2 seconds
256 megabytes
standard input
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.
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.
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.
7 8
6 2 9 2 7 2 3
6 5
1 2 4 6 7
6 4
2 2 2 3 3 3
2 3 1 2 3/*题目大意:给定n大小的数组a,求数组a的最长子序列的最小公倍数不超过m。要求输出其满足要求的任意子序列
*算法分析:先将小于等于m的数择出来,然后从后向前筛
*/
#include <bits/stdc++.h>
using namespace std; typedef long long int llint;
const int maxn = 1e6 + 100;
llint a[maxn], b[maxn]; int main() {
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
llint n, m, flag = 0;
scanf("%I64d%I64d",&n, &m);
for (llint i = 0; i<n; i++) {
scanf("%I64d",&a[i]);
if (a[i] <= m) {
b[a[i]] ++ ;
flag = 1;
}
}
if (!flag) cout << "1 0" << endl;
else {
for (llint i = m; i>=1; i--) {
for (llint j = 2*i; j<=m; j+=i) {
b[j] += b[i];
}
}
llint max = 0, lcm = 0;
for (llint i = 1; i<=m; i++) {
if (b[i] > max) {
max = b[i];
lcm = i;
}
}
cout << lcm << " " << max << endl;
for (llint i = 0; i<n; i++) {
if (lcm%a[i] == 0) cout << i+1 << " " ;
}
cout << endl;
}
return 0;
}