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 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.

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 10000007
typedef 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;
}