大意: 给定$n$个数, 求选择最少的数满足积为$k$的倍数, 并且和最小
刚开始想着暴力维护$k$的素因子向量, 用map转移, 结果T了. 看了下别的dala0题解, 不需要考虑素因子, 我们考虑k的所有因子, 用map预处理一下每个因子再转移就好了.
总的复杂度是$O(n\sigma_0(k)logk)$, 1e12以内除数函数最大值是6720, 应该是可以过的, 但这题太卡时限了, long long 的gcd跑太慢. 但是可以发现每次只对k求gcd, 可以优化到$O(n\sigma_0(k)primes(k))$, 或者提前对a数组取一下gcd, 可以优化一下....
#include <iostream>
#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'
using namespace std;
typedef long long ll;
typedef pair<int,ll> 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;}
//head const int N = 1e3+10, M = 7e3+10;
int n, sz;
ll k, a[N], b[N];
vector<ll> A;
map<ll,int> S;
pii dp[N][M]; int main() {
scanf("%d%lld", &n, &k);
REP(i,1,n) scanf("%lld", a+i),b[i]=gcd(a[i],k);
if (k==1) return printf("1\n%d\n",int(min_element(a+1,a+1+n)-a)),0;
int mx = sqrt(k+0.5);
REP(i,1,mx) if (k%i==0) {
A.pb(i);
if (k/i!=i) A.pb(k/i);
}
sort(A.begin(),A.end());
sz = A.size();
REP(i,0,sz-1) S[A[i]]=i;
REP(i,1,sz-1) dp[0][i]=pii(n+1,0);
REP(i,1,n) REP(j,0,sz-1) {
ll pre = S[A[j]/gcd(A[j],b[i])];
dp[i][j] = pii(dp[i-1][pre].x+1,dp[i-1][pre].y+a[i]);
dp[i][j] = min(dp[i][j], dp[i-1][j]);
}
if (dp[n][sz-1].x==n+1) return puts("-1"),0;
printf("%d\n", dp[n][sz-1].x);
PER(i,1,n) if (dp[i][S[k]]!=dp[i-1][S[k]]) {
printf("%d ", i);
k /= gcd(k,b[i]);
} hr;
}