首先对于所有数求gcd并求出这个gcd含有的质因子,那么在所有数中,只有这一些质因子会对答案产生影响,而且对于所有的数,每一个质因子只会在一个数中被删去。
质因子数量不会超过(11),所以考虑暴力的状压DP:设(f_{i,j,k})表示前(i)个数中选出(j)个数并删去了集合为(k)的质因子的最小代价,转移枚举子集并计算转移是否合法(即第(i)个数中当前集合质因子的乘积是否超过(K)),复杂度(O(nm3^m))
考虑优化:对每一个数将不能对答案造成影响的因子全部丢掉。对于剩下的数相同的一些数,最多只会选按权值从小到大排序之后前质因子个数个产生贡献。
再加上一些奇怪的优化就可以快很多了。时间复杂度(O(?m3^m))
#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<map>
//This code is written by Itst
using namespace std;
#define int long long
inline int read(){
int a = 0;
char c = getchar();
while(!isdigit(c))
c = getchar();
while(isdigit(c)){
a = a * 10 + c - 48;
c = getchar();
}
return a;
}
inline int gcd(int a , int b){
int r = a % b;
while(r){a = b; b = r; r = a % b;}
return b;
}
#define st first
#define nd second
const int MAXN = 1e6 + 7;
struct zt{
int num , w;
bool operator <(const zt a)const{return w < a.w;}
}now[MAXN];
vector < int > in;
int N , K , allGcd , dp[12][1 << 11] , calc[1 << 11];
map < int , vector < int > > lsh;
signed main(){
#ifndef ONLINE_JUDGE
freopen("in","r",stdin);
//freopen("out","w",stdout);
#endif
N = read(); K = read();
for(int i = 1 ; i <= N ; ++i){
now[i].num = read();
allGcd = gcd(allGcd , now[i].num);
}
for(int i = 1 ; i <= N ; ++i) now[i].w = read();
if(allGcd == 1) return puts("0") , 0;
for(int i = 2 ; i <= 1e6 && 1ll * i * i <= allGcd ; ++i)
if(allGcd % i == 0){
in.push_back(i);
while(allGcd % i == 0) allGcd /= i;
}
if(allGcd - 1) in.push_back(allGcd);
int M = in.size();
for(int i = 1 ; i <= N ; ++i){
int z = 1;
for(auto j : in)
while(now[i].num % j == 0){
now[i].num /= j;
z *= j;
}
lsh[z].push_back(now[i].w);
}
memset(dp , 2 , sizeof(dp));
dp[0][0] = 0;
for(auto t : lsh){
int x = t.st;
sort(t.nd.begin() , t.nd.end());
if(t.nd.size() > M) t.nd.resize(M);
vector < int > num;
for(int i = 0 ; i < 1 << M ; ++i){
int times = 1 , p = x;
for(int j = 0 ; j < M ; ++j)
if(i & (1 << j))
while(p % in[j] == 0){
p /= in[j];
times *= in[j];
}
num.push_back(times);
}
for(auto p : t.nd){
bool flg = 0;
for(int i = M - 1 ; i >= 0 ; --i)
for(int j = 0 ; j < 1 << M ; ++j)
if(dp[i][j] <= 1e12){
int all = (1 << M) - 1 - j;
for(int k = all ; k ; k = (k - 1) & all)
if(num[k] <= K)
if(dp[i + 1][j | k] > dp[i][j] + p){
dp[i + 1][j | k] = dp[i][j] + p;
flg = 1;
}
}
if(!flg) break;
}
}
int minN = 1e18;
for(int i = 1 ; i <= M ; ++i)
minN = min(minN , dp[i][(1 << M) - 1] * i);
if(minN > 1e12) cout << -1;
else cout << minN;
return 0;
}