CF：402D - Upgrading Array 数分解为素数之积滴判定

D. 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.

Sample test(s)

input

5 24 20 34 10 102 5

output

-2

input

4 52 4 8 163 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.

这题有点神啊……题意看了好久都没太理解，又研究队友那时过的代码，也好久了才懂什么意思，太神了，以前知道怎么分解一个数为素数之积，不过没写过代码，这题就作为模板了吧……

#include <iostream>#include <cstdio>#include <fstream>#include <algorithm>#include <cmath>#include <deque>#include <vector>#include <list>#include <queue>#include <string>#include <cstring>#include <map>#include <stack>#include <set>#define PI acos(-1.0)#define mem(a,b) memset(a,b,sizeof(a))#define sca(a) scanf("%d",&a)#define pri(a) printf("%d\n",a)#define MM 1000000000#define MN 5005#define INF 10000007typedef long long ll;using namespace std;int n,m,bad[MN],a[MN],is[MN*200],ans,prime[MN*2],gcd[MN];void get_prime(){    int i,j,k=sqrt(MM+0.5);    for(i=2;i<=k;i++)        if(!is[i])        {            prime[ans++]=i;            for(j=i*i;j<=k;j+=i) is[j]=1;        }}int fenjie(int x){    int cnt=0,i,t,pos;    for(i=0;i<ans&&x>1;i++)        if(!(x%prime[i]))        {            t=0;            while(!(x%prime[i])) x/=prime[i],t++;            pos=lower_bound(bad,bad+m,prime[i])-bad;            if(pos<m&&bad[pos]==prime[i]) cnt-=t;            else cnt+=t;        }    if(x>1) //说明x比prime中最大的数还大，那么就在bad里面找了    {        pos=lower_bound(bad,bad+m,x)-bad;        if(pos<m&&bad[pos]==x) cnt--;        else cnt++;    }    return cnt;}int main(){    ll sum=0;    int i,j;    get_prime();    scanf("%d%d%d",&n,&m,&a[0]);    gcd[0]=a[0];    for(i=1;i<n;i++)        sca(a[i]),gcd[i]=__gcd(gcd[i-1],a[i]);    for(i=0;i<m;i++) sca(bad[i]);    for(i=0;i<n;i++) sum+=fenjie(a[i]);    for(i=n-1;i>=0;i--)        if(gcd[i]>0)        {            int d=fenjie(gcd[i]);            if(d<0)            {                for(j=0;j<=i;j++)                    a[j]/=gcd[i],gcd[j]/=gcd[i];                sum-=d*(i+1);            }        }    cout<<sum<<endl;    return 0;}