大意: 给定序列$a$, 求选出最长的一个子序列, 使得lcm不超过m.
刚开始想复杂了, 想着枚举gcd然后背包, 这样复杂度就是$O(\sum\limits_{i=1}^m \frac{m\sigma_0(i)}{i})$...... 估计了一下1e6大概只有1e8, 感觉剪个枝应该就可以过了, 打到最后才发现似乎不能输出方案...
看题解后发现就是个沙茶题, 直接枚举lcm即可.
#include <iostream>
#include <random>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif
int n, m, sum[N], a[N], f[N];
int main() {
scanf("%d%d", &n, &m);
REP(i,1,n) {
scanf("%d", a+i);
if (a[i]<=m) ++f[a[i]];
}
REP(i,1,m) if (f[i]) {
for (int j=i; j<=m; j+=i) sum[j]+=f[i];
}
int ans = 0, pos = 0;
REP(i,1,m) if (sum[i]>ans) ans=sum[i],pos=i;
if (!pos) return puts("1 0"),0;
printf("%d %d\n", pos, ans);
REP(i,1,n) if (pos%a[i]==0) printf("%d ", i);hr;
}