B. Upgrading Array
time limit per test 1 second
memory limit per test 256 megabytes
input standard input
output standard output

You have an array of positive integers a[1], a[2], …, a[n] and a set of bad prime numbers b1, b2, …, bm. The prime numbers that do not occur in the set b are considered good. The beauty of array a is the sum , where function f(s) is determined as follows:

f(1) = 0;
Let’s assume that p is the minimum prime divisor of s. If p is a good prime, then , otherwise .
You are allowed to perform an arbitrary (probably zero) number of operations to improve array a. The operation of improvement is the following sequence of actions:

Choose some number r (1 ≤ r ≤ n) and calculate the value g = GCD(a[1], a[2], …, a[r]).
Apply the assignments: , , …, .
What is the maximum beauty of the array you can get?

Input
The first line contains two integers n and m (1 ≤ n, m ≤ 5000) showing how many numbers are in the array and how many bad prime numbers there are.

The second line contains n space-separated integers a[1], a[2], …, a[n] (1 ≤ a[i] ≤ 109) — array a. The third line contains m space-separated integers b1, b2, …, bm (2 ≤ b1 < b2 < … < bm ≤ 109) — the set of bad prime numbers.

Output
Print a single integer — the answer to the problem.

Examples
input
5 2
4 20 34 10 10
2 5
output
-2
input
4 5
2 4 8 16
3 5 7 11 17
output
10

Note
Note that the answer to the problem can be negative.

The GCD(x1, x2, …, xk) is the maximum positive integer that divides each xi.

从a[1]…a[r]同时除以最大公因数，当i < j时，GCD(j) | GCD(i) ，那么如果在i处进行操作，那么 j 操作将无法执行。
所以应该倒着考虑，如果当前 r 处不执行操作，那么如果在后面执行操作，(i,j] 这部分的GCD没有对答案进行贡献，所以倒着考虑时如果可以操作，并且操作后对答案有贡献那么就在当前位置进行操作，这样可以保证答案不会更劣。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<string>
#include<iomanip>
#include<ctime>
#include<climits>
#include<cctype>
#include<algorithm>
#define AUTO "%I64d"
using namespace std;
#define smax(x,tmp) x=max((x),(tmp))
#define smin(x,tmp) x=min((x),(tmp))
#define maxx(x1,x2,x3) max(max(x1,x2),x3)
#define minn(x1,x2,x3) min(min(x1,x2),x3)
typedef long long LL;
LL gcd(LL a,LL b) { return !b?a:gcd(b,a%b); }
const int N = sqrt(1e9)+1;
const int maxp = 100000;
const int maxpp = 10000;
int prime[maxpp],tot;
bool no[maxp];
void get_prime()
{
    no[1]=true;
    for(int i=2;i<=N;i++)
    {
        if(!no[i]) prime[++tot] = i;
        int j=1;
        LL to = (LL)i*prime[j];
        while(j<=tot && to<=N)
        {
            no[to] = true;
            if(i%prime[j]==0) break;
            to = (LL) i * prime[++j];
        }
    }
}
set <int> bad;
int f(int num)
{
    int ret = 0;
    for(int i=1;i<=tot&&num^1;i++) if(num%prime[i]==0)
        while(num%prime[i]==0)
        {
            num /= prime[i];
            if(bad.find(prime[i]) == bad.end()) ret++;
            else ret--;
        }
    if(num^1)
        if(bad.find(num) == bad.end()) ret++;
        else ret--;
    return ret;
}
const int maxn = 5005;
int a[maxn],GCD[maxn];
int n,m;
int main()
{
#ifndef ONLINE_JUDGE
    freopen("array.in","r",stdin);
    freopen("array.out","w",stdout);
#endif
    get_prime();
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%d",a+i),GCD[i]=gcd(GCD[i-1],a[i]);
    for(int i=1;i<=m;i++)
    {
        int tmp;
        scanf("%d",&tmp);
        bad.insert(tmp);
    }
    int div=1;
    for(int i=n;i>=1;i--)
    {
        int tmp=f(GCD[i]/div);
        if(tmp<0) div = GCD[i];
        a[i] /= div; // everyone is supposed to be divided once!!
    }
    LL ans = 0;
    for(int i=1;i<=n;i++)
        ans += f(a[i]);
    printf(AUTO,ans);
    return 0;
}