CodeForces - 757B . Bash's Big Day - 贪心+暴力...也可因子数分解

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.
InputThe 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.
OutputPrint single integer — the maximum number of Pokemons Bash can take.
ExampleInput

3
2 3 4

Output

2

Input

5
2 3 4 6 7

Output

3

Notegcd (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.
题意：找到n个数中有公因数（除了1）的最多的数的个数 例2 3 4 6 7，，，因数是2的有2 4 6，且三个数是最多的
题解：用一个数组记录各个数的因子，找到最大数组元素即可#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 100100
using namespace std;
int vis[maxn];
int main()
{
int n;
while (scanf("%d", &n) != EOF)
{
memset(vis, 0, sizeof(vis));
for (int i = 0; i < n; i++)
{
int num;
scanf("%d", &num);
for (int i = 1; i*i <= num; i++)
{
if (num % i == 0)
{
vis[i]++;
if (num / i != i)
vis[num / i]++;
}
}
}
int ans = 1;
for (int i = 2; i < maxn; i++)
ans = max(ans, vis[i]);
printf("%d\n", ans);
}
return 0;
}
当然   也有别的思路 对于每个数来说 一一找寻因子，用
（2，maxn），，maxn==这组数中的最大的那个数，取有相同的因子的个数的最大值即可#include<iostream>
#include<cstring>
using namespace std;
int s[100005],b[100005];
int main()
{
int n;
while(cin>>n)
{
memset(b,0,sizeof(b));
int maxn=0,ans=1;
for(int i=0;i<n;i++)
{
cin>>s[i];
b[s[i]]++;
maxn=max(maxn,s[i]);//因子的最大可能性值
}
for(int i=2;i<=maxn;i++)
{
int sum=0;
for(int j=i;j<=maxn;j+=i) //依次求是否i及i的倍数是否为因子
sum+=b[j];
ans=max(ans,sum);
}
cout<<ans<<endl;
}
return 0;
}