Codeforces 632D Longest Subsequence 2016-09-28 21:29 37人阅读 评论(0) 收藏

D. Longest Subsequence

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given array a with n elements
and the number m. Consider some subsequence of a and
the value of least common multiple (LCM) of its elements. Denote LCM as l. Find any longest subsequence of a with
the value l ≤ m.

A subsequence of a is an array we can get by erasing some elements of a.
It is allowed to erase zero or all elements.

The LCM of an empty array equals 1.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 106)
— the size of the array a and the parameter from the problem statement.

The second line contains n integers ai (1 ≤ ai ≤ 109)
— the elements of a.

Output

In the first line print two integers l and kmax (1 ≤ l ≤ m, 0 ≤ kmax ≤ n)
— the value of LCM and the number of elements in optimal subsequence.

In the second line print kmax integers
— the positions of the elements from the optimal subsequence in the ascending order.

Note that you can find and print any subsequence with the maximum length.

Examples

input

7 8
6 2 9 2 7 2 3

output

6 5
1 2 4 6 7

input

6 4
2 2 2 3 3 3

output

2 3
1 2 3

题目的意思是输入n个数和一个m，求n个数里面能组成小于m的最小公倍数的用到数字最多是多少，并输出最小公倍数，用到的数的量，及每一个数的下标；

开始想想毫无头绪，但发现m比较小，才1e6,就想到开一个p[i]数组记录数字i作为最小公倍数会用到多少数字，先记下原序列中每个值出现的次数，然后k倍的这个值加上次数，因为很明显的是以a为因子的最小公倍数一定在ka中，然后搜一边p数组找到最大值就好了，要注意的是要取第一个最大值，因为答案的最小公倍数的整数倍也会是最大值。

最后在遍历一遍输出下标

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
using namespace std;
#define inf 0x3f3f3f3f

int n,m;
int a[1000006];
int p[1000006];
int cnt[1000006];

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(cnt,0,sizeof(cnt));
        memset(a,0,sizeof(a));
        memset(p,0,sizeof(p));
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            if(a[i]<=m)
            {
                cnt[a[i]]++;
            }
        }

        for(int i=1; i<=m; i++)
            for(int j=1; j*i<=m; j++)
            {
                p[i*j]+=cnt[i];
            }

        int mx=-1;
        int u=-1;
        for(int i=1;i<=m;i++)
        {
            if(mx<p[i])
            {
            u=i;
            mx=p[i];
            }
        }
        printf("%d %d\n",u,mx);
        int q=0;
        for(int i=1;i<=n;i++)
        {
        if(u%a[i]==0)
        {
        if(q++)
        printf(" ");
        printf("%d",i);
        }
        }
        printf("\n");

    }

    return 0;
}