Codeforces-632D Longest Subsequence

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

题目大意：

给你一个具有n个元素的序列a以及一个数m，找出能够使得所有子序列中最长的一个子序列，并使其满足该子序列的最小公倍数<=m。

解题思路：

这个题目在codeforces里面被归在了暴力(brute force),数学(math),数论(number theory)下，其实感觉就是一个思维题，不过是挺暴力的= =对于整个序列里的每一个元素，只要是大于m的就不需要考虑直接跳过，因为两个数a和b的最小公倍数lcm，必然是大于或者等于a或b的，那么对于所有小于等于m的元素，只需要把所有小于m的元素的因子找到，并且更新因子的个数，找到最终小于m的元素中，具有相同因子个数最多的元素，就是所找到的l，然后再对原数组遍历一遍，找到他的因子们的位置，就可以了。

#include 
#include 
using namespace std;
const int maxn = 1e6 + 5;
int num[maxn], a[maxn], sum[maxn];
int main()
{
	int n, m;
	scanf("%d%d", &n, &m);
	memset(num, 0, sizeof(num));
	memset(sum, 0, sizeof(sum));
	for(int i = 0; i < n; ++i){
		scanf("%d", &a[i]);
		if(a[i] <= m){
			++num[a[i]];
		}
	}
	for(int i = 1; i <= m; ++i){
		if(num[i]){
			for(int j = i; j <= m; j += i){
				sum[j] += num[i];
			}
		}
	}
	int max_sum = 0, max_val = 0;
	for(int i = 1; i <= m; ++i){
		if(sum[i] && max_sum < sum[i]){
			max_val = i;
			max_sum = sum[i];
		}
	}
	if(max_sum == 0){
		printf("1 0\n\n");
	}else{
		int cas = 0;
		printf("%d %d\n", max_val, max_sum);
		for(int i = 0; i < n; ++i){
			if(max_val % a[i] == 0){
				if(cas){
					printf(" ");
				}
				printf("%d", i + 1);
				++cas;
			}
		}
		puts("");
	}
	return 0;
}

如有错误，还请各位菊苣指出！

转载于:https://www.cnblogs.com/wiklvrain/p/8179482.html