codeforces632D. Longest Subsequence (最小公倍数)

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

题意：给你n个数，让你找出最长的子序列，使得子序列的最小公倍数小于等于m，并把序列顺序输出来。

思路：这题容易想到用dp[i][j]表示当前遍历到第i个数，最小公倍数为j的最长子序列的长度是多少。状态转移为dp[i][lcm(a[i],k)]=dp[i-1][k]+1，但是这样的复杂度就爆了，所以要换种思路。我们可以枚举1~m中的每一个数作为最小公倍数，然后数出n个数中多少个数是它的因子，然后取最大就行了，这里要把a[i]的个数记下来，不然时间复杂度就爆。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
#define inf 99999999
#define pi acos(-1.0)
#define maxn 1000050
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef long double ldb;
int len[maxn],a[maxn],cnt[maxn];

int main()
{
    int n,m,i,j;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=1;i<=m;i++)len[i]=0,cnt[i]=0;
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
            if(a[i]<=m)cnt[a[i] ]++;
        }
        for(i=1;i<=m;i++){
            for(j=i;j<=m;j+=i){
                len[j]+=cnt[i];
            }
        }
        int maxx=0;int maxlen=0;
        for(i=1;i<=m;i++){
            if(len[i]>maxlen){
                maxx=i;
                maxlen=len[i];
            }

        }
        if(maxx==0){
            printf("1 0\n");continue;
        }

        printf("%d %d\n",maxx,maxlen);
        int flag=1;
        for(i=1;i<=n;i++){
            if(maxx%a[i]==0){
                if(flag){
                    flag=0;printf("%d",i);
                }
                else printf(" %d",i);
            }
        }
        printf("\n");
    }
    return 0;
}

转载于:https://www.cnblogs.com/herumw/p/9464536.html