

The next line contains n space separated integers, where the i-th of them denotes si (1 ≤ si ≤ 105), the strength of the i-th Pokemon.

Output
Print single integer — the maximum number of Pokemons Bash can take.

Examples
input
3
2 3 4
output
2
input
5
2 3 4 6 7
output
3
Note
gcd (greatest common divisor) of positive integers set {a1, a2, …, an} is the maximum positive integer that divides all the integers {a1, a2, …, an}.

In the first sample, we can take Pokemons with strengths {2, 4} since gcd(2, 4) = 2.

In the second sample, we can take Pokemons with strengths {2, 4, 6}, and there is no larger group with gcd ≠ 1.

【中文题意】给你n个数，让你在这n个数中找出k个数，这k个数有不为1的公因子，求k的最大值。特别注意是：假如只有1个数，结果为1。

【思路分析】直接筛法求一下素数，然后对每个数进行素因子分解，记录每个素因子出现的次数，最后求一下最大值就可以了。
【AC代码】

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<stack>#include<queue>using namespace std;#define ll long longint a[100005];bool isprime[100005];int prime[100005],nprime=0;void doprime(){    memset(isprime,true,sizeof(isprime));    isprime[1]=false;    for(int i=2; i<=100000; i++)    {        if(isprime[i])        {            prime[nprime++]=i;            for(int j=2*i; j<=100000; j+=i)            {                isprime[j]=false;            }        }    }}int num[100005];int main(){    int n;    doprime();    while(~scanf("%d",&n))    {        memset(num,0,sizeof(num));        int maxnnum=0;        for(int i=1; i<=n; i++)        {            scanf("%d",&a[i]);            maxnnum=max(maxnnum,a[i]);            if(isprime[a[i]])            {                num[a[i]]++;            }            else            {                int nn=a[i];                for(int j=0; j<nprime; j++)                {                    if(a[i]%prime[j]==0)                    {                        num[prime[j]]++;                        while(a[i]%prime[j]==0)                        {                            a[i]/=prime[j];                        }                    }                    if(a[i]==1)                    {                        break;                    }                }            }        }        int maxn=1;        for(int i=2;i<=maxnnum;i++)        {            maxn=max(maxn,num[i]);        }        printf("%d\n",maxn);    }    return 0;}