misaka and last order SCU - 4489 (筛法的灵活应用)

Time Limit: 1000 MS Memory Limit: 131072 K

Description

Misaka Mikoto is a main character of the Animation “To Aru Majutsu no Index” and “To Aru Kagaku no Railgun”.

She was enrolled into the Academy City to train herself and had been working very hard until she was ranked Level 5,

which is regarded as the highest ranking around the city.

By the way, there are only seven people who reached that level, and she won the third place. Now, she is the best “Electromaster” in Academy City.

Mikoto’s unique skill is to shoot metal objects such as coins at a high speed, destructively.

Hence, she finally gained a nickname, “Railgun” (超電磁砲 (レールガン) .

Mikoto attended Tokiwadai Middle School,which is a famous girls’ secondary school for the rich and elites and was acknowledged

as one of the five famous schools in the Academy City.

To most people, she is considered a typical lady, but in reality, she is short-tempered, prideful, and greatly interested in some childish things.

Misaka has many sisters, they are Misaka’s clones. Unlike Mikoto, they seem rather quiet and emotionless,

their voices are rather monotonous and computer-like, and they speak in third person by adding a description of themselves after every sentence.

In addition, they wear a special set of goggles that enables them to see electron beams and magnetic field lines for the reason that,

unlike Mikoto, they do not have that ability. Misaka clone No. 20001(Last order, 打ち止め, 最终之作).

She is the administrator of the Misaka Network and the fail-safe mechanism in case the Sisters go out of control.

She is not designed to function independently and survive for long outside a culture container and appears to be only about ten years old.

Misaka finds that all her sisters were trapped in a laboratory. Accelerator(一方通行 ,アクセラレータ) is going to kill them!!

She must take actions to save them. Last order tells her that she may not take all the sisters away.

Each sister has an electronic value a. If the GCD(Greatest Common Divisor) of some sisters’ value is greater than m

(if only one sister was chosen, the GCD is the electronic value of herself), then these sisters can be taken away.

Misaka wants to save as many sisters as possible. Please help her!!

Input

In the first line, there is an integer T, indicating the number of test cases.

The next T cases follow. For the first line, there are two integer n and m, n is the number of the sisters, m is the GCD limit.(1 <= n, m <= 1000000)

Then another line shows n integer, represents all the electronic values of all the sisiters.(1 <= values <= 1000000)

Output

Output one line for each test case.

Just print the max number of sisters which Misaka can save.

Sample Input

3

3 1

1 2 3

3 2

1 2 4

3 4

1 3 5

Sample Output

3

2

1

Hint

For the third case, Misaka can only choose the third sister.

#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

const int INF=1e9+7;
const int maxn=1010000;
typedef long long ll;

//(筛法的活用)
//首先这一道题讲的是n个数与m的关系,仔细想想
//如果某几个数的gcd是m,那么这几个数肯定是m的倍数,这就是筛法的巧妙之处
//既可以筛出素数,也可以筛出一个数的倍数,放在这一题中,从最小的m开始找,
//把从** m **到*** n个数中的最大数为终点 **范围内的所有m的倍数都找出来
//之前用桶排的思想把出现的数某个的** 次数 **记录下来,如果某个数
//** 是m的倍数,且在这n个数中 **,则把这个数的频数加起来,最后结果就是答案
int n,m,t;
int vis[maxn];
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        int maxx=0,a;
        scanf("%d%d",&n,&m);
        for(int i=0; i<n; i++)
        {
            scanf("%d",&a);
            vis[a]++;
            maxx=max(maxx,a);
        }
        int ans=0;
        for(int i=m; i<=maxx; i++)
        {
            int cnt=0;
            for(int j=1; j<=maxx; j++)
            {
                int temp=i*j;
                if(temp>maxx)
                    break;
                cnt+=vis[temp];
            }
            ans=max(cnt,ans);
        }
        printf("%d\n",ans);
    }
    return 0;
}