Codeforces 632D Longest Subsequence 【求因子和变形】

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 86 2 9 2 7 2 3

output

6 51 2 4 6 7

input

6 42 2 2 3 3 3

output

2 31 2 3

fuck，偷懒用cin>>，T了。。。

题意：给定一个n元素构成的序列和一个数m，让你找到一个最长的子序列c[]使得lcm(c[]) <= m。输出序列的lcm 和 长度以及 组成这个序列的元素下标。

思路：发现m最大为10^6，可以直接暴力搞出i(1 <= i <= m)在序列中的因子数sum[i]，就是求因子和的变形。

然后我们找到最大的sum[i]，这个就是满足题意的子序列的最大长度。此时对应的i并不一定是子序列的lcm，我们就从末尾扫一次，对符合条件的元素求一次lcm即可。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <map>#include <string>#include <vector>#include <queue>#include <stack>#define CLR(a, b) memset(a, (b), sizeof(a))#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;const int MOD = 1e9+7;const int MAXN = 1e6+10;const int INF = 0x3f3f3f3f;void add(LL &x, LL y) {x += y; x %= MOD;}map<LL, int> fp;LL gcd(LL a, LL b) {    return b == 0 ? a : gcd(b, a%b);}LL lcm(LL a, LL b) {    return a / gcd(a, b) * b;}LL a[MAXN]; int sum[MAXN];int main(){    int n, m; cin >> n >> m; fp.clear();    for(int i = 1; i <= n; i++) scanf("%lld", &a[i]), fp[a[i]]++;    for(int i = 1; i <= m; i++)    {        if(i == 1)            sum[i] = fp[1];        else            sum[i] = fp[i] + fp[1];    }    for(int i = 2; i*i <= m; i++)    {        for(int j = i; i*j <= m; j++)        {            if(i == j)                sum[i*j] += fp[i];            else                sum[i*j] += fp[i] + fp[j];        }    }    int ans = 0; LL Wlcm = 1LL;    for(int i = m; i >= 1; i--)    {        if(sum[i] > ans)        {            Wlcm = i;            ans = sum[i];        }        //num[i] = 0;    }    LL Tlcm = 1LL; stack<int> S;    for(int i = n; i >= 1; i--)    {        if(Wlcm % a[i] == 0)        {            Tlcm = lcm(Tlcm, a[i]);            S.push(i);        }    }    cout << Tlcm << " " << ans << endl;    int num = 0;    while(!S.empty())    {        if(num) cout << " ";        cout << S.top(); S.pop(); num++;    }    cout << endl;    return 0;}